Class 11 Engineering Physics - Work,Energy And Power

A particle moves from a point $\vec{r_1} = (2 \hat{i} + 3 \hat{j})$ to another point $\vec{r_2} = (3 \hat{i} + 2 \hat{j})$ during which a certain force $\vec{F} = (5 \hat{i} + 5 \hat{j})$ acts on it. Calculate work done by the force on the particle during this displacement.

$\\vec{R}=(3\\hat{i}+2\\hat{j}-2\\hat{i}-3\\hat{j}$ )=

$(\\hat{i}-\\hat{j})$

The work done by force is

$\\overrightarrow{F}.\\overrightarrow{R}=(5\\hat{i}+5\\hat{j}).(\\hat{i}-\\hat{j})=0$

Force ,F= $6{x^2} + 4x + 7$ . If $x = 0$ to $x=2$ ,calculate W.D.

We know that

$W=\int Fdx=2x^3+2x^2+7x|^2_0=38J$

Define the following and give one example.
Hybrid rocket propellants.

A hybrid propellant rocket is a rocket with a rocket motor that uses rocket propellants in two different phrases. one solid and the other either gas or liquid. The hybrid rocket concept can be traced back at least 75 years.

example: AE524 Rocket Engine Propulsion.

Spring of force content 100N/m is extended to 240 cm and released. Find out the work done when the extension has reduced to 10 cm.

The work done is stored as elastic potential energy in the spring.

It is given by

$W = \dfrac12 × k ( x_2^2 - x_1^2)$

$x_1= 240 cm$

$x_2 = 10 cm$

Force constant,

$k = 100 N/m = \dfrac{100}{100}\ N/cm = 1\ N/cm$

$W=\dfrac12 \times 1 (10^2- 240^2)= -28750\ N\ cm$

Work done ,

$W = -28750N\ cm = -287.5 N\ m = -287.5\ J$

A particle moves along X -axis from $x=0$ to $x=1m$ under the influence of a force given by $F=3x^2+2x-10$ . Work done in the process is

A particle moves along X -axis from

$x=0$ to

$x=1m$ under the influence of a force given by

$F=3x^2+2x-10$ .

Work done in the process is,

$W=\int _0^1 F\cdot dx = \int _0^1 (3x^2 +2x-10)dx$

$\implies W=[x^3+x^2- 10x] _0 ^1 = -8 J$

A string can bear a maximum tension of $3.16kg-wt$ . What maximum number of revolutions per second can be made by a stone of mass $49g$ tied to one end of a $1m$ long piece of this string so that the string may not break? ( $\pi =22/7$ )

length of string,

$r=1m$

the maximum tension in the string

$T=3.16kg-wt$

$=3.16\times 10$

$T=31.6N$

mass of stone

$m=49g=0.049kg$

the maximum tension in the string,

$T=\dfrac { { m{ v^{ 2 } } } }{ r } +mg$

$\begin{array}{l} \Rightarrow 31.6=\dfrac { { 0.049\times { v^{ 2 } } } }{ { 12 } } +0.049\times 10 \\ \Rightarrow 31.6=0.049{ v^{ 2 } }+0.49 \\ \Rightarrow { v^{ 2 } }=\dfrac { { 31.6-0.49 } }{ { 0.049 } } \\ \Rightarrow { v^{ 2 } }=\dfrac { { 31.11 } }{ { 0.049 } } \\ \Rightarrow v=\sqrt { 634.89 } \\ \Rightarrow v=25.19m/s \end{array}$

$\begin{array}{l} v=rw \\ w=\dfrac { v }{ r } =\dfrac { { 25.19 } }{ 1 } =25.19 \\ w=2\pi n \\ n=\dfrac { w }{ { 2\pi } } =\dfrac { { 25.19 } }{ { 2\times 3.14 } } =\dfrac { { 25.19 } }{ { 6.25 } } =4.01 \\ n=4.01 \end{array}$

A single force acts on a 3.0 kg particle-like object whose position is given by $x = 3.0 t - 4.0 t^2 + 1.0 t^3$ , with x in meters and t in seconds. Find the work done by the force from t = 0 to t = 4.0 s.

We use the work-kinetic energy theorem. To find theinitial and final kinetic energies, we need the speeds, so

$v = \frac{dx}{dt} = 3.0 - 8.0 t + 3.0 t^2$

in SI units.

Thus, the initial speed is

$v_i = 3.0 m/s$ and the speed at t = 4 s is

$v_f = 19 m/s$ The change in kinetic energy for the object of mass m = 3.0 kg is therefore

$\Delta K = \frac{1}{2} m (v_f^2 - v_i^2) = 528 J$

which we round off to two figures and (using the work-kinetic energy theorem) conclude that the work done is

$W = 5.3 \times 10^2 J$ .

Fill in the blanks
Work done when a body of mass 100 kg lifted up to a height of 1 mater is ..........

Mass =100 kg

Height = 1 m

W.D

$=mhg=100\times 1\times 10=1000\ J$

Show that a force that does no work must be a velocity deependent force.

Force doesn't work . Means , work done = 0 by acting of force .

Let force is , displacement due to force act on body is .

So, W =

we also know,

⇒W = , but dt ≠ 0

so, ,

It is possible only when Force is perpendicular upon velocity . hence, force is dependent upon velocity. If velocity change the direction then force should be change the direction in such a way that angle is maintained at 90° .

An object of mass 1000 kg is traveling with a velocity 72 km/h. Calculate the work done to bring it rest.

$m = 1000kg$

$v=72km/h=20m/s$

Work = different of kinetic energy

$W=\dfrac{1}{2}mv^{2}$

$W=\dfrac{1}{2}\times 1000\times 20\times 20=400000J$

Estimate the work done on an object of mass 80 kg to change its velocity from 5m /s to 10 m /s.

$m = 80 kg$

Work = different of kinetic energy

$W=\dfrac{1}{2}m(v_{2}^{2}-v_{1}^{2})$

$W=\dfrac{1}{2}\times 80(10^{2}-5^{2})=3000J$

A 20 kg object is being lifted through a height of ______ m when 484 J of work is done on it.

$m = 20 kg ; h = ?$

$PE = mgh ; h = \displaystyle \dfrac{484}{20\times 9.8}= 2.469m$

What should be the change in velocity of a body, if its mass is increased by four times for the same K.E of the body?

Let,

Initial velocity = $u$ and initial mass = $m$
Final velocity = $v$ and final mass = $4 \times m$

K.E remains same.

$K.E_{initial} = K.E_{final}$

$\dfrac{1}{2}mu^2 = \dfrac{1}{2}(4 \times m)v^2$

$u^2 = 4 \times v^2$

$v^2 = \dfrac{u^2}{4}$

$v = \dfrac{u}{2}$

$\therefore$ The velocity of a body should be halved, if its mass is increased by four times for the same K.E. of the body.

Calculate the work done to raise a body of the mass $30 kg$ to a height of $50 m$ $(g = 10 m s^{-2})$

$W= F \times s = mgh$

$= 30 \times 10 \times 50 = 15000$
Work done

$= 15 \times 10^3 joule$

Two bodies A and B of equal masses are kept at heights of h and 2h respectively. What will be the ratio of their potential energy?

Mass of body A,

$m_1 =m$
Mass of body B,

$m_2 =m$
Height

$h_1 =h$
Height

$h_2 = 2h$

Acceleration due to gravity,

$g_1 = g$
Acceleration due to gravity,

$g_2 = g$

$\displaystyle \frac{Potential Energy (P.E_1)}{Potential Energy (P.E_2)} = ?$

$P.E = mgh$

$\Rightarrow P.E \alpha h$
(

$\therefore$ 'm' and 'g' are constants)

$\displaystyle \Rightarrow \frac{P.E_1}{P.E_2} = \frac{h_1}{h_2}$

$\Rightarrow \displaystyle \frac{P.E_1}{P.E_2} = \frac{h}{2h} \Rightarrow \frac{P.E_1}{P.E_2} = \frac{1}{2}$
Therefore, the ratio of potential energy is of the two bodies is 1 : 2.

Compute the work which must be performed (in $Kgf-m$ ) to slowly pump water out of a hemispherical reservoir of radius $R=0.6 m$

Since the center of mass of hemisphere from the top is

$h=\dfrac{3R}{8}$

Mass of water

$m=\dfrac{2\pi R^3\rho}{3}$

So work done by pump

$W=mgh=\dfrac{2g\pi R^3\rho\times 3R}{3\times 8}=\dfrac{g\pi R^4\times 1000}{4}=102\ N$

A spring does not obey Hooke's law. Rather it follows, $F = kx - bx^2 + cx^3$ . If the spring has natural length $l$ and compressed length $l'$ then find the work done.

Work done in compression=

$^x_0\int Fdx$

$=^x_0\int (kx-bx^2+cx^3)dx$

$=\dfrac{1}{2}kx^2-\dfrac{1}{3}bx^3+\dfrac{1}{4}cx^4$

$=\dfrac{1}{2}k(l-l')^2-\dfrac{1}{3}b(l-l')^3+\dfrac{1}{4}c(l-l')^4$

A bob of mass $m$ is suspended from point $O$ by a massless string of length $l$ as shown. At the bottommost point it is given a velocity $\displaystyle u= \sqrt{12gl}$ for $l=1m$ and $m=1 kg$ , match the following two columns when string becomes horizontal $\displaystyle \left ( g= 10ms^{-2} \right )$

Given :

$u = \sqrt{12 gl}$

$m = 1 kg$

$l =1$ m

$g =10 m/s^2$

Let the velocity of the bob be

$v$ when the string becomes horizontal.

Using work-energy theorem :

$W =\Delta K.E$

$\therefore$

$-mgl = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2$

$-2 gl = v^2 - (12 gl)$

$\implies v =\sqrt{10 gl}$

$\therefore$

$v = \sqrt{10 \times 10 \times 1} =10$ m/s

Radial acceleration,

$a_r = \dfrac{v^2}{l} = \dfrac{10^2}{1} =100$

$m/s^2$

Tension in the string,

$T = ma_r =1 \times 100 = 100$

$N$

Tangential acceleration,

$a_t = mg =1 \times 10 = 10$

$N$

$\therefore$ Total acceleration,

$a = \sqrt{a_r^2 + a_t^2} =\sqrt{(100)^2 + (10)^2} =10 \sqrt{101} = 100.5$

$m/s^2$

Find work done by a force $\vec{F} = x \hat{i} + xy \hat{j}$ acting on a particle to displace it from point O(0, 0) to C(2, 2).

$W=\int_{a}^{b}$

$\vec{F}$ .

$\vec{dr}$

$\vec{dr}$ =

$dx\hat{i}$ +

$dy\hat{j}$ +dz

$\hat{k}$

$W$ =

$\int_{0}^{2}$ xdx +

$\int_{0}^{2}$ xydy

On substituting the limits we get,

$W=(2+4x)J$

A stone of mass m tied to the end of a string revolves in a vertical circle of radius R.The net forces at the lowest and highest points of the circle directed vertically downwards are : [Choose the correct alternative]

Lower Point Highest Point
a. mg - $T_1$ mg + $T_2$
b. mg + $T_1$ mg - $T_2$
c. mg + $T_1 - (mv^2_1) / R$ mg - $T + (mv^2_1) / R$
d. mg - $T_1 - (mv^2_1) / R$ mg + $T_2 + (mv^2_1) / R$
$T_1$ and $v_1$ denote the tension and speed at the lowest point. $T_2$ and $v_2$ denote corresponding values at the highest point

In the given figure shown here

$L$ and

$H$ show the lowest and highest point respectively.

At point L:

$T_1$ acts towards the centre of the circle and

$mg$ acts vertically downwards.

Now net force on the stone at the lowest point in the downward direction

$=mg-T_1$

At the point : Both

$T_2$ and

$mg$ act vertically downward towards the centre of the vertical circle.

Again the net force on the stone at the highest point in the downward direction

$=T_2+ mg$

Hence option (a) is the correct alternative.

A molecule in a gas container hits a horizontal wall with speed 200 m $s^{-1}$ and angle $30^0$ with the normal, and rebounds with the same speed. Is momentum conserved in the collision ? Is the collision elastic or inelastic ?

Since the molecule approaches the wall of the container at a particular speed and then rebounds from the container at the same speed. Since there is no loss in the velocity due to any factor and the mass is same. So, the momentum of the system remains conserved.

The molecule approach and rebound with same speed. So, the kinetic energy of approach and rebound remains the same. Therefore it will an elastic collision.

An object of mass, m is moving with a constant velocity, $v$ . How much work should be done on the object in order to bring the object to rest?

2170367_464567_ans_448d67437149468cb0a7f1719c8687b6.jpeg

Certain force acting on a $20\ kg$ mass changes its velocity from $5\ ms^{-1}$ to $2\ ms^{-1}$ . Calculate the work done by the force.

Given : Initial velocity

$u = 5$

$m/s$

Final velocity

$v = 2$

$m/s$

The work done is equal to the change in the kinetic energy of an object.

$W = \dfrac{1}{2}m(v^2 - u^2)$

$W = \dfrac{1}{2}\times 20 \times (2^2 - 5^2)$

$\implies W = -210$

$J$

Calculate the work required to be done to stop a car of $1500\ kg$ moving at the velocity of $60\ km/h$ ?

When a object is moving with a constant velocity, it possess K.E

$K.E = \dfrac{1}{2} \times m \times v^{2}$

So, in order to bring the object to rest i.e its same magnitude of energy is required so, as the final final energy comes to zero.

$v =60 km/hr = 16.66 m/s$

So, magnitude of work that needs to be done

$= \dfrac{1}{2} \times m \times v^{2}$

Hence, work required to stop the car

$= \dfrac{1}{2} \times 1500 \times 16.66^{2}$

$=208166.7 J = 208.17 kJ$

A force is applied on a body of mass $20 kg$ moving with a velocity of $40 m{s}^{-1}$ . The body attains a velocity of $50 m{s}^{-1}$ in $2$ seconds. Calculate the work done to attain this velocity.

Force

$= m \dfrac{\left(v - u\right)}{t}=m \cdot a$

Displacement

$=\dfrac{\left({v}^{2}-{u}^{2}\right)}{2a}$

Word done (W)

$=$ Force

$\times$ Displacement

$= m \cdot a \cdot \dfrac{\left({v}^{2}-{u}^{2}\right)}{2a}$

$= \dfrac{1}{2} m \left({v}^{2}-{u}^{2}\right)$

$= \dfrac{1}{2} \times 20 \times \left(2500 - 1600\right)$

$= \dfrac{1}{2} \times 20 \times 900$

$= 9000J$

Therefore, the work done to attain the final velocity is

$9000J$

A moving body weighing $400N$ possesses $500 J$ of kinetic energy. Calculate the velocity with which the body is moving. $(g= 10 ms^{-1})$

Let

$W=mg$

$m=\dfrac{400}{10}=40 kg$

We know that K.E.

$=\dfrac{1}{2}mv^2$

$500=\dfrac{1}{2}\times 40 \times v^2$

$v^2=25$

$v=5 ms^{-1}$

Thus, velocity with which the body is moving =

$5 ms^{-1}$

Define kinetic energy.

Kinetic energy is the energy associated with the movement of objects. The kinetic energy of an object depends on both its mass and velocity, with its velocity playing a much greater role. Let a body of mass

$M$ moving with velocity

$V$ .

$K.E=\dfrac12MV^2,$ its SI unit is

$Jule$

You did $150 \ joules$ of work by lifting a $120 \ Newton$ backpack.

a) How high did you lift the backpack?
b) How much did the backpack weigh in pounds?

(a) Work done is equal to the change in potential energy of the backpack

Hence,

$W=mgh$

where, weight of object

$= mg$ , and

$h$ is the height to which the backpack is lifted

Hence,

$150=120\times h$

$h=\dfrac{150}{120}=\dfrac{5}{4}=1.25 \ m$

(b) Since,

$1\ pound = 4.448 \ N$

Let

$120 \ N=x$ pounds

$=x\times 4.448 \ N$

$x=\dfrac{120}{4.448}=26.98$

Hence

$120 \ N = 26.98 \ pounds$

Look at the two diagrams given below. Calculate the work done in compressing each spring completely, assuming that the force applied remains constant.

Work done by hand = Force* displacement

As the force is remaining constant.

So, W - 10N* 4cm = 0.4 J

A particle is suspended vertically from a point O by an inextensible massless string of length L. A vertical line AB is at a distance $\displaystyle{L}{8}$ from O as shown in figure. The object is given a horizontal velocity u. At some point, its motion ceases to be circular and eventually the object passes through the line AB. At the instant of crossing AB. Its velocity is horizontal. Find u.

The particle has circular path before it passes $AB$ second time during its upward motion. Let the particle have circular path at $'P'$ and $V$ be its velocity at $P$ .

Applying conservation of energy at $P$ and at the lowest point gives:

$\cfrac { 1 }{ 2 } m{ u }^{ 2 }=\cfrac { 1 }{ 2 } m{ v }^{ 2 }+mgL\left( 1+\sin { \theta } \right)$

At point $P$

$T+mg\sin { \theta } =\cfrac { m{ v }^{ 2 } }{ 2 }$

and when $T=0$

$2g\sin { \theta } ={ v }^{ 2 }$

After moving in a circular path, the particle moves along a parabolic trajectory. Since the velocity of the particle when it crosses $AB$ is horizontal. Therefore, it is passing through its highest point in the parabolic path. Thus, half of the range of this parabolic path $Lcos\theta=\cfrac{L}{8}$

Projection angle of the particle at $P$

$=90°-\theta$

$\therefore$ minimum range

$=\cfrac { { v }^{ 2 }\sin { 2\left( 90-\theta \right) } }{ g } \\ \therefore \cfrac { 1 }{ 2 } \cfrac { { v }^{ 2 }\sin { 2\left( 90-\theta \right) } }{ g } =L\cos { \theta } -\cfrac { L }{ 8 }$

$\Rightarrow \cos { \theta =\cfrac { 1 }{ 2 } } \Rightarrow \theta =60°$

Using the above equations:

$u=\sqrt { \left( \cfrac { 4+3\sqrt { 3 } }{ 2 } \right) gL }$

1031696_694130_ans_0c075052cd284f4fa17dab6aeb6bf1de.png

A particle is moving along a vertical circle of radius R. The velocity of particle at P will be (assume critical condition at C).

$T-mg\cos\theta=\cfrac{mV^2}{R}$

Assuming critical condition:

$\implies T_C=0$

$\implies \cfrac{mV^C}{R}+mg\cos(180^o)=0$

$\implies V_C^2=gR$

$\implies V_C=\sqrt{gR}$

From energy conservation:

$E_A=E_C$

$\implies \cfrac{1}{2}mV_A^2=\cfrac{1}{2}mV_C^2+2mgR$

$\implies \cfrac{V_A^2}{2}=\cfrac{gR}{2}+2gR$

$\implies V_A=\sqrt{5gR}$

Similarly, from energy conservation:

$E_A=E_P$

$\implies \cfrac { 1 }{ 2 } mV_{ A }^{ 2 }=\cfrac { 1 }{ 2 } mV_{ P }^{ 2 }+(mgR+mgRcos 37^ o)$

$\implies \cfrac{5gR}{2}=\cfrac{1}{2}V_P^2+gR+0.8gR$

$\implies \cfrac{1}{2}V_P^2=\cfrac{5}{2}gR-gR-0.8gR$

$\implies \cfrac{1}{2}V_P^2=0.7gR$

$\implies V_P^2=1.4gR$

$\implies V_P=1.18\sqrt{gR}$

A particle of mass $m$ moving with a speed $v$ hits elastically another stationary particle of mass $2m$ in a fixed smooth horizontal circular tube of radius $r$ . Find the time when the next collision will take place?

By momentum conservation.

${ v }_{ 1 }=\dfrac { 2m-m }{ 3m } v=\dfrac { v }{ 3 } \\ { v }_{ 2 }=\dfrac { 2m }{ 3m } v=\dfrac { 2 }{ 3 } v$

after collision relatiive velocity

$={ v }_{ 1 }+{ v }_{ 2 }$

$=\dfrac { v }{ 3 } +\dfrac { 2v }{ 3 } =v$

Now time taken to cover 2

$\pi$ r distance is time of next collision.

$t = \dfrac{2\pi r}{v}$

Two springs have force constants $k_{1}$ and $k_{2}(k_{1} > k_{2})$ . On which spring is more work done, if (i) they are stretched by the same force and (ii) they are stretched by same distance?

Work done is given by:

$W=\dfrac{1}{2}kx^2$ where k is spring constant and x is displacement.

If we have two springs :

$W_1=\dfrac{1}{2}k_1x_1^2$

$W_2=\dfrac{1}{2}k_2x_2^2$

Force is given by

$F=-kx$

(i)If force is same for both:

$k_1x_1=k_2x_2$

$\dfrac{W_1}{W_2}=\dfrac{\dfrac{1}{2} \dfrac{(k_1x_1)^2}{k_1}}{\dfrac{1}{2} \dfrac{(k_2x_2)^2}{k_2}}$

$=\dfrac{1}{2}\dfrac{(k_1x_1)^2}{k_1} \times \dfrac{2k_2}{(k_2x_2)^2}=\dfrac{k_2}{k_1}$

$W_2>W_1$

(ii) If stretched by same distance:

$\dfrac{W_1}{W_2}=\dfrac{\dfrac{1}{2}k_1x_1^2}{\dfrac{1}{2}k_2x_2^2}$

$=\dfrac{1}{2} k_1x_1^2\times \dfrac{2}{k_2 x_2^2}$

And

$x_1=x_2$ ,

$\dfrac{W_1}{W_2}=\dfrac{k_1}{k_2}$

$W_1>W_2$

The bob of a simple pendulum of length $1m$ has mass $100g$ and a speed of $1.4m/s$ at the lowest point in its path. Find the tension in the string at this instant.

Mass,

$m=100$

$g=0.1$

$kg$

Length,

$l=1$

$m$

at its lowest point, the only force acting on it is tension only.

$\therefore T=\cfrac{mv^2}{l}$

$\implies T=\cfrac{0.1\times 1.4\times 1.4}{1}$

$\implies T=0.196$

$N$

A particle of mass $m$ is fixed to one end of a light rigid rod of length $l$ and rotated in a vertical circular path about its other end. The minimum speed of the particle at its highest point must be:

Since, the string has to be taut for the mass to complete its circular motion, the tension in the string at its topmost point can be at minimum equal to zero.

$\therefore T=\cfrac{mV_B^2}{l}+mg\cos(180^o)=0$

$\implies \cfrac{mV_B^2}{l}=mg$

$\implies V_B^2=gl$

$\implies V_{B_{min}}=\sqrt{gl}$

This is the minimum speed the string of rod, must have at its highest point.

Suppose the amplitude of a simple pendulum having a bob of mass $m$ is ${ \theta }_{ 0 }$ . Find the tension in the string when the bob is at its extreme position.

The velocity is zero at the extreme position. The particle reaches this extreme position stays there for a very short period of time and comes back downwards due to gravitational force. In that short period of time, the force of tension is exactly balanced by the

$mg cos{ \theta }_{ 0 }$ component.

$\therefore T=mgcos{ \theta }_{ 0 }$

1033812_794179_ans_a221ff3f962f42bca619778936be817d.png

Two capacitors ${ C }_{ 1 }$ and ${ C }_{ 2 }$ have equal amount of energy stored in them. What is the ratio of potential differences across their plates.

$E_1=\cfrac{1}{2}C_1V_1^2\\ E_2=\cfrac{1}{2}C_2V_2^2\\ E_1=E_2$

So,

$C_1V_1^2=C_2V_2^2\\ \cfrac{V_2}{V_1}=\sqrt{\cfrac{C_2}{C_1}}$

A stone tied to an inextensible string of length $l = 1 \,m$ is kept horizontal. If it is released, find the angular speed of the stone when the string makes an angle $\theta =30^o$ with horizontal.

When the stone is released, it moves down executing a circular motion. At any instant, the stone accelerates towards (w.r.t.) the center of its revolution 0, that is, centripetal acceleration a, and simultaneously accelerates down with an acceleration, that is, gravitational acceleration g;

$\vec{a}_r = \omega ^2r$ . Resolving

$\vec{a}_r$ , and along the tangent, we obtain the tangential acceleration of the stone as

$a_t =g\,\cos\,\theta$

Angular acceleration,

$\displaystyle \alpha = \frac{a_t}{r}$

$\displaystyle \Rightarrow\,\,\,\alpha = \frac{g\,\cos\,\theta}{l} (since\,r=l)$

Putting

$\displaystyle \alpha = \frac{\omega d \omega}{d \theta}$ , we obtain

$\displaystyle \frac{\omega d \omega}{d \theta} = \frac{g\,\cos\,\theta}{r}$

$\Rightarrow \,\,\,\omega d \omega =(g/l)cos\,\theta\,d\theta.$

Integrating both the sides, we obtain

$\displaystyle \int_{0}^{\omega} \omega d \theta = \frac{g}{l} \int_{0}^{\theta} \cos\,\theta \,d\theta$

$\displaystyle \Rightarrow \,\,\,\ \frac{\omega ^2}{2} = \frac{g}{l} \sin\,\theta \,or\,\omega =\sqrt {\frac{2g}{l} \sin\,\theta}$

Putting

$l = 1\,m$ and

$\theta = 30^o$ , we obtain

$\displaystyle \omega =\sqrt{\frac{2(10)\sin\,30^o}{(1)}}= \sqrt{10} \,\,\Rightarrow \,\omega\,\cong 3.1\,rad\,s^{-1}$

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The displacement $x$ of a body of mass 1 kg on smooth horizontal surface as a function of time t is given by $x=\dfrac { { t }^{ 3 } }{ 3 }$ (where x is in metres and t is in seconds). Find the work done by the external agent for the first one second.

Given,

$mass=1kg$ and

$X=\dfrac{t^3}{3}$ Where x is in meter and t in second.

So,

$X=\dfrac{t^3}{3}$

By differentiating with respect to its derivatives.

$dx = t^2 dt$

$\dfrac{dx}{dt} = v = t^2$

$\dfrac{d^2x}{dt^2} = a = 2 t$

Force acting on the body

$= ma = 20 t$ Newtons

So work done

$W= \int _{t=0}^{t=t} {F} dx$

$=\int _{0}^t {20t \times t^2}\ dt$

$=\dfrac{20}{4}[t^4]_0^t$

$=5\ t^4$

Thus the work done in the last second is

$W(t=1sec)=5 Joules$

A bicycle chain of length 1.6 m and of mass 1 kg is lying on a horizontal floor. What is the work done in lifting it with one end touching the floor and the other end 1.6 m above the floor?
(take g= ${ 10 }ms^{ -2 }$ )

Given,

Mass of bicycle

$m = 1Kg$

Length

$h = 1.6 m$

$g = 10 m/s^2$

So,

Whenever the object is in contact with some other object or walls at the endpoints. its mass is divided into two equal parts.

So work done

$W = \dfrac{mgh}{2}$

Therefore,

$W = \dfrac{(1)(10)(1.6)}{2} = 8 J$

Thus, the work is done in lifting it with one end touching the floor and the other end

$1.6 m$ above the floor is

$8J$

An object of mass $0.5$ kg attached to a rod of length $0.5$ m is whirled in a vertical circle at constant angular speed. If the maximum tension in the string is $5$ kg wt., Calculate
(i) speed of stone
(ii) maximum number of revolutions it can complete in a minute.

$\begin{array}{c}1{\rm{ rpm}} = \frac{{2\pi }}{{60}}{\rm{rad/sec}}\\{\rm{1 rad/sec}} = \frac{{60}}{{2\pi }}{\rm{rpm}}\\{\rm{13}}{\rm{.28 rad/sec}} = \frac{{60}}{{2\pi }} \times 13.28{\rm{ rpm}}\\{\rm{ = 126}}{\rm{.7 rpm}}\end{array}$

What is work done in lifting a body of mass $20$ kg at a height of $2$ m above the ground? $(g=10m/s^2)$

Work done to lift an object at a height above earth surface is given by formula,

$W = mgh$ ,

where

$m=20 \ kg$ = mass of object,

$h=2 \ m$ = height

On putting values, we get

$W = 20 \times 10 \times 2 = 400 \ Joule$

Two springs $A$ and $B$ are identical but $A$ is harder than $B\ (k_{A} > k_{B})$ . On which spring more work will be done if: (a)They are stretched through the same distance (b)They are stretched by same force?

Given

$k_A > k_B$

Case (a) Stretched through same distance

$x$ .

Work done by spring

$A$ ,

$W_A=\dfrac{1}{2}k_Ax^2$

Work done by spring

$B$ ,

$W_B=\dfrac{1}{2}k_Bx^2$

$\therefore W_A>W_B$

Case (b) Stretched by same force

$F$

Let spring

$A$ is stretched by

$x_A$ and spring

$B$ is stretched by

$x_B$ .

Force

$F= k_A x_A= k_B x_B$

$\Rightarrow x_A< x_B$

Work done by spring

$A$ ,

$W_A= Fx_A$

Work done by spring

$B$ ,

$W_B= Fx_B$

$\therefore W_A<W_B$

A stone of mass $1kg$ is tied to a light string of length $l=10m$ is whirling in a circular path in the vertical plane. If the ratio of maximum to minimum tensions in the string is $3$ , find the speeds of the stone at the lowest and highest points.

${ T }_{ H }+mg=\dfrac { m{ v }^{ 2 } }{ r } \\ { T }_{ L }-mg=\dfrac { m{ u }^{ 2 } }{ r } \\ Energy\quad conservtion\\ \dfrac { 1 }{ 2 } m{ u }^{ 2 }=\dfrac { 1 }{ 2 } m{ v }^{ 2 }+mg(2r)\\ \dfrac { { T }_{ L } }{ { T }_{ H } } =\dfrac { 3 }{ 1 } \\ \dfrac { \dfrac { m{ v }^{ 2 } }{ r } +mg }{ \dfrac { m{ u }^{ 2 } }{ r } -mg } =3\\ as\quad { u }^{ 2 }={ v }^{ 2 }+4gr\\ \dfrac { { v }^{ 2 }+4gr+gr }{ { v }^{ 2 }-gr } =3\\ { v }^{ 2 }+5gr=3{ v }^{ 2 }-3gr\\ { 2v }^{ 2 }=8gr\\ { v }^{ 2 }=4gr\\ v=\sqrt { 4\times 10\times 10 } =20m/sec\\ u^{ 2 }={ v }^{ 2 }+4gr\\ =800\\ u=20\sqrt { 2 } m/sec$

980085_1022499_ans_e8a41080ed86463fa881d5255d344405.png

Position-time variation of a particle of mass $1\ kg$ is $x=(2t^{3}+3)m$ . Calculate the work done by the force in first $2\ seconds$ .

Given,

$x = 2t^3+3$

$\Rightarrow \; dx = 6t^2$

$\Rightarrow a= \dfrac{\mathrm{d^2 x} }{\mathrm{d} t^2}$

$\Rightarrow a= 12t \; And, Force = 1 \times 12t = 12t$

So, the word done in first two second

$= \int Fdx$

$\Rightarrow W = \int_{0}^{2} 72t^3$

$= 288 J$

A particle moves along the x-axis from $x=0$ to $x=5m$ under the influence of a force $F$ (in N) given by $F=3{x}^{2}-2x+7$ . Calculate the work done by this force

The work done in moving the particle from

$x=0$ to

$x=5$ is given by,

$W=\int{dW}=\int F dx$

$W=\int _{ 0 }^{ 5 }{( { 3x }^{ 2 }-2x+7) }$

On integrating

$W=\left[\dfrac{3x^3}{3}-\dfrac{2x^2}{2}+7x\right]_0^5$

$W=[5^3-5^2+7\times5]-0$

$W=125-25+35$

$W=135J$

A block of mass m= 0.1 kg is released from a height of 4 m a curved smooth surface. On the horizontal surface path AB is smooth and path BC offers coefficient of friction $\mu =0.1$ . If the impact of block with vertical wall at C be perfectly elastic, find the total distance covered by the block on the horizontal surface before coming to rest. (Take g=10 $\frac{m}{s^{2}}$ )

Given:

Mass =

$0.1 \ kg$

Height

$= 4\ m$

Length of the path AB

$= 1 \ cm$

Length of the path BC

$= 2\ cm$

Impact at

$C$ is Perfectly elastic

The total distance covered by the block on the horizontal surface before coming to rest can be given by,

$mgh = μmgd$

$d = \dfrac hμ = \dfrac4{0.1}$

$d = 40 \ m$

Number of times,

$n = \dfrac{40}2 = 20$ times.

$= 20$ times

$=19$ times

Distance

$= 20(2)+19(1) = 59\ m$

Hence, the distance is 59 m.

Two identical copper spheres $A$ and $B$ equally charged, repel each other with a force of $4\times {10}^{-5}N$ . A third similar copper sphere $C$ carrying no charge is now brought into contact with $A$ and placed mid-way between $A$ and $B$ .what force would $C$ experience? $(4\times {10}^{-5})N$
(Comment: charge on $A$ and $C$ is $q/2$ each and on $B$ it is $q$ . It $r$ is the distance between $A$ and $B$ , $q/r$ is $2\times {10}^{-7})$

in the first situation

$F=\dfrac{kq^2}{r^2}=4\times10^{-5}N$

Now in the 2nd situation when identical sphere

$C$ touched with sphere

$A$ then there is transfer of charge from

$A$ to

$C$ till both have same potential , in this situation both will acquired equal charge this gives charge on both sphere will be

$q/2$ (since total charge should be equal to

$q$ due to conservation of charge )

so, from the figure we can see that net force on sphere

$C$ will be effective of both by

$A$ and

$B$

$F_C=\vec F_{CA}+\vec F_{CB}$

$=\dfrac{kq^2/4}{r^2/4}-\dfrac{k(q/2)(q)}{r^2/4}$

$=\dfrac{kq^2}{r^2}(1-2)$

$=-\dfrac{kq^2}{r^2}=-4\times10^{-5}N$

this shows that force experienced by sphere C will be

$4\times10^{-5}N$

991586_1029080_ans_76ab0168857d4827a8d8972d89db92ba.png

Calculate the kinetic energy of an object weighing $600 \ kg$ moving at a velocity of $30 \ m/s$

Given, mass

$m = 600 \ kg$ , velocity

$v = 30 \ m/s$

Kinetic energy is given by,

$K.E=\dfrac{1}{2} m v^2$

$=\dfrac{1}{2}\times 600\times (30)^2$

$=27 \times 10^{4} \ J$

A $4\ kg$ swing in a vertical circle at the end of the chord $1\ m$ long. Find out the maximum speed at which it can swing if chord can sustain maximum tension of $140\ N$ .

We know that maximum tension in vertical circular motion is at the bottom position where we get

$T-Mg=F_C=\dfrac{Mv^2}{r}$ , here

$F_c$ =centripetal force

$\dfrac{(T-Mg)\times r}{M}=v^2$

$v^2=\dfrac{(140-40)\times1}{4}=25$

$v=5m/s$

A ball moving with a momentum of $5kgm/s$ strikes against a wall at angle of ${45^ \circ }$ and is deflected at the same angle. Calculate the change in momentum.

Given,

$P=5 kg m/s$

$\theta=45^0$

from the figure given above, there is no change in perpendicular component and cancel each other.

The initial momentum along the x-axis,

$P_i=P cos\theta$

The final momentum along the x-axis,

$P_f=-Pcos\theta$

So, change in momentum,

$\Delta P=P_f-P_i$

$\Delta P=-2Pcos\theta$

$\Delta P= -2\times 5\times cos45^0$

$\Delta P=-7.071 kg m/s$

1423253_1033216_ans_f2b00eb0a44944b5adf1b44a431ce0d7.png

A force $F={x}^{2}+2x+1$ acts on a particle in the x direction, where $F$ is in newton and $x$ in metre. Find the work done by this force during a displacement from $x=-1m$ to $x=2.0m$

Work done =

$\int_{x_{1}}^{x_{2}} F.dx$

Here

$F$ =

$x^{2}+2x+1$

Therefore ,

Work done =

$\int_{-1}^{2}(x^{2}+2x+1)dx$

$\frac{x^{3}}{3}+x^{2}+x$

$(\frac{8}{3}+4+2)-(\frac{-1}{3}+1-1)$

$9$

$J$

An object is displaced from position vector $r_{1} = (2\hat{i} + 3\hat{j})\ m$ to $r_{2} = (4\hat{i} + 6\hat{j})\ m$ under a force $F = (3x^2 \hat{i} + 2y\hat{j})\ N$ . Find the work done by this force.

Work done

$W = \int\vec{F}.(dx\hat{i}+dy\hat{j})$

Work done

$W = \int^{4}_2 \ 3x^2 \ dx + \int^6_3 \ 2y dy$

$W = x^3\bigg|^4_2+y^2\bigg|^6_3$

$W= 4^3-2^3 + 6^2-3^2 = 85 \ J$

A particle is moving in a circular path of radius a under the action of an attractive potential $U=-\dfrac{K}{2r^2}$ . Its total energy is

$U=\dfrac{-K}{2r^2}$

By the concept of Force and potential energy gradient,

$F$

$=-\dfrac{-dU}{dr} = \dfrac{K}{r^3}$ ...... (1)

Since it is performing circular motion,

$F=\dfrac{mv^2}{r}$ ...... (2)

by eq (1) and (2), we get

$\dfrac{mv^2}{r}=\dfrac{K}{r^3}$

$mv^2=\dfrac{K}{r^2}$

Hence,

kinetic energy

$=\dfrac{mv^2}{2} =$

$\dfrac{k}{2r^2}$

now total energy

$= P.E. + K.E.$

$=\dfrac{-k}{2r^2}$

$+ \dfrac{k}{2r^2}$

$=0$

This is the required solution.

A position-dependent force $f=7-2x+3x^2$ newton acts on a small body of mass $2$ kg and displaces from $x=0$ to $x=5$ . Calculate the work done by the force.

Given force

$f=7-2x+3{x}^{2}$

And we know,

Work done

$W=force\times displacement=fx$

Now integrating the above equation with respect to x from limit

$x=0$ to

$x=5$ , we get,

$\int{dW}=\int_{0}^{5}({7-2x+3{x}^{2}})dx$

$=|{ 7x-\dfrac { 2{ x }^{ 2 } }{ 2 } }+\dfrac { { 3x }^{ 3 } }{ 3 } |_{ 0 }^{ 5 }$

$=35-25+125=135$ J

Find the ratio of the weights, as measured by a spring balance, of a $1\,kg$ block of iron and a $1\,kg$ block of wood.
Density of iron = $7800\,kg/m^3$ , density of wood = $800\,kg/m^3$ and density of air= $1.293\,kg\,m^3$

$T=mg$
Tension=spring balance reading
both have

$mg=10$
so both have same reading

$\therefore \frac{{{T_1}}}{{{T_2}}} = 1$

When the bucket filled with water is resolved fast in a vertical circle, the water does not fall even when the bucket is inverted.

When the bucket is inverted, a pseudo force acts on the water upwards. Thus water does not fall even when bucket is inverted.

The relation between displacement $x$ and time $t$ for a body of mass $2 Kg$ moving under the action of a force is given by $x=\dfrac{t^3}{3}$ , where $x$ is in meter and $t$ in second, calculate the work done by the body in first $2$ seconds.

Given that,

Mass $M=2\,kg$

Displacement = $x$

Time $t=2\,s$

Now, the displacement is

$x=\dfrac{{{t}^{3}}}{3}$

On differentiate

$\dfrac{dx}{dt}=\dfrac{3{{t}^{2}}}{3}dt$

$v={{t}^{2}}dt$

Again differentiate

$\dfrac{dv}{dt}=2tdt$

$a=2tdt$

Now, the force is

$F=ma$

$F=2\times 2t$

$F=4t$

Now, the work done is

$dW=F\centerdot dx$

$dW=4t\times {{t}^{2}} dt$

$dW=4{{t}^{3}} dt$

Work done during t = 0 to 2 sec is found by integration

$\int{dW=\int\limits_{0}^{2}{4{{t}^{3}}}} dt$

$W=\left[ {{\left( 2 \right)}^{4}}-0 \right]$

$W=16\,J$

Hence, the work done is $16\ J$

Calculate the work done in moving the object from $x=2$ to $x=3$ m from the given graph.

Work done = Area under F-x curve

Work done = Area of ABCD

$= \cfrac{1}{2} \times 1 \times (40+60)$

$=50J$

978290_1074234_ans_29ca104a054948709938dcba73e093f6.PNG

A 2 kg block slides on a horizontal floor with a speed of 4 m/s. It strikes an uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is 15 N and spring constant is 10,000 N/m. The spring compresses by :-

Change in kinetic energy=work done by the force

$0.5 mv^2=0.5kx^2$

$x=v\times \sqrt\frac{m}{k}$

$x=4\times \sqrt\frac{m2}{10^4}$

$=5.657$ cm.

The work done by an applied variable force $F=x+x^3$ from $x=0m$ to $x=2m$ ,where $x$ is displacement ,is

Given that,

$F=x+x^3$

Work done by force ,

$F= \int _0 ^2(x+x^3)dx= [\dfrac{x^2}2 +\dfrac{x^4}4]_0^2=2+4= 6J$

If force on a block is $F=8/x^2$ where $x$ is displacement ,then find total work done in moving block from from $x=2m$ to $x=6m$

Given that,

Force on the block ,

$F=\dfrac8{x^2}$

Work done in moving the block from

$x=2m$ to

$x=6m$ ,

$W= \int _2^6 F\cdot dx=\int _2^6 (\dfrac8 {x^2})dx= [-\dfrac{8}{x}]_2^6= (-4+\dfrac86)= \dfrac{-8}3 J$

Two ball having masses $m$ and $2m$ are fastened to two light strings of same length $l$ (figure 9-E18). The other ends of the strings are fixed at $O$ . The strings are kept in the same horizontal line and the system is released from rest. The collision between the balls is elastic. (a) Find the velocities of the balls just after their collision. (b) How high will the ball rise after the collision?

(a) According to Work energy theorem

for light mass ball,velocity ${v}_{1}$

$\Delta {E}_{K}=W$ where $\Delta {E}_{K}$ =Kinetic energy and W=work done.

$\dfrac{1}{2}m{v}_{1}^{2}-\dfrac{1}{2}m{0}^{2}=mgl$

${v}_{1}=\sqrt{2gl}$

Similarly for heavy ball,velocity ${v}_{2}=\sqrt{2gl}$

Hence ${v}_{1}={v}_{2}=x$ ----------(A)

As per law of conservation of momentum,

$m{v}_{1}-2m{v}_{2}=m{v}_{1}+2m{v}_{2}$

$mx-2mx=m{v}_{1}+2m{v}_{2}$

$-x={v}_{1}+2{v}_{2}$ -----------------(B)

Again

${v}_{1}-{v}_{2}=-({v}_{1}-{v}_{2})$

${v}_{1}-{v}_{2}=-2x$ as ${v}_{1}={v}_{2}=x$ ---------------(C)

Subtracting B and C, we get

$3{v}_{2}=x$

${v}_{2}=\dfrac{x}{3}=\dfrac{\sqrt{2gl}}{3}$ ---------------(D)

Substituting the above value in c we get,

${v}_{1}={-2x}+\dfrac{x}{3}=\dfrac{-5x}{3}=\dfrac{-5\sqrt{2gl}}{3}$ ----------(E)

(b). Again from work energy theorem we get,

$\dfrac{1}{2}2m{0}^{2}-\dfrac{1}{2}2m{v}_{2}^{2}=2mgh$

Putting the value of D in the above equation,

$\dfrac{gl}{9}=gh$

$h=\dfrac{ l}{9}$

The above h=height for the heavy ball

Similarly for light ball,

$\dfrac{1}{2}m{0}^{2}-\dfrac{1}{2}m{v}_{1}^{2}=mgh$

Putting the value of E in the above equation,

${h}_{1}=\dfrac{25}{9}l$

Explain why water stored in a dam has potential energy.

The water held behind the dam has PE because of its position water in the reservoir is on height and when something is on a certain height the P.E. increases.

A body of mass $2$ kg starts with an initial velocity $5$ m/s. If the body is acted upon by a time dependent force (F) as shown in figure, then work done on the body in $20$ s is?

The graph is given between force and time.

Area under Force-time graph gives Impulse.

So, Total area in 20s is

${ A }_{ 2 }=\dfrac { 1 }{ 2 } \times 5\times 20\\ \quad \quad =50{ m }^{ 2 }$

${ A }_{ 1 }=\quad 20\times 5=\quad 100 { m }^{ 2 }$

${ A }_{ 3 }=\dfrac { 1 }{ 2 } \times 10\times (-20)=-100{ m }^{ 2 }$

Therefore total area A=

${ A }_{ 1 }{ +A }_{ 2 }+{ A }_{ 3 }$

A= 50

${ m }^{ 2 }$

Work done W=

$F\times s\\ =(F\times t)\times v$

Initial velocity is given, v=$$5m/s$$

Work done,

$W = Ft. v$

W= 50 x 5 J

= 250J

A car moves on a flat circular track of $200$ m radius with a constant speed of $20$ if the car weighs $400$ kg the centripetal force acting on the car is:

We know centripetal force =mass

$\times$ radius

$\times$

${\omega}^{2}$ -------------(1)

Given mass of the car=400 kg and radius =200 m and

$\omega=\dfrac{v}{r}=\dfrac{20}{200}=0.1 rad/s$

Now putting the above values in 1 we get,

${F}_{c}=400\times 200 \times {0.1}^{2}=800 N$

A stone of mass 1 Kg, tied to the end of a string of length 30cm, is whirled in a vertical circle such that its angular velocity when the string is horizontal is 10 rad/s. If the tension in the string at that instant is 5T, find the value of T

Given that,

Mass $m = 1\ kg$

Length $l = 30\ cm =0.30\ m$

Angular velocity $\omega = 10\ r/s$

We know that,

$T=m\dfrac{{{v}^{2}}}{r}$

$T=m{{\omega }^{2}}r$

$5T=1\times 100\times 0.30$

$5T=30$

$T=6\,N$

Hence, the value of $T$ is $6\ N$

A stone is tied to a string of length l and is whirled in a vertical circle with the other end of the string as the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of the change in velocity as it reaches a position where the string is horizontal (g being acceleration due to gravity) is?

let the direction of counter clockwise initial velocity

$=ui$

i is the directional of initial velocity

when the string is horizontal the body is covered the quarter of revolution the velocity is v j

j is the direction of velocity towards y axis

change in the velocity is

$vj-ui$

magnitude of the change in velocity

$\sqrt { { v }^{ 2 }-{ u }^{ 2 } }$

by using the conservation of energy

energy of the body is

$=1/2{ mu }^{ 2 }$

when it goes to quarter position KE spent is doing work mgh and remaing KE

$=1/2{ mv }^{ 2 }$

$1/2m{ u }^{ 2 }=1/2{ mv }^{ 2 }+mgh$

$h=l$

$1/2m{ u }^{ 2 }=1/2{ mv }^{ 2 }+mgh$

${ v }^{ 2 }-{ u }^{ 2 }=-2gl$

${ v }^{ 2 }-{ u }^{ 2 }+2{ u }^{ 2 }={ 2u }^{ 2 }-2gl$

${ v }^{ 2 }-{ u }^{ 2 }+2{ u }^{ 2 }={ 2u }^{ 2 }-gl$

$=\sqrt { { v }^{ 2 }+{ u }^{ 2 } } =\sqrt { 2 } \left( { u }^{ 2 }-gl \right)$

You may have seen in a circus a mototcyclist driving in vertical loops inside a 'death well' (a hollow spherical chamber with holes, so the spectators can watch from outside).Explain clearlywhy the mototrcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is $25m$ . ?

The motor cyclist drop down when he is at the upper most point if antifungal is greater or equal –

${{m{v^2}} \over R} \ge mg$

$v \ge \sqrt {rg}$

${V_{\min }} = \sqrt {rg}$

$r = 25m,g = 9.8m{s^{ - 2}}$

${V_{\min }} = \sqrt {25 \times 9.8}$

$= 15.7m{s^{ - 1}}$

Find the velocity of a body of mass $100g$ having a kinetic energy of $20J$ .

Let

$v$ be the velocity of object ,

Given that mass of object ,

$m= 100\ g= \dfrac{100}{1000} kg = 0.1\ kg$

& kinetic energy of object

$=20 \ J$

K.E = $\dfrac{1}{2}$ $mv ^{2}$

$20 = \dfrac12 \times 0.1 v^2$

$\implies v^2= 400$

$\implies v= 20\ m/s$

Seen the displacement X and time t for a body of mass 2 Kg moving $x=\dfrac { t^ 3 }{ 3 }$ , where x is in meter and t in sec. Calculate the work done by body in first 2 second.

Given that,

Mass $m=2\,kg$

Displacement $x=\dfrac{{{t}^{3}}}{3}$

Now, the velocity is

$x=\dfrac{{{t}^{3}}}{3}$

On differentiating

$v=\dfrac{dx}{dt}$

$v=\dfrac{d\left( \dfrac{{{t}^{3}}}{3} \right)}{dt}$

$v={{t}^{2}}$

$dx={{t}^{2}}\,dt$

Again on differentiating

The acceleration is

$a=\dfrac{dv}{dt}$

$a=\dfrac{d\left( {{t}^{2}} \right)}{dt}$

$a=2t$

Now, the force is

$F=ma$

$F=2\times 2t$

$F=4t$

Now, the work is

$dW=F\centerdot dx$

$dW=4tdx$

$dW=4t\times {{t}^{2}}dt$

$dW=4{{t}^{3}}dt$

Now, the work done during t =0 to t= 2 s

So,

$\int{dW=\int\limits_{0}^{2}{4{{t}^{3}}\,dt}}$

$W=\left[ {{t}^{4}} \right]_{0}^{2}$

$W=\left[ {{\left( 2 \right)}^{4}}-0 \right]$

$W=16\,J$

Hence, the work done is $16\ J$

A stone of mass 1 kg is tied up with an inextensible string of length l = 10/3 m. It is rotated in a vertical circle of l radius. If a ratio of maximum and minimum tension in the string is 4 and the value of g is 10 $m/s^2$ then calculate a speed of stone at the highest point of a circle.

Given that,

Mass $m=1\,kg$

Length $L=\dfrac{10}{3}m$

Radius $r=1\,m$

Ratio of tension $\dfrac{{{T}_{2}}}{{{T}_{1}}}=4$

Now,

The ratio of maximum ${{T}_{2}}$ to the minimum tension is

$\dfrac{{{T}_{2}}}{{{T}_{1}}}=4$

${{T}_{2}}=4{{T}_{1}}$

Now, the difference between the two tensions should be 6 mg

Now,

${{T}_{2}}-{{T}_{1}}=6mg$

$4{{T}_{1}}-{{T}_{1}}=6mg$

$3{{T}_{1}}=6mg$

${{T}_{1}}=2mg$

Now, the tension at the top of the circle is

${{T}_{1}}+mg=\dfrac{mv_{1}^{2}}{r}$

${{T}_{1}}+mg=\dfrac{mv_{1}^{2}}{L}$

$2mg+mg=\dfrac{mv_{1}^{2}}{\dfrac{10}{3}}$

$3mg=\dfrac{3mv_{1}^{2}}{10}$

$v_{1}^{2}=\dfrac{30mg}{3m}$

${{v}_{1}}=\sqrt{10\times 10}$

${{v}_{1}}=10\,m/s$

Hence, the speed of stone at highest point of circle is 10 m/s

A porter lift a luggage of $15\ kg$ from the ground and puts it on her head $2\ m$ above the ground. Calculate the work done by her on the luggage? Take $g=10\ ms^{-2}$

W = mgh = 15 x 10 x 2 = 300 J

A pendulum bob has a speed of $3ms^{-1}$ at its lowest position. The pendulum is $0.5m$ long. What is the speed of the bob, when the length makes an angle of $60^\circ$ to the vertical? $(g = 10ms^{-2})$

We know that,

$m{{v}^{2}}=mv_{1}^{2}+2mgh$

$9=v_{1}^{2}+2\left( 0.5-0.5\cos {{60}^{0}} \right)\times 10$

${{v}_{1}}=\sqrt{9-5}$

${{v}_{1}}=2\,m/s$

Hence, the speed of the bob is $2\,m/s$

A position dependent force $F = 7-2x + 3x^2$ newton (where x is in meter) on a small body of mass 2 Kg & displace it from $x = 0$ to $x = 5$ m. Calculate the work done.

$\textbf{Step 1: Work done}$

Work done for small displacement can be defined as:

$dW=Fdx$ $....(1)$

$\textbf{Step 2: Integrate the equation}$

Total work done can be calculated by integrating the equation $(1)$ with limits $x=0$ to $x=5m$

$\int{dW}=\int\limits_{0}^{5}{Fdx}$

$\int{dW}=\int\limits_{0}^{5}{\left( 7-2x+3{{x}^{2}} \right)dx}$

$\Rightarrow W=\left[ 7x-\dfrac{2{{x}^{2}}}{2}+\dfrac{3{{x}^{3}}}{3} \right]_{0}^{5}$

$\Rightarrow W=35-25+125= 135\ J$

Hence, the work done is $135\ J$

A force that object A exerts on object B is observed over a 10 second interval,as shown on the graph.How much work did object A do during that 10 s?

Work done by the force taht object A exerts ,

$W= \text{ Area Under F-x curve} = \dfrac12 \times 5 \times 10 = 25\ J$

The car $A$ of mass $1500kg$ travelling at $40m/s$ collides with another car $'B'$ of mass $1250kg$ travelling at $25m/s$ in the same direction. After collision the velocity of car $'A'$ becomes $30m/s$ . calculate the velocity of car $'B'$ after the collision.

BY LAW OF CONSERVATION OF MOMENTUM WE HAVE

1500 * 40 + 1250* 25 = 30 * 1500 + 1250* Vb

60000 + 31250 = 45000 + 1250*Vb

91250 - 45000 = 1250Vb

46250 = 1250Vb

therefore Vb = 46250 / 1250

= 37 m/sec

A sphere is suspended by a thread of length L. What minimum horizontal velocity to has to be imparted to it so that if completes a full vertical revolution?

Tension is the string parties makes angle

$\theta$ with vertical, during the circular motion is :

$T=\dfrac{mu^{2}}{L}- 2 mg (1- \cos \theta )+ mg \cos \theta =\dfrac{mu^{2}}{L}-mg (2-3 \cos \theta)$

Particle completes a circle only with tension in

$\le 0$ at highest point

$(\theta = 180^{o})$

so,

$\dfrac{mu^{2}}{L}- mg (2-3 \cos (180^{o}) ) \ge 0$

$\Rightarrow \dfrac{mu^{2}}{L} - 5 mg \ge 0 \Rightarrow u \ge \sqrt{5 gL}$

$\therefore$ A minimum velocity of

$u=\sqrt{5 gL}$ has to be imparted at the lowest point.

Two objects having equal masses are moving with uniform velocities of 2 m/s and 6m/s respectively Calculate the ratio of their kinetic energies.

1664737_1175582_ans_d315a999361c4821a6fe10299fa95767.jpg

The bob of a simple pendulum at rest position is given a velocity V in horizontal direction so that bob describes verticle circle of radius equal to length of pendulum L . If the tension in string is 4 times weight of of bob when the string is horizontal , the velocity of bob :

The bob has a velocity

$'v'$ as in lowest point and it describe a vertical circle of radius

$=L.$

Tension in the string when it makes angle

$\theta$ with vertical is given by :

$T= \dfrac{mv^{2}}{L}- mg(2-3\cos \theta)$

Given that , at horizontal

$(\theta = 90^{o}), T=4\ mg$ , so

Gmg

$=\dfrac{mV^{2}}{L}- mg (2-3 \cos 90^{o})$

$=\dfrac{mV^{2}}{L}- 2 mg$

$\Rightarrow 6 mg= \dfrac{mV^{2}}{L} \Rightarrow V= \sqrt{6gL}$

A ball of mass 100 g is suspended a string 40 cm long. Keeping the string taut , the ball describes a horizontal circle of radius 10 cm Find the angular speed.

Given that,

Mass $m=100\,g$

Length $l=40\,cm$

Radius $r=10\,cm$

Now, by force equilibrium equation in vertical direction

${{F}_{T}}\cos \theta =mg.....(I)$

By horizontal direction force equation

${{F}_{T}}\sin \theta =m{{\omega }^{2}}r....(II)$

Now, the ratio of above equations

$\tan \theta =\dfrac{{{\omega }^{2}}r}{g}....(III)$

We know that,

$\sin \theta =\dfrac{1}{4}$

$\theta =14.47$

$\theta =14.5$

Now, from equation (III)

$\tan \theta =\dfrac{{{\omega }^{2}}r}{g}$

$\omega =\sqrt{\dfrac{g\tan \theta }{r}}$

$\omega =\sqrt{\dfrac{10\times \tan 14.5}{10}}$

$\omega =4.949$

$\omega =5.1\,rad/s$

Hence, the angular speed is 5.1 rad/s

If the velocity of the body increases 3 times how does its K.E changes?

K.E = $\dfrac{1}{2}$ $mv ^{2}$

New velocity = $3v$

New K.E = $\dfrac{1}{2}$ $m\ (3v) ^{2}$ = 9 ( $\dfrac{1}{2}$ $mv ^{2}$ ) = 9 x K.E

Hence, new kinetic energy becomes 9 times.

A boy weighing $50\, kg$ climbs up a vertical height of $100\,m$ . Calculate the amount of work done by him. How much potential energy does he gain ( $g=9.8\, m/{s}^{2}$ )

Given,

mass of boy = m = 50 kg

height = h = 100m

since boy does not move anything with his force, work done by him is zero

work done on the boy = gain in potential energy

$P.E = mgh$

$P.E = 50 \times 9.8 \times 100$

$P.E = 49 KJ$

so, potential energy gained by boy is 49 KJ

A block of mass $m$ welded with a light spring of stiffness $k$ is in equilibrium on an smooth inclined plane with angle of inclination $\theta .$ If a variable external force is applied slowly till the spring comes to its relaxed position, find the work done by spring by spring force.
ferring to the $FBD$ as shown in figure

At equilibrium

$Kx=mg\sin \theta$

$x=\cfrac{mg\sin ]theta}{K}$

So,

${U}_{i}=\cfrac{K{x}^{2}}{2}=\cfrac{{m}^{2}{g}^{2}\sin ^{ 2 }{ \theta } }{2K}$

At relaxe position

${U}_{f}=K{(0)}^{2}=0$

As the mass is moved slowly

$\Delta KE=0$ so

spring

$=-\Delta U$

$=-{U}_{f}+{U}_{i}=\cfrac{{m}^{2}{g}^{2}\sin ^{ 2 }{ \theta } }{2K}$

1368580_1177180_ans_1590c77a238347f79ecafd4acd4dfc9a.PNG

If the speed of a body is halved, what will be the change in its kinetic energy?

Kinetic energy of the body

$K.E = \dfrac{1}{2}mv^2$

So,

$K.E\propto v^2$

If speed of the body is halved, its kinetic energy will be reduced by a factor of

$4$ .

That is

$New\ K.E = \dfrac{1}{2}m(\dfrac{v}{2})^2$ =

$\dfrac{K.E}{4}$

Certain force acting on a $20\ kg$ mass changes its velocity from $5 m/{s}$ to $2 m/{s}$ . Calculate the work done by the force.

1973482_1211431_ans_153a29b185fa4a1a950ad8a4a3ee9537.png

A man weighing 70 kg carries a weight of 10 kg to top of a tower 100 m high. Calculate the work done by the man

The work done by the man is in termns of the potential energy change.

The change in the potential energy is given as:

$w=mg\Delta h$

Substitute the values in above formula:

$w=800\times9.8\times100$

$=7.84\times10^5J$

What is Kinetic Energy?

$\textbf{Explanation}$

Energy associated with a particle due to virtue of its motion is called kinetic energy

It is given by the expression

$K=\dfrac 12 mv^2$

A car of mass $2000\ kg$ changes its speed from $18\ km/h$ to $90\ km/hr$ . Find the work done by the engine.

Given,

$m=2000\ kg$

$u =18\ km/hr=18 \times \dfrac{5}{18} = 5\ m/s$

$v =90\ km/hr=90 \times \dfrac{5}{18} =25\ m/s$

Work done by the engine

$=$ Change in kinetic energy

$W = \dfrac{1}{2}m(v^2-u^2) = \dfrac12\times 2000(25^2-5^2)=600000\ J$

$W=600\ kJ$

What is the work done to increase the velocity of a car from $30 \ km { h }^{ -1 }$ to $60 \ km { h }^{ -1 }$ , if the mass of the car is $1500 \ kg$ ?

$u =30\ km/hr=30 \times \dfrac{5}{18} = 8.33\ m/s$

$v =60\ km/hr=60 \times \dfrac{5}{18} = 16.67\ m/s$

Initial energy of the car

$=\dfrac 12 {m}{u^2}$

Final energy of the car

$=\dfrac 12 {m}{v^2}$

Work done

$W = change\ in\ energy =\dfrac{1}{2}m(v^2-u^2)$

$W = \dfrac{1}{2} \times 1500 \times (16.67^2-8.33^2)$

$W = 156375\ J$

To what height should a box of mass 150 kg be lifted, so that its potential energy may become 7350 joules? $\left( {g = 9.8\;m/{s^2}} \right).$

Given,

mass of box

$= m = 150 Kg$

P.E

$= 7350 J$

WE know that

$P.E = mgh$

$7350 = 150 \times 9.8 \times h$

$h = \dfrac{7350}{1470}$

$h = 5 m$

A ball of mass $100g$ is thrown with a speed of $15$ m/s. Calculate its kinetic energy.

Given,

Mass of ball,

$m=100\ g=0.1\ kg$

Speed of ball ,

$v=15\ m/s$

Kinetic energy of ball,

$K= \dfrac12 mv^2=\dfrac12 \times 0.1 \times 15^2= 11.25\ J$

In a conical pendulum, a string of length 120 cm is fixed at rigid support and carries a mass of 150g at its free end. If the mass is revolved in a horizontal circle of radius 0.2m around a vertical axis, calculate tension in the string $\left ( g=9.8m/s^{2} \right )$

$l = 120cm = 1.2m$

$r = 0.2m$

$m = 150gm = 0.15kg$ .

tension in supporting thread (T)

By pythagoras theorem

$l^2 = r^th^2 = Dh^2 = l^2-r^2$

$Dh^2 = 1.44 - 0.04 = 1.4$

$\therefore h = 1.183m$ .

The weight of Bob is balanced by vertical component of tension T

$\therefore T \cos \theta = mg$

$\cos \theta =\dfrac{h}{l} = \dfrac{1.183}{1.2} = 0.9858$

$\therefore T = \dfrac{mg}{\cos \theta} = \dfrac{0.15\times 9.3}{0.9858} = 1.491N$ .

2086598_1359750_ans_c2b35fd5a7124d9cab6577b3a54623ef.png

Two masses $m$ and $2m$ are dropped from height $h$ and $2h$ . On reaching the ground, which will have greater kinetic energy and why?

$\large\textbf{Step 1: Kinetic Energy for mass m, dropped from height h}$

By equation of motion

${v^2} = {u^2} + 2gH$

Or ${v^2} = {0^2} + 2gh = 2gh$

$Kinetic\ {\rm{ }}energy{\rm{ }} = \frac{1}{2}m{v^2} = \frac{1}{2}m(2gh)\ (from\,above\,\,equation)$

$Kinetic\ {\rm{ }}energy,K{\rm{ }} = mgh$

$\large\textbf{Step 2: Kinetic Energy for mass 2m, dropped from height 2h}$

We know,

By equation of motion

${v^2} = {u^2} + 2gH'$

Or ${v^2} = {0^2} + 2g(2h) = 4gh$

$Kinetic\ {\rm{ }}energy{\rm{ }} = \frac{1}{2}m'{v^2} = \frac{1}{2}(2m)(4gh) = 4mgh$

$Kinetic\ {\rm{ }}energy\,K'{\rm{ }} = 4mgh$

Hence, on reaching the ground, object of mass $2m$ have more kinetic energy,

A force 30 N is applied at the small plunger which moves down 10 m then work done by the applied force is

2046907_1393039_ans_3e7f73ef8ba44d0983ca1899a38a8d44.jpg

A bullet of mass 40 g has its kinetic energy equal to 200 J. Find the speed of the bullet.

Given,

Mass of the bullet

$, m = 40g = 0.04 kg$

Kinetic Energy of the bullet

$, KE = 200J$

we know,

Kinetic energy

$= \frac{1}{2} m v^2$

$200J = \frac{1}{2}(0.04\ kg )v^2$

solving for

$v= 100 m/s$

Hence the speed of the bullet is 100 m/s

Name the form of energy stored in a wound up spring of a watch .

The spring have stored potential energy.

What happens to potential energy of a body when its height is doubled?

we know ,

$P.E = mgh$

$P.E \propto h$

$New\ P.E = mg(2h) = 2(mgh) = 2 \times P.E$

So, when height doubles potential energy also doubles

In loading a trunk, a man lifts boxes of 100 N each through a height f 1.5 m,
(a) How much work does he do in lifting oe box?
(b) How much energy is transferred when one box is lifted ?
(c) if the man lifts 4 boxes per minutes, at what power is he working ? $(g = 10 m s^{-2})$

1358105_1559957_ans_02e70b28bfac4a3681c4e794ba03f263.jpg

Find the mass of a body which has $10\, J$ of kinetic energy while moving at a speed of $3\, m/s$ .

$KE=10J$ $m=?$ $v=3m/s$

$K.E = \dfrac{1}{2}$ $mv ^{2}$

$10 = \dfrac{1}{2} m(3)^2$

$\implies m = \dfrac{10 \times 2 }{9} = 2.22\ kg$

What is the kinetic energy of a body of mass $1\,kg$ moving with a speed of $2\,m/s$ ?

$Mass = 1\,kg$

$Velocity = 2\,m/s$

$KE = \dfrac12 \,mv^2$

$=\dfrac12\times1\times2^2$

$= 2\,J$

Write an expression for the kinetic energy of a body of mass m moving with a velocity v.

The kinetic energy of a body is given by:

$KE = \dfrac12 mv^2$

Where,

$m$ is the Mass and

$v$ is the Velocity.

State whether the following statement is true or false :
The potential energy of a body of mass $1\,kg$ kept at a height of $1\,m$ is $1\,J.$

The potential energy of the body is given as:

$PE = mgh$

$= (1) (9.8) (1)$

$= 9.8\,J$

The amount of potential energy in the body is not

$1\ J$ . So, the statement is False.

How does the kinetic energy of a moving body depends on its
a) speed
b) mass

a) Kinetic energy is directly proportional to the square of speed.
b) Kinetic energy is directly proportional to the mass.

K.E = $\dfrac{1}{2}$ $mv ^{2}$

On what factors does the kinetic energy of a body depend ?

K.E = $\dfrac{1}{2}$ $mv ^{2}$

Following are the factors on which the kinetic energy of a body depends:
a) mass of the body ,

$m$ .
b) square of velocity of the body ,

$v^2.$

Which would have a greater effect on the kinetic energy of an object: doubling the mass or doubling the velocity ?

K.E = $\dfrac{1}{2}$ $mv ^{2}$

If mass is doubled, K.E becomes doubled $(\times 2)$ .

If velocity is doubled, K.E becomes quadrupled $(\times 4)$ .

Doubling the velocity would have a greater effect on the kinetic energy of an object because the formula contains the square of velocity.

A weightlifter is lifting weights of mass $200\,kg$ up to a height of $2$ meters. If $g = 9.8\,m/s^2 ,$ calculate :
a) potential energy acquired by the weight
b) Work done by the weightlifter

Mass ,

$m = 200\,kg$
Height ,

$h = 2\,m$
Acceleration due to gravity ,

$g = 9.8 \,m/s^2$

a) Potential energy

$= mgh = 3920\,J$

b) Work done against gravity = potential energy gained by the weight
Therefore , work done ,

$W = mgh = 3920\,J$

How fast should a man of $50\,kg$ run so that his kinetic energy will be $625\,J$ ?

$Mass = 50\,kg$
Kinetic energy

$= 625\,J$

$v = ?$

Using the relation
:

$KE = \dfrac12 \,mv^2$

$625=\dfrac12\times50\times v^2$

$v^2=25$

$v = 5 \,m/s$

If a car of mass $1200 \ kg$ is moving along a highway at $120 \ km/h$ , what is the car’s kinetic energy as determined by someone standing alongside the highway?

Given , :

$v = 120 km/h = 120 \times \dfrac{1000}{3600} \ m/s = 33.33 m/s$ ,

mass

$m = 1200 \ kg$

Calculating Kinetic energy,

$K = \dfrac{1}{2} \ mv^2$

$= \frac{1}{2} (1200 \ kg) (33. 33 \ m/s)^2$

$= 6.67 \times 10^5 J$ .

What is the spring constant of a spring that stores $25\, J$ of elastic potential energy when compressed by $7.5 \,cm$ ?

The potential energy stored by the spring is given by

$U= \dfrac {1}{2} kx^2,$ where

$k$ is the spring constant and

$x$ is the displacement of the end of the spring from its position when the spring is in equilibrium. Thus

$k = \dfrac{2U}{x^2} = \dfrac{2(25\,J)}{(0.075\,m^2)} = 8.9 \times 10^3\,N/m.$

The angular acceleration of a wheel is $\alpha =6.0t^{4}-4.0t^{2}$ , with $\alpha$ in radians per second-squared and t in seconds. At time t=0, the wheel has an angular velocity of + 2.0 rad/s and an angular position of +1.0 rad. Write expressions for (a) the angular velocity (rad/s) and (b) the angular position (rad) as functions of time (s).

1751167_1765960_ans_1ec82d19a731460e933698911c75b6b4.jpg

A spring and block are in the arrangement of given figure.When the block is pulled out to $x = +4.0 \ cm$ , we must apply a force of magnitude $360 \ N$ to hold it there. We pull the block to $x = 11 \ cm$ and then release it. How much work does the spring do on the block as the block moves from $x_i = + 5.0 \ cm$ (a) $x = +3.0 \ cm$ , (b) $x = -3.0 \ cm$ , (c) $x = -5.0 \ cm$ , and (d) $x = -9.0 \ cm$ ?

From the given figure, we see that the work done by the spring force is given by

$W_s = \frac{1}{2}k(x_i^2 - x_f^2)$ .

The fact that 360 N of force must be applied to pull the block to x = + 4.0 cm implies that the spring constant is

$k = \frac{360 N}{4.0 cm} = 90 N/cm = 9.0 \times 10^3 N/m$

(a) When the block moves from

$x_i = + 5.0 \ cm$ to x = + 3.0 cm, we have

$W_s = \frac{1}{2}(9.0 \times 10^3 N/m) [(0.050 m)^2 - (0.030 m)^2 ] = 7.2 J$

(b) Moving from

$x_i = + 5.0 \ cm$ to x = - 3.0 cm, we have

$W_s = \frac{1}{2}(9.0 \times 10^3 N/m) [(0.050 m)^2 - (-0.030 m)^2 ] = 7.2 J$

$x_i = + 5.0 \ cm$ to x = - 5.0 cm, we have

$W_s = \frac{1}{2}(9.0 \times 10^3 N/m) [(0.050 m)^2 - (-0.050 m)^2 ] = 0 J$

(d) Moving from

$x_i = + 5.0 \ cm$ to x = - 9.0 cm, we have

$W_s = \frac{1}{2}(9.0 \times 10^3 N/m) [(0.050 m)^2 - (-0.090 m)^2 ] = -25 J$

In Fig. $8-42$ , a block of mass $m = 3.20\,kg$ slides from rest a distance $d$ down a frictionless incline at angle $\theta = 30.0^\circ$ where it
runs into a spring of spring constant $431 \,N/m.$ When the block momentarily stops, it has compressed the spring by $21.0\, cm.$ What are (a) distance $d$ and (b) the distance between the point of the first blockspring contact and the point where the blocks speed is greatest?

(a) The (final) elastic potential energy is

$U = \dfrac{1}{2} kx^2 = \dfrac{1}{2} (431\,N/m)(0.210\,m)^2 = 9.50\,J.$

Ultimately this must come from the original (gravitational) energy in the system

$mgy$ (where we are measuring y from the lowest “elevation” reached by the block, so

$y = (d + x) sin (30^\circ).$

Thus,

$mg(d + x) sin (30^\circ) = 9.50\,J \Rightarrow d = 0.396\,m.$

(b) The block is still accelerating (due to the component of gravity along the incline,

$mgsin(30^\circ)$ ) for a few moments after coming into contact with the spring (which exerts the Hooke’s law force kx), until the Hooke’s law force is strong enough to cause the block to begin decelerating.

This point is reached when

$kx = mg \,sin\,30^\circ$

which leads to

$x = 0.0364\,m = 3.64\,cm;$

this is long before the block finally stops (

$36.0 \,cm$ before it stops).

How much work is done by a force $\vec{F} = (2x \ N)\hat{i} + (3 \ N)\hat{j}$ , with x in meters, that moves a particle from a position $\vec{r_i} = (2 \ m)\hat{i} + (3 \ m) \hat{j}$ to a position $\vec{r_f} = -(4 \ m)\hat{i} - (3 \ m)\hat{j}$ ?

One approach is to assume a “path” from

$\vec{r_i}$ to

$\vec{r_f}$ and do the line-integral accordingly.

Another approach is to simply it, which we demonstrate:

$W = \int_{x_i}^{x_f} \ F_x dx + \int_{y_i}^{y_f} \ F_y dy = \int_{-4}^{2} (2x) dx + \int_{-3}^{3} (3) dy$

with SI units understood. Thus, we obtain

$W = 12 J – 18 J = – 6 J$ .

A can of sardines is made to move along the x-axis from x = 0.25 m to x = 1.25 m by a force with a magnitude given by $F = exp(-4x^2)$ , with x in meters and F in newtons. (Here exp is the exponential function.). How much work is done on the can by the force?

We know,

$\displaystyle W = \int_{0.25}^{1.25} e^{-4x^2}\ dx = 0.21 J$

where the result has been obtained numerically. Many modern calculators have that capability, as well as most math software packages that a great many students have access to.

Figure gives spring force $F_x$ versus position $x$ for the spring–block arrangement. The scale is set by $F_s = 160.0 \ N$ We release the block at $x = 12 \ cm$ . How much work does the spring do on the block when the block moves from $x_i = +8.0 \ cm$ to (a) $x = +5.0 \ cm$ , (b) $x = -5.0 \ cm$ , (c) $x = -8.0 \ cm$ , and (d) $x = -10.0 \ cm$ ?

The work done by the spring force is given by

$W_s = \frac{1}{2}k(x_i^2 - x_f^2)$ . Since

$F_x = -kx$ , the slope in corresponds to the spring constant k. Its value is given by k = 80 N/cm =

$8.0 \times 10^3 N/m$ .

(a) When the block moves from

$x_i = +8.0 cm$ to x = + 5.0cm, we have

$W_s = \frac{1}{2}(8.0 \times 10^3 N/m)[(0.080 m)^2 - (0.050 m)^2] = 15.6 J = 16 J$ .

(b) Moving from

$x_i = +8.0 cm$ to x = - 5.0cm, we have

$W_s = \frac{1}{2}(8.0 \times 10^3 N/m)[(0.080 m)^2 - (-0.050 m)^2] = 15.6 J = 16 J$ .

$x_i = +8.0 cm$ to x = - 8.0cm, we have

$W_s = \frac{1}{2}(8.0 \times 10^3 N/m)[(0.080 m)^2 - (-0.080 m)^2] = 0 J$ .

(d) Moving from

$x_i = +8.0 cm$ to x = - 10.0cm, we have

$W_s = \frac{1}{2}(8.0 \times 10^3 N/m)[(0.080 m)^2 - (-0.10 m)^2] = -14.4 J = -14 J$ .

During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with large catapults that are made with surgical hose mounted on a window frame. A balloon filled with dyed water is placed in a pouch attached to the hose, which is then stretched through the width of the room. Assume that the stretching of the hose obeys Hookes law with a spring constant of $100 \ N/m$ . If the hose is stretched by $5.00 \ m$ and then released, how much work does the force from the hose do on the balloon in the pouch by the time the hose reaches its relaxed length?

The spring constant is k = 100 N/m and the maximum elongation is

$x_i = 5.00 m$ .

with

$x_f = 0$ ,

The work is found to be

$W = \frac{1}{2}kx_i^2 = \frac{1}{2}(100 N/m)(5.00m)^2$

$= 1.25 \times 10^3 J$ .

An automobile with passengers has weight $16 \,400 \,N$ and is moving at $113\, km/h$ when the driver brakes, sliding to a stop. The frictional force on the wheels from the road has a magnitude of $8230\,N$ . Find the stopping distance.

The initial kinetic energy of the automobile of mass m moving at speed

$v_i$ is

$K_i = \dfrac{1}{2} mv^2_i,$ where

$m = 16400 / 9.8 = 1673 \,kg.$ Using Eq.

$8-31$ and Eq.

$8-33,$ this relates to the effect of friction force f in stopping the auto over a distance d by

$K_i = fd,$

where the road is assumed level (so

$\Delta U = 0).$ With

$\displaystyle v_{i} = (113 \,km/h) (1000 \,m/km)(1 \,h / 3600 \,s) = 31.4 \,m/s,$

we obtain

$d = \dfrac{K_i}{f} = \dfrac{mv^2_i}{2f} = \dfrac{(1673 \,kg)(31.4 \,m/s)^2 }{2(8230 \,N)} = 100 \,m.$

A $91 \,kg$ man lying on a surface of negligible friction shoves a $68 \,g$ stone away from himself, giving it a speed of $4.0\, m/s.$ What speed does the man acquire as a result?

No external forces with horizontal components act on the man-stone system and the sum of the vertical forces to zero, so the total momentum of the system is conserved. Since the man and the stone are initially at rest, the total momentum is zero both before and after the stone is kicked. Let

$m_s$ be the mass of the stone and

$\nu_s$ be its velocity after it is kicked; let

$m_m$ be the mass of the man and

$\nu_m$ be his velocity after he kicks the stone. Then

$m_s \, \nu_s + m_m \, \nu_m = 0 \, \rightarrow \, \nu_m = - m_s \nu_s / m_m$ .

We take the axis to be positive in the direction of the motion of the stone. Then

$\nu_m = -\dfrac{(0.068\, kg)(4.0\, m/s)}{91 \, kg} = - 3.0 \times 10^{-3} \, m/s$ .

or

$| \nu_m| = 3.0 \times 10^{-3} \, m/s$ . The negative sign indicates that the man moves in the direction opposite to the direction of motion of the stone .

A $6100\, kg$ rocket is set for vertical firing from the ground. If the exhaust speed is $1200\, m/s,$ how much gas must be ejected each second if the thrust (a) is to equal the magnitude of the gravitational force on the rocket and (b) is to give the rocket an initial upward acceleration of $21\, m/s^2$ ?

(a) The thrust is

$R_{rel}$ where

$\nu_{rel} = 1200 \, m/s$ . For this to equal the weight

$Mg$ where

$M = 6100 \,kg$ , we have

$R = (6100)(9.8) / 1200 \approx 50 \,kg/s$ .

(b) Using Eq.

$9-42$ with the additional effect due to gravity , we have

$R \nu_{rel} - Mg = Ma$

so that requiring

$a = 21 \, m/s^2$ leads to

$R = (6100)(9.8 + 21) / 1200 = 1.6 \times 10^2 \, kg/s$ .

As a particle moves along an x axis, a force in the positive direction of the axis acts on it. Figure shows the magnitude F of the force versus position x of the particle.The curve is given by $F = a/x^2$ with $a = 9.0 \ N.m^2$ .
Find the work done on the particle by the force
as the particle moves from x = 1.0 m to x = 3.0 m by
(a) estimating the work from the graph
(b) integrating the force function.

(a) To estimate the area under the curve between x = 1 m and x = 3 m (which should yield the value for the work done),

one can try “counting squares” (or half-squares or thirds of squares) between the curve and the axis.

Estimates between 5 J and 8 J are typical for this (crude) procedure.

(b) Since we know that,

$\int_{1}^{3} \ \frac{a}{x^2} dx$

$= \frac{a}{3} - \frac{a}{1} = 6 \ J$

where

$a = –9 \ N·m^2$ is given in the problem statement.

1678066_1766759_ans_a4d660d0896647b384985eb8c7a7afa3.png

A $140 \,g$ ball with speed $7.8 \,m/s$ strikes a wall perpendicularly and rebounds in the opposite direction with the same speed. The collision lasts $3.80\, ms.$ What are the magnitudes of the (a) impulse and (b) average force on the wall from the ball during the elastic collision?

(a) The magnitude of the impulse is equal to the change in momentum .

$J = m \nu - m(-\nu) = 2 m \nu = 2 ( 0.140 \, kg ) (7.80 \, m/s) = 2.18 \, kg \cdot m/s$

(b) Since in the calculus sense the average of a function is the integral of it divided by the corresponding interval, then the average force is the impulse divided by the time

$\Delta t$ .Thus, our result for the magnitude of the average force is

$2 m \nu / \Delta t$ With the given values, we obtain

$F_{avg} = \dfrac{2(0.140\,kg)(7.80 \, m/s)}{0.00380 \, s} = 575 \, N$

Calculate the kinetic energy of a body of mass $50 \,kg$ moving with a velocity of $50 \,cms^{-1}$

Given,
Mass

$m = 50 \,kg$

$Velocity = 50 \,cms^{-1}$

$1 \,cm = 0.01 \,m$
So,

$velocity = 50 \times 0.01 \,ms^{-1}$

$v = 0.5 \,ms^{-1}$
We know,
Kinetic energy

$= \dfrac{1}{2} mv^2$
Substituting value of m and v in the equation:
Kinetic energy

$= \dfrac{1}{2} \times 50 \times 0.5^2$

$= 6.25 \,J$

A $4.0 \,kg$ mess kit sliding on a frictionless surface explodes into two $2.0\, kg$ parts: $3.0\, m/s,$ due north, and $5.0\,m/s, 30^{\circ}$ north of east .What is the original speed of the mess kit?

Our +x direction is east and +y direction is north . The linear momenta for the two

$m = 2.0 \, kg$ parts are then

$\vec{p}_1 = m \vec{\nu}_1 = m \vec{\nu}_1 \hat{j}$

where

$\nu_1 = 3.0 \, m/s$ , and

$\vec{p}_2 = m \vec{\nu}_2 = m \left ( \nu_{2x} \hat{i} + \nu_{2y} \hat{j} \right ) = m \nu_2 ( cos \theta \hat {i} + sin \theta \hat{j} )$

where

$\nu_2 = 5.0 \, m/s$ and

$\theta = 30^{\circ}$ . The combined linear momentum of both parts is then

$\vec{P} = \vec{p}_1 + \vec{p}_2 = m \nu_1 \hat{j} + m \nu_2 ( cos \theta \hat{i} + sin \theta \hat{j} ) = (m \nu_2 \, cos \theta ) \hat{i} + ( m \nu_1 + m \nu_2 \, sin \theta ) \hat{j}$

$= (2.0 \, kg)(5.0 \, m/s) (cos \, 30^{\circ} ) \hat{i} + (2.0 \,kg) ( 3.0 \, m/s + ( 5.0 \, m/s)(sin \, 30^{\circ}))\hat{j}$

$= ( 8.66 \hat{i} + 11 \hat{j} ) \, kg \cdot m/s$ .

From conservation of linear momentum we know that this is also the linear momentum of the whole kit before it splits. Thus the speed of the

$4.0-kg$ kit is

$\nu = \dfrac{P}{M} = \dfrac{\sqrt{P_{x}^{2} + P_{y}^{2}}}{M} = \dfrac{\sqrt{(8.66 \, kg \cdot m/s)^2 + (11 \, kg \cdot m/s)^2}}{4.0 \, kg} = 3.5 \, m/s$

A force $\vec{F}$ in the positive direction of an x axis acts on an object moving along the axis. If the magnitude of the force is $F = 10e^{-x/2.0} \ N$ , with x in meters, find the work done by $\vec{F}$ as the object moves from x = 0 to x = 2.0 m by (a) plotting F(x) and estimating the area under the curve and (b) integrating to find the work analytically.

(a) The plot of the function (with SI units understood) is shown below.

Estimating the area under the curve allows for a range of answers.

Estimates from 11 J to 14 J are typical.

(b) Evaluating the work analytically , we have

$W = \int_{0}^{2} \ 10 e^{-x/2} dx$

$=-20 e^{-x/2} |_{2}^{0}$

$= 12.6 \ J = 13 \ J$

1677892_1766738_ans_c4406e3ac8a84579a999ea00ba16cd64.png

Numerical integration. A breadbox is made to move along an x axis from x = 0.15 m to x = 1.20 m by a force with a magnitude given by $F = exp(-2x^2)$ , with x in meters and F in newtons. (Here exp is the exponential function.) How much work is done on the breadbox by the force?

The problem indicates that SI units are understood, so the result is in joules.

Done numerically, using features available on many modern calculators,

the result is roughly

$0.47 \ J$ .

For the interested student it might be worthwhile to quote the“exact” answer (in terms of the “error function”):

$\int_{.15}^{1.2} \ e^{-2x^2} dx$

$= 1/4 \sqrt{2\pi} \ [erf(6\sqrt{2} / 5) - erf(3\sqrt{2} / 20)]$ .

Does a stationary body possess kinetic energy ?

Kinetic energy of a body is due to its motion or velocity, if the body is stationary then velocity

$= 0$
Kinetic energy

$= \dfrac{1}{2} mv^2$
Substituting value of velocity v in the equation we get kinetic energy

$= 0$
So, a stationary body does not possess kinetic energy.

A raindrop of mass $1.00\ g$ falling from a height of $1\ km$ hits the ground with a speed of $50\ m/s$ . Calculate the gain in $K.E.$ of the drop.

Givrn,

$m=1\, g=0.001\ kg$ ,

$h=1\ km=1000\ m$
Speed of

$v=50 \ m/s$

Since, initial velocity of the drop is

$0$ .

Gain in

$KE=\dfrac 12 mv^2$

$=\dfrac 12 \times 0.001\times 50\times 50=1.25$
Gain in

$KE=1.25\ J$

Find the energy possessed by an object of mass $10 \,kg$ when it is at a height of $6m$ above the ground.
Take, $g = 9.8 \,ms^{-1}$

Given,
Mass of object

$m = 10\, kg$ .
Height

$h = 6\, m$ .

Acceleration due to gravity

$g = 9.8\, m/s^{2}$

Potential Energy of object

$P.E = m\times g\times h$

$\Rightarrow mgh =10\times9.8\times6=98\times6$

$= 588\, J$
Energy possessed by the object is

$588\, Joule$ .

To what height should a body of mass 5 kg be raised so that its potential energy is 490 J?
Take, $g = 9.8 \,m/s^2$

The potential energy of a body due to gravitation is given by:

$P.E = m\times g\times H$

Where, m = mass,

g = acceleration due to gravity,
H = height,

Given,

$mass = 5\,kg, P.E = 490\,J, g = 9.8\,m/s^2$ ,

$H =?$

$490=5\times 9.8\times h$

$\Rightarrow h=\dfrac{490}{9.8\times 5}=\dfrac{490}{49}=10 \,m$
Hence, the body must be raised by

$10\, m$ .

A ball is drop from a height of $10\ m$ . If the energy of the ball reduced by $40\%$ after striking the ground, how much height can the ball bounce back?

$mgh =m\times 10\times 10=100 mJ$
Energy is reduced by

$40\%$ then the remaining energy is

$60\ mJ$
Therefore,

$60m=m\times 10\times h$ or

$h'-6m$

Name the type of energy (kinetic or potential) possessed by the following:
A stone at rest on the top of a building.

A Stone at rest on the top of a building possesses POTENTIAL ENERGY ,

$\therefore$ Due to height above ground.

The energy possessed by a body due to its position is called.............. energy.

The energy possessed by a body due to its position is called

$potential$ energy.

Two toy-cars $A$ and $B$ of masses $200g$ and $500g$ respectively are moving with the same speed. Which of the two has grater kinetic energy?

Kinetic energy

$K=\dfrac{1}{2} mv^2$

$\therefore K\propto m$

Since both bodies have different masses but have the same speed.

Hence, toy car

$B$ has greater kinetic energy as it has greater mass.

Name the type of energy (kinetic or potential) possessed by the following:
A stretched rubber band.

When we stretch a rubber band. The energy transferred to the band is stored as potential energy.

The potential energy possessed by the object is the energy present in it by virtue of its position or configuration.

The energy of a body by virtue of its motion is known as kinetic energy.

Energy possessed by a compressed spring is ......... energy.

Energy possessed by a compressed spring is

$Elastic$

$potential$ energy.

Kinetic energy= $1/2 \times mass \times$ ______ .

Kinetic energy=

$1/2 \times mass \times (velocity)^2$

Name the type of energy (kinetic or potential) possessed by the following:
A compressed spring.

A compressed spring possesses potential energy by virtue of its configuration. The work done by a person gets stored in it as elastic potential energy of the spring. This energy has "potential" to do work once released.

A cyclist doubles his speed. How will his kinetic energy change: increase, decrease or remain same?

Kinetic energy $K.E. = \dfrac{1}{2}$ $mv ^{2}$

$K.E.\propto v^2$

$\dfrac{K.E._1}{K.E._2}=\dfrac{(v)^2}{(2v)^2}= \dfrac{v^2}{4v^2}=\dfrac{1}{4}$

When speed

$v$ doubles, the K.E. will increase four times.

What do you understand by the kinetic energy of a body?

A body in motion is said to possess kinetic energy. The energy possessed by a body by virtue of its state of motion is called kinetic energy.

$KE = \dfrac{1}{2} mv^2$

m; the mass of the object

v: velocity of the object

A body of mass m is moving with a uniform velocity u. A force is applied on the body due to which its velocity changes from u to v. How much work is being done by the force.

The initial kinetic energy of the body is

$K_1=\dfrac{1}{2}mu^2$ ,

$u$ is the initial velocity.

The final kinetic energy of the body is

$K_2=\dfrac{1}{2}mv^2$ ,

$v$ is the final velocity.

$\text{Work done by the force}=\text{Change in kinetic energy}$

$\therefore W =\dfrac{1}{2}m(v^2-u^2)$

A car is moving with a speed of $15 km h^{-1}$ and another identical car is moving with a speed of $30 km h^{-1}$ . Compare their kinetic energy.

Two identical cars means , they have equal mass

$K.E. = \dfrac{1}{2} mv^2$

Let

$m$ be the mass of each car

Kinetic energy of car A =

$\dfrac {1}{2} m \times (15)^2 = \dfrac {225}{2}m$

Kinetic energy of car B =

$\dfrac{1}{2}m \times ( 30)^2 = 450m$

$\dfrac {K.E. of A }{K.E. of B } = \dfrac { 225 m}{ 2 \times 450 m} = \dfrac {1}{4} = 1 : 4$

K.E. of car B is

$4$ times K.E. of car A.

Find the kinetic energy of a body of mass $1\, kg$ moving with a uniform velocity of $10\, m\, s^{-1}$

Mass,

$m = 1\,kg$

Velocity,

$v = 10\, m/s$

Kinetic energy

$= \dfrac {1}{2} \times mass \times (velocity)^2$

$= \dfrac {1}{2}\times 1 \times (10)^2 = \dfrac {1}{2} \times 1 \times 100$

$= 50\, J$

If the speed of a car is halved, how does its kinetic energy change?

$KE=\dfrac{1}{2}mv^2$

$KE \propto v^2$

$v\rightarrow v/2$

$KE\rightarrow KE/4$

If the speed is halved (keeping the mass same), the kinetic energy decreases, it becomes one-fourth (since kinetic energy is proportional to the square of velocity).

The potential energy of a body of mass $0.5\ kg$ increases by $100\ J$ when it is taken to the top of the tower from the ground. If forces of gravity on $1\ kg$ is $10\ N$ , what is the height of the tower?

$\text{Potential energy} = ( mg) h$

$\Rightarrow PE = \text{Force} \times \text{height}$

According to the question potential energy is

$100\ J$ ,

$\Rightarrow 100 =( 0.5 \times 10) \times h$

$\Rightarrow h = \dfrac{PE}{mg}$

If mass is

$1\ kg$ then force is

$10\ N$ so force on

$0.5\ kg$ will be

$5\ N$

$\Rightarrow h = \dfrac{100}{5}$

$\Rightarrow h = 20\ m$

A body of mass $60$ kg is moving with a speed $50 ms^{-1}$ . find its kinetic energy .

Given:-

$m = 60 kg$

$Speed,\ v =50 m/s$

$K.E. = \dfrac{1}{2} mv^2 = 75000\ J$

It takes $20 s$ for a girl $A$ to climb up the stairs while girl $B$ takes $15 s$ for the same job. compare :
$(i)$ : the work done and
$(ii)$ : the power spent by then .

(i) As height is same for both girls A and B work done is same (irrespective of time)

∴ Work done by A : Work done by B = 1 : 1.

Power =

$\dfrac{work \, done}{Time \,taken}$ =

$\dfrac{Energy}{Time}$

$\Rightarrow$ work done

$\times$ Time taken = Energy

$\times$ Time

=Power of A : Power of B = Energy

$\times t_1$ : Energy

$\times t_2$ =

$t_1 : t_2$ =

$15 : 20$ =

$3 : 4$

Two bodies of equal masses are moving with uniform velocities $v$ and $2v$ . Find the ratio of their kinetic energies.

Given velocity of first body

$v_1 =v$

And velocity of second body,

$v_2 = 2v$

$KE=\dfrac{mv^2}{2}$

Since masses are same, kinetic energy is directly proportional to the square of the velocity

$(KE$

$\alpha$

$v^2)$

Hence, ratio of their kinetic energies is:

$\dfrac{KE_1}{KE_2}= \dfrac{v_1^2}{v^2 _2}= \dfrac{v^2}{(2v)^2}= \dfrac{v^2}{4v^2}= \dfrac{1}{4}= 1:4$

A box of weight $150\, kgf$ has gravitational potential energy stored in it equal to $14700\, J$ . Find the height of the box above the ground. (Take $g = 9.8\, N \,kg^{-1}$ )

Gravitational potential energy

$14700\, J$
Force of gravity

$= mg = 150 \times 9.8\, N/kg = 1470\,N$
Gravitational potential energy

$= mgh$

$14700 =1470 \times h$

$h = 10\, m$

A car is running at a speed of $15\, km\, h^{-1}$ while another similar car is moving at a speed of $30\, km \,h^{-1}$ . Find the ratio of their kinetic energies.

Given, velocity of first car,

$v_1=15 km/h$

And velocity of second car,

$v_2$ =

$30 \, km/h$

Cars are similar so masses are same.

Kinetic energy

$KE=\dfrac{1}{2}mv^2$

Kinetic energy is directly proportional to the square of the velocity

$(K$

$\alpha$

$v^2)$

Hence, ratio of their kinetic energies is:

$\dfrac{k_1}{k_2}= \dfrac{v_1^2}{v^2 _2}= \dfrac{(15)^2}{(30)^2}= \dfrac{15\times 15}{30\times 30}= \dfrac{1}{4}= 1:4$

Name the energy stored when a rubber band is stretched.

When we stretch a rubber band. The energy transferred to the band is stored as potential energy. The potential energy possessed by the object is the energy present in it by virtue of its position or configuration.

Name the form of energy present in the following situations:
(a) stretched bow and arrow
(b) a falling coconut

(a) A stretched bow and arrow has potential energy as it is in the condition of strain.

(b) A falling coconut has kinetic energy as it is in motion.

A machine moves an object of 40 kg mass to 10 m height. Find the amount of work done.

Here mass = 40 kg, h = 10 m and

$g = 9.8 m/s^2$

Work done can be calculated as follows:

$W = F.s = mgh$

$W= 40 kg \times 9.8 ms^2 \times 10$

$W= 3920 J = 3.92 kJ$

An object is moving with velocity u. Its velocity becomes v when force F is applied on it. If the object covers a distance s during this, then calculate the increase in kinetic energy of object.

According to third equation of motion;

$V^2 = u^2 + 2as$

Or,

$a=\dfrac{v^2-u^2}{2s}$

As per Newton's Second Law of Motion:

$F = ma$

Or,

$F=m(\dfrac{v^2-u^2}{2s})$

or,

$F.s=\dfrac{1}{2}m(u^2-u^2)$

Since,

$W=F.s$

Hence,

$W=\dfrac{1}{2}m(u^2-u^2)$

or,

$W=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$

This shows that work done is equal to change in kinetic energy of object.

Derive the formula for kinetic energy.

Let an object of mass m, move with uniform velocity u.

Let us displace it by s, due to constant force F, acting on it.

Work done on the object of mass 'm' is

$W=F \times s$ ...(i)

Due to the force, velocity changes to v and the acceleration produced is 'a'. Relationship between v, u, a and s can be given by formula

$v^2- u^2 = 2as$

$\therefore s=\dfrac{v^2-u^2}{2a}$ ..........(ii)

$F=ma$ ............(iii)

Substituting (ii) and (iii) in (i) we get

$W =F \times s$

$=ma\times \dfrac{v^2-u^2}{2a}$

$W=\dfrac{1}{2}m(v^2-u^2)$

If u=0,(object starts at rest)

$W=\dfrac{1}{2}mv^2$

Work done=Change in kinetic energy

$\therefore E_k=\dfrac{1}{2}mv^2$

Which energy is stored in the spring while winding a clock? In which energy it gets converted when the clock is working?

Potential energyand and Kinetic energy.

Stressing the rubber in sling-shot, the………………… energy is stored in it.

Stressing the rubber in a sling-shot, the potential energy is stored in it. Because in this situation, work is done on the rubber by an external agent.

Define kinetic energy. Give an example of a body possessing kinetic energy.

The energy possessed by a body due to its motion is called kinetic energy.

Ex:- A bullet fired from a gun can penetrate a target due to its kinetic energy.

$K.E = \dfrac{1}{2}mv^2$

where

$m$ is mass and

$v$ is velocity of object.

A man having a mass of 70kg is riding scooter of mass 80kg. What is the total energy if the velocity of the scooter is 10 m/s?

$m = 70 kg + 80 kg = 150kg$

$v=10m/s$

$K=\dfrac{1}{2}mv^{2}=\dfrac{1}{2}\times 150\times 10^{2}=7500J=7.5kJ$

Three equal masses are placed on three corners of equilateral triangle. What is work done in doing this ?

When

$m_{a}$ is brought no work is done when

$m_{a}$ is brough work dine.

Will the potential energy increase or decrease in pressing a solid or stretching a wire?

During pressing or stretching, work has done on the Solid. So The potential energy will increase in both situations.

State whether the energy possed in the following is kinetic or potential.
Bird resting on the branch of a tree

Potential energy is the energy of an object by virtue of its position.

Since, the Bird resting on the branch of tree is at a height from the ground. It possesses potential energy.

What is kinetic energy ? Mention the factors that the kinetic energy of a body depends.

The energy possessed of a body by virtue of its motion is called kinetic energy.

$KE = \dfrac{1}{2} mv^2$
Kinetic energy depends on

the mass of the body and
velocity with which it is moving

A car of mass $1500\ kg$ is moving at a velocity of $20\ m/s$ . What is its kinetic energy?

Given,

Mass,

$m = 1500 kg$
Velocity,

$v = 20 m/s$

We know,

Kinetic energy,

$KE = \dfrac{1}{2}mv^{2}$

$KE=\dfrac 12 \times 1500\times 20^{2}$

$KE=300000\ J$

$KE = 300\ kJ$

_______ is the energy possessed by an object due to its motion.

Kinetic energy.

the thread tension at the moment when the vertical component of the sphere's velocity is maximum;

Vertical component of velocity

$v_{y}=v\sin \theta$

So,

$v_{y}^{2}=v^{2}\sin^{2}\theta=2g l\cos\theta\sin^{2}\theta$ (using

$1$ )

For maximum

$v_{y}$ or

$v_{y}^{2}, \dfrac{d(\cos \theta\sin^{2}\theta)}{d\theta}=0$

which yields

$\cos\theta=\dfrac{1}{\sqrt{3}}$

Therefore from

$(2) T=3mg\dfrac{1}{\sqrt{3}}=\sqrt{3}mg$

1792904_1851905_ans_1bce1385bbf94a2b826ce87b895d1bda.png

Calculate the kinetic energy of an athlete of mass 60 kg running with a velocity 10 m /s
$m=60kg$
$v = 10 m/s$

Kinetic energy K =

$K=\dfrac{1}{2}mv^{2}=\dfrac{1}{2}\times 60\times 10^{2}=3000J$

A bird of mass $0.5 \ kg$ is flying at the height of $5 \ m$ . In this state, if its kinetic energy and potential energy are equal. What is the potential energy of the bird?

Given, mass

$m = 0.5 kg$

height

$h = 5 \ m$

$g=10m/s^{2}$

Given that Potential energy = Kinetic energy

Hence,
Potential energy

$U=\dfrac12m$

$u^2=mgh$

$U=0.5\times 10\times 5=25 \ J$

A $40.0kg$ child swings in a swing supported by two chains, each $3.00 m$ long. The tension in each chain at the lowest point is $350 N$ . Find (a) the childs speed at the lowest point and (b) the force exerted by the seat on the child at the lowest point. (Ignore the mass of the seat.)

We first draw a force diagram that shows the forces acting on the child-seat system and apply Newton’s second law to solve the problem. The child’s path is an arc of a circle, since the top ends of the chains are fixed. Then at the lowest point the child’s motion is changing in direction: He moves with centripetal acceleration even as his speed is not changing and his tangential
acceleration is zero.
(a) Below figure shows that the only forces acting on the system of child + seat are the tensions in the two chains and the weight of the boy:

$\sum F=F_{net}=2T-mg=ma=\dfrac{mv^{2}}{r}$
with

$F_{net}=2T-mg=2(350N)-(40.0kg)(9.80m/s^{2})=308N$
solving for

$v$ gives,

$v=\sqrt{\dfrac{F_{net}r}{m}}=\sqrt{\dfrac{(308N)(3.00m)}{40.0kg}}=4.81m/s$
(b) The normal force from the seat on the child accelerates the child in the same way that the total tension in the chain accelerates the child-seat system. Therefore,

$n = 2T = 700 N$
1801979_1860363_ans_12fe46c1566f444083c670a0a1a7753a.png

Let the mass of the flower pot 15 g and the height of the sunshade 4 m. When the flower pot is on the sunshade, what is its potential energy ? $(g = 10 m/s^2)$

Given

mass,

$m=\dfrac{15}{1000}=0.015\ kg$

height,

$h=\ 4m$

$PE=mgh$

$=0.015\times 10\times 4$

$=0.6\ J$

A bird of mass 0.5 kg is flying the same speed at the same height of 5 m. In this state, if its Kinetic energy and potential energy are equal. What is the velocity of the bird ?

$m = 0.5 kg$

$g=10m/s^{2}$

$U=\dfrac12m$

$u^2=mgh$

$U=0.5\times 10\times 5=25J$

Hence

$K = 25 J (\therefore U=K)$

$\dfrac{1}{2}\times 0.5\times v^{2}=25$

$v^{2}=\dfrac{25\times 2}{20.5}$

$=\dfrac{50}{0.5}$

$V^2=100\\\therefore v\sqrt{100}$

$V=10m/s$

A boy of mass 50 kg is riding a bicycle with a speed 2m /s . The bicycle has a mass of 10kg. Calculate the total kinetic energy?

$m = 50kg + 10kg = 60 kg$

$v = 2m/s$

$k=\dfrac{1}{2}mv^{2}$

$1/2\times 60\times 2^{2}=120J$

Due to the external vertical force $F_x=F_0 \cos \omega t$ a body suspended by a spring performs forced steady-state oscillations according to the law $x=a \cos (\omega t- \phi)$ . Find the work performing by the force $F$ during one oscillation period.

The work done in one cycle is

$A= \displaystyle \int F dx= \int_0^ T F v dt =\int _0 ^T F_0 \cos \omega t(- \omega a \sin (\omega t- \phi))dt$

$=\int_0^T F_0 \omega a (- \cos \omega t \sin \omega t \cos \phi+ \cos^2 \omega t \sin \phi )dt$

$=\dfrac{1}{2}F_0 \omega a \dfrac{T}{2} \sin \phi = \pi a F_0 \sin \phi$

A body of mass $20 \ kg$ is moving with a velocity of $5 \ m /s$ . Calculate its kinetic energy?

Given,

$mass \ m = 20 \ kg$ ,

$velocity \ v = 5 \ m/s$

Kinetic energy

$KE=\dfrac{1}{2}mv^{2}$

Hence

$KE=\dfrac12\times 20\times 5^{2}=\dfrac{1}{2}\times 20 \times 25=250 \ J$

Calculate the potential energy of a body of mass 1 kg at a height of 6m from the ground ?

Mass m = 1 kg, Acceleration due to gravity

$g = 10m/s^{2}$

$h=6m$

$U=mgh=1\times 10\times 6=60J$

Jane, whose mass is $50.0-kg$ needs to swing across a river (having width $D$ )filled with person-eating crocodiles to save Tarzan from danger. She must swing into a wind exerting constant horizontal force $\vec F$ on a vine having length $L$ and initially making an angle $\theta$ with the vertical Fig Take $D=50.0\ m, F=110\ N,\ L=40.0\ m$ and $\theta=50.0^o$ With what minimum speed must Jane begin her swing to just make it to the other side?

1978992_1865342_ans_9717399b204b4d1097f8d9d04432b67c.jpg

A child of mass m swings in a swing supported by two chains, each of length $R$ . If the tension in each chain at the lowest point is $T$ , find (a) the childs speed at the lowest point and (b) the force exerted by the seat on the child at the lowest point. (Ignore the mass of the seat.)

Follow the figure below:
(a)

$\sum F=2T-Mg=\dfrac{Mv^{2}}{R}$

$v^{2}=(2T-Mg)\left ( \frac{R}{M} \right )$

$v=\sqrt{(2T-Mg)\left ( \frac{R}{M} \right )}$
(b)

$n-Mg=F=\frac{Mv^{2}}{R}$

$n=Mg+\dfrac{Mv^{2}}{R}$
1802064_1860374_ans_992b0b73ec4d44d0bac053044234abc3.png

Jane, whose mass is $50.0-kg$ needs to swing across a river (having width $D$ )filled with person-eating crocodiles to save Tarzan from danger. She must swing into a wind exerting constant horizontal force $\vec F$ on a vine having length $L$ and initially making an angle $\theta$ with the vertical Fig Take $D=50.0\ m, F=110\ N,\ L=40.0\ m$ and $\theta=50.0^o$ Once the recure is complete, Tarzan and Jane must swing back across the river. With what minimum speed must they begin their swing? Assume tarzan has a mass of $80.0$ kg

1978993_1865344_ans_c2784e29bd8545c38689331b8e3e7ada.jpg

A flexible chain weight $40.0N$ hangs between two hooks located at the same height $Fig.P12.15$ . At each hook, the tangent to the chain makes an angle $\theta =42.0^o$ with the horizontal. Find
The tension in the chain at its midpoint.
suggestion: make a force diagram for half of the chain.

1978983_1865331_ans_cde5dc50293040cca884d36d1e2ae0a7.jpg

A particle is subject to a force $F_{x}$ that varies with position as shown in above figure. Find the work done by the force on the particle as it moves (a) from $x = 0$ to $x = 5.00 m$ , (b) from $x = 5.00 m$ to $x = 10.0 m$ , and (c) from $x = 10.0 m$ to $x = 15.0 m$ . (d) What is the total work done by the force over the distance $x = 0$ to $x = 15.0 m$ ?

We use the graphical representation of the definition of work.

$W$ equals the area under the force-displacement curve. This definition is still written

$W =\int F_{x}dx$ but it is computed
geometrically by identifying triangles and rectangles on the graph.
(a) For the regoin

$0\leq x\leq 5.00m$

$W=\dfrac{(3.00N)(5.00m)}{2}=7.50J$
(b) For the region

$5.00\leq x\leq 10.0,W=(3.00N)(5.00m)=15.0J$
(c) For the regoin

$10.00\leq x\leq 15.0,W=\dfrac{(3.00N)(5.00m)}{2}=7.50J$
(d) For the region

$0\leq x\leq 15.0,W=(7.50+7.50+15.0)J=30.0J$
1804925_1862248_ans_151a687c16344f26bf5263ebd71cf323.png

A $4.00kg$ particle is subject to a net force that varies with position as shown in above figure. The particle starts from rest at $x = 0$ . What is its speed at (a) $x = 5.00 m$ , (b) $x = 10.0 m$ , and (c) $x = 15.0 m$ ?

(a)

$\Delta k=k_{f}-k_{i}=\dfrac{1}{2}mv_{f}^{2}-0=\sum W$ = (area under curve from

$x=0$ to

$x=5.00m$ )

$v_{f}=\sqrt{\dfrac{2(area)}{m}}=\sqrt{\dfrac{2(7.50J)}{4.00kg}}=1.94m/s$
(b)

$\Delta k=k_{f}-k_{i}=\dfrac{1}{2}mv_{f}^{2}-0=\sum W$ = (area under curve from

$x=0$ to

$x=10.0m$ )

$v_{f}=\sqrt{\dfrac{2(area)}{m}}=\sqrt{\dfrac{2(22.5J)}{4.00kg}}=3.35m/s$
(c)

$\Delta k=k_{f}-k_{i}=\dfrac{1}{2}mv_{f}^{2}-0=\sum W$ = (area under curve from

$x=0$ to

$x=15.0m$ )

$v_{f}=\sqrt{\dfrac{2(area)}{m}}=\sqrt{\dfrac{2(30.0J)}{4.00kg}}=3.87m/s$

An adventurous archeologist $(m = 85.0 kg)$ tries to cross a river by swinging from a vine. The vine is $10.0 m$ long, and his speed at the bottom of the swing is $8.00 m/s$ . The archeologist doesnt know that the vine has a breaking strength of $1 000 N$ . Does he make it across the river without falling in?

Let the tension at the lowest point be

$T$ . From Newton’s second law

$\sum F=ma$ and

$T-mg=ma_{c}=\dfrac{mv^{2}}{r}$

$T=m\left ( g+\dfrac{v^{2}}{r} \right )$

$T=(85.0kg)\left [ 9.80m/s^{2}+\dfrac{(8.00m/s)^{2}}{10.0m} \right ]$

$=1.38kN>1000N$
He doesn’t make it across the river because the vine breaks.
1802089_1860434_ans_f88e3c87c5f2422c843156b3b9939654.png

A ball of mass $m=300 g$ is connected by a strong string of length $L=80.0\ cm$ to a pivot and held in p[lace with the string vertical. A wind exerts constant force $F$ to the night on the ball as shown Figure The ball is released from rest. The wind makes it swing up to attain maximum height $J$ above its starting point before it swings down again as $F$ approaches zero

1896540_1865365_ans_c7aab9e4d74d4feb93d30acf87786375.jpg

Two bodies A and B have masses in the ratio $5:1$ and their kinetic energies are in the ratio $125:9$ . Find the ratio of their velocities.

$\textbf{Step 1: Kinetic energy of bodies}$

$K_A=\dfrac{1}{2}m_Av_A^2$

$....(1)$

$K_B=\dfrac{1}{2}m_Bv_B^2$

$....(2)$

$\textbf{Step 2: Ratio of velocities}$

Ratio of masses

$\dfrac{m_A}{m_B} = \dfrac{5}{1}$

Ratio of Kinetic Energies

$\dfrac{K_A}{K_B} = \dfrac{125}{9}$

Dividing

$(1)$ by

$(2)\Rightarrow$

$\dfrac{K_A}{K_B}=\dfrac{m_A}{m_B}\dfrac{v_A^2}{v^2_B}$

$\Rightarrow \dfrac{v_A}{v_B}=\sqrt{\dfrac{K_Am_B}{K_Bm_A}}=\sqrt{\dfrac{125}{9}\times \dfrac{1}{5}}=\sqrt{\dfrac{25}{9}}$

$\Rightarrow \dfrac{v_A}{v_B}=\dfrac{5}{3}$

A car weighing $1200$ kg is uniformly accelerated from rest and covers a distance of $40$ m in 5 s. Calculate the work done by the engine of the car during this time. What is the final kinetic energy of the car?

Hint: Work done is the product of force and displacement.

Kinetic energy is given by = $\frac{1}{2}m\times v^2$

$\textbf{Step 1 : Applying equations of motion to find Acceleration and final velocity.}$

$x=ut+\dfrac{1}{2}at^{2}$

$\Rightarrow 40=0\times t+\dfrac{1}{2}\times a\times(5)^{2}$

$\Rightarrow \dfrac{80}{25}m/sec^{2}=a$

$\text{Now,} Force= Mass \times Acceleration$

$=1200\times \dfrac{80}{25}=3840N$

$\text{Also, } v=u+at$

$\Rightarrow v=0+\dfrac{80}{25}\times 5$

$\Rightarrow v= 16m/sec$

$\textbf{Step 2: Calculate Workdone and Kinetic Energy}$

$Workdone=Force\times Displacement$

$=Mass\times Acceleration \times Displacement$

$=1200\times \dfrac{80}{25}\times 40=153600J$

$Kinetic \ Energy=\dfrac{1}{2}MV^{2}$

$=\dfrac{1}{2}\times 1200 \times (16)^{2}=153600J$

$\textbf{Hence, Work done and Kinetic Energy are 153600J and 153600J Respectively.}$

A torch bulb is rated $2.5\ V$ and $750\ mA$ . Calculate (i) its power, (ii) its resistance, and (iii) the energy consumed, if this bulb is lighted for four hours.

Given :

$V = 2.5\ V$ ,

$I = 750\ mA$ ,

$\Rightarrow I= 750 \times 10^{-3}\ A$

(i) Power

$P = VI$

$\Rightarrow P = 2.5 \times 750 \times 10^{-3}$

$\Rightarrow P = 1.875\ W$

(ii) Resistance

$V = IR$

$\Rightarrow R = \dfrac{V}{I}$

$\Rightarrow R = \dfrac{2.5}{750 \times 10^{-3}}$

$\Rightarrow R = 3.33\ \Omega$

(iii) Energy consumed in four hour by the bulb

$E = P \times t$

$\Rightarrow E = 1.875 \times 4$

$\Rightarrow E = 7.5\ Wh$

$\Rightarrow E = 7.5 \times 10^{-3}\ kWh$

Determine $\tan { \theta }$ , and find $v_{ 2 }$ in terms of $v_{ 0 }$

For

$X$ components,

The momentum of system before collision is

$P_{m_1}=mv_0$

$P_{m_2}=m(0)=0$

The momentum of system after collision is

$P_{m_1}=0$

$P_{m_2}=(2m)(v_2cos\theta)$

Thus from conservation of momentum,

$mv_0=2mv_2cos\theta$

For

$Y$ components,

The momentum of system before collision is

$P_{m_1}=0$

$P_{m_2}=0$

The momentum of system after collision is

$P_{m_1}=-m\dfrac{v_0}{2}$

$P_{m_2}=(2m)(v_2sin\theta)$

Thus from conservation of momentum,

$0=2mv_2sin\theta-\dfrac{mv_0}{2}$ .

Thus from both the equations,

$v_2=\dfrac{mv_0}{2mcos\theta}=\dfrac{v_0}{2(\dfrac{2}{\sqrt{5}})}$

$=\dfrac{\sqrt{5}}{4}v_0$

A system consists of two springs connected in series and having the stiffness coefficients $k_1$ and $k_2$ . Find the minimum work to be performed in order to stretch this system by $\Delta l$ .

Two springs are connected in series has a effective spring constant K given by,

$K= \frac {1}{k_1}+\frac {1}{k_2}$ ..............(1)

So the minimum work to be performed in order to stretch this system by an amount

$\Delta l$ is =

$1/2\times K \times \Delta l^2$ , where K is given by equation (1).

Write down the $x$ and $y$ - components of the equation of conservation of momentum for the collision.

For

$X$ components,

The momentum of system before collision is

$P_{m_1}=mv_0$

$P_{m_2}=m(0)=0$

The momentum of system after collision is

$P_{m_1}=0$

$P_{m_2}=(2m)(v_2cos\theta)$

Thus from conservation of momentum,

$mv_0=2mv_2cos\theta$

For

$Y$ components,

The momentum of system before collision is

$P_{m_1}=0$

$P_{m_2}=0$

The momentum of system after collision is

$P_{m_1}=-m\dfrac{v_0}{2}$

$P_{m_2}=(2m)(v_2sin\theta)$

Thus from conservation of momentum,

$0=2mv_2sin\theta-\dfrac{mv_0}{2}$ .

The simple pendulum $A$ of mass ${ m }_{ A }$ and length $l$ is suspended from the trolley $B$ of mass ${ m }_{ B }$ . If the system is released from rest at $\theta=0$ , determine the velocity ${ v }_{ B }$ of the trolley and tension in the string when $\theta={ 90 }^{ \circ }$ . Friction is negligible.

The frictional forces are zero, therefore the energy of system will be conserved.
Assuming zero potential energy at

$\theta = 0$ , total energy at

$\theta = 0$ is

$0$
When the pendulum is at

$\theta = 90$
Total energy

$=\frac{1}{2} m_A v_A^{2}+\frac{1}{2} m_B v_B^{2}-m_A gl$ ...(i)
where

$v_A,v_B$ are velocities of pendulum and trolley.

As there is no external force acting in horizontal direction on the system, the momentum of system will be conserved.

$m_Av_A+m_Bv_B=0$ ...(ii)

Kinetic energy can be written in form of momentum

$K.E= \cfrac{1}{2} m v^{2}\times \cfrac{m}{m}= \cfrac{P^2}{2m}$ ...(iii)

From equation (iii)

$\dfrac{P_{A}^{2}}{2m_A}+ \dfrac{P_{B}^{2}}{2m_B}=m_Agl$

From equation (ii)

$P_A=P_B$ ...(iv) substituting in equation (iii)

$P_{B}^{2}{\dfrac{1}{m_A}+\dfrac{1}{m_B}}=2m_Agl$

$\Rightarrow { v }_{ B }=\cfrac { { m }_{ A } }{ { m }_{ B } } \sqrt { \cfrac { 2gl }{ 1+ { m }_{ A }/{ m }_{ B }} }$

A small steel ball A is suspended by an inextensible thread of length $l = 15 m$ from O. Another identical ball is thrown vertically downwards such that its surface remains just in contact with thread during downward motion and collides elastically with the suspended ball. If the suspended ball just completes vertical circle after collision, calculate the velocity (in cm/s) of the falling ball just before collision : $(g = 10 m s^{-2})$

Since the ball just completes a vertical circle, it must have a velocity of

$\sqrt{5gl}$

From geometry, the angle between vertical and line of collision between the balls is

$30^{\circ}$

Thus, from conservation of momentum of both the balls and the condition of elasticity,

velocity of ball A along the horizontal direction after collision

$=0.8v_0\cos30^{\circ}=\sqrt{5gl}$

$\implies v_{0}=12.5m/s$

A heavy particle hanging from a fixed point by a light inextensible string of length l is projected horizontally with speed $\displaystyle \sqrt{gl}.$ Find the speed of the particle and the inclination of the string to the vertical at the instant of the motion when the tension in the string is equal to the weight of the particle.

Let T=mg at angle

$\displaystyle \theta$ as shown in figure.

$\displaystyle h= l\left ( 1-\cos \theta \right )$ ...(i)
Applying conservation of mechanical energy between points A and B, we get

$\displaystyle \frac{1}{2}m\left ( u^{2}-v^{2} \right )= mgh$
Here,

$\displaystyle u^{2}= gl$ ...(ii)
and v=speed of particle in position B

$\displaystyle \therefore v^{2}= u^{2}-2gh$ ...(iii)
Further,

$\displaystyle T-mg\cos \theta = \frac{mv^{2}}{l}$
or

$\displaystyle mg-mg\cos \theta = \frac{mv^{2}}{l}$ (T=mg)
or

$\displaystyle v^{2}= gl\left ( 1-\cos \theta \right )$ ...(iv)
Substituting values of

$\displaystyle v^{2},u^{2}$ and h from Eqs. (iv), (ii) and (i) in Eq. (iii), we get

$\displaystyle gl\left ( 1-\cos \theta \right )= gl-2gl\left ( 1-\cos \theta \right )$
or

$\displaystyle \cos \theta = \frac{2}{3}$
or

$\displaystyle \theta = \cos ^{-1}\left ( \frac{2}{3} \right )$
Substituting

$\displaystyle \cos \theta = \frac{2}{3}$ in Eq. (iv), we get

$\displaystyle v= \frac{\sqrt{gl}}{3}$

A simple pendulum is constructed by attaching a bob of mass m to a string of length L fixed at its upper end. The bob oscillates in a vertical circle. It is found that the speed of the bob is v when the string makes an angle $\displaystyle \alpha$ with the vertical. Find the tension in the string and the magnitude of net force on the bob at that instant.

(i)The forces acting on the bob are
(a)the tension T
(b)the weight mg
AS the bob moves in a circle of radius L with centre at O. A centripetal force of magnitude

$\displaystyle \frac{mv^{2}}{L}$ is required towards O. This force will be provided by the resultant of T and mg

$\displaystyle \cos \alpha .$ Thus,
or

$\displaystyle T-mg\cos \alpha = \frac{mv^{2}}{L}$

$\displaystyle T= m\left ( g\cos \alpha +\frac{v^{2}}{L} \right )$
(ii)

$\displaystyle \left | \vec{F}_{net} \right |= \sqrt{\left ( mg\sin \alpha \right )^{2}+\left ( \frac{mv^{2}}{L} \right )}= m\sqrt{g^{2}\sin ^{2}\alpha +\frac{v^{4}}{L^{2}}}$

A ring of radius R is placed such that it lies in a vertical plane. The ring is fixed, A bead of mass m is constrained to move along the ring without any friction.One end of the spring is connected with the mass m and other end is rigidly fixed with the topmost point of the ring. Initially the spring is in un-extended position and the bead is at a vertical distance R from the lowermost point of the ring.The bead is now released from rest.
a) What should be the value of spring constant K such that the bead is just able to reach bottom of the ring.
b) The tangential and centripetal accelerations of the bead at initial and bottommost position for the same value of spring constant k.

Natural length of the spring is

$\sqrt{2}R$ . When the bead just reaches the bottom of the ring, it has zero velocity and:

$(a)$

$kx=mg$

$\implies k(2R-\sqrt{2}R)=mg$

$\implies k=\cfrac{mg}{\sqrt{2}R(\sqrt{2}-1)}$

$\implies k=\cfrac{mg(\sqrt{2}+1)}{3\sqrt{2}R}$

$(b)$ Initially, when the bead is at

$A$ , there is no spring force acting on it. Therefore, centripetal acceleration initially

$=$ zero.

Tangentially acceleration:

$\alpha=\cfrac{\tau}{I}$

$A$ , the torque is due to gravity:

$\therefore \alpha=\cfrac{mgR}{mR^2}$

$\therefore \alpha=\cfrac{g}{R}$

$B$ , the bead just reaches. Therefore, its centripetal acceleration is zero, and since at this point no tangential forces are acting on it, the tangential acceleration is also zero.

A ball of mass m is attached with a light string of length $\ell$ and released from positionString becomes taut at position 2 and reaches the lowest position 3, then find the velocity of the ball after the strings become taut.

The mass

$=m$

Length

$=l$

The acceleration of the string is

$\cfrac{v^2}{r}$ and the force due to tension

$T$ is

$T\cfrac{dl}{r}$ where

$dl$ is the length of an element of string. Both directed towards the centre of curvature

$T\cfrac{dl}{r}=m\cfrac{dl\times v^2}{r}$

(m being mass/unit length)

then there is no need for the tube.

$v^2=\cfrac{T}{m}$

1218667_769107_ans_fca61acc587f4df3abaceae2a9f1ab81.png

A particle of mass m is tied to one end of a string of length l. The particle is held horizontal with the string taut. It is then projected upward with a velocity u. The tension in the string is $\frac{mg}{2}$ when it is inclined at an angle $30^0$ to the horizontal. Find the value of $u$ .

$T=\frac { mg }{ 2 } \\ but\quad T=\frac { m{ u }^{ 2 } }{ r } \\ \therefore \quad \frac { mg }{ 2 } =\frac { m{ u }^{ 2 } }{ r } \\ ={ u }^{ 2 }=\frac { rg }{ 2 } \\ using\quad energy\quad balance,\\ \frac { 1 }{ 2 } m{ v }^{ 2 }=\frac { 1 }{ 2 } m{ u }^{ 2 }+mg(rsin{ 30 }^{ \circ })\\ \therefore \quad { v }^{ 2 }={ u }^{ 2 }+2gr\times \frac { 1 }{ 2 } \\ =\frac { rg }{ 2 } +rg=\frac { 3rg }{ 2 } \\ \therefore \quad v=\sqrt { \frac { 3rg }{ 2 } }$

A particle of mass $m_{1}$ moving at a speed $v_{1}$ allong the positive $x\ direction$ , collides with another particle of mass $m_{2}$ at rest. After the collision, the particles have velocities $v_{1}'$ and $v_{2}'$ in the $xy$ -plane in the directions of $\theta_{1}$ and $\theta_{2}$ with the x-axis as shown in the figure. There are no external forces acting on the particles. Express your answer in terms of $m_{1}, m_{2}, v_{1}, \theta_{1}$ and $\theta_{2}$ . What is the total momentum of the system before the collision (direction and magnitude)?

Total momentum of system before collision

$=m_1v_1\uparrow +0$

$\Rightarrow \vec { p } =m_1v_1\uparrow$

$\Rightarrow \left( \vec { p } \right) =m_1v_1$

Hence, the answer is

$m_1v_1.$

A block of mass $m$ initially has a velocity ${v}_{0}$ when it just touching a spring. The block moves through a distance $I$ before it stops after compressing the spring. The spring constant is $k$ and the coefficient of kinetic friction between block and table is $\mu$ . As the block moves the distance $l$
(a) what is the work done on it by the spring force? Are there other forces acting on the block, and if so, what work do they do?
(b) what is the total work done on the block?
(c) use the work energy theorem to find the value of $l$ in terms of ${v}_{0}, \mu, g$ and $k$

(a) The net force acting on the block by the spring is equal to

${ F }_{ sprinf }=kx$
Work done by the spring

$\int { { \overrightarrow { F } }_{ spring } } .d\overrightarrow { s }$

$=\int { { F }_{ sprinf } } ds\cos { { 180 }^{ o } } =-\int { { F }_{ sprinf } } .ds=-\int { kx } dx=\cfrac { -k{ l }^{ 2 } }{ 2 }$
(b) The total work done

$W=\Delta KE=0-\left( \cfrac { 1 }{ 2 } \right) m{ v }_{ 0 }^{ 2 }=-\left( \cfrac { 1 }{ 2 } \right) m{ v }_{ 0 }^{ 2 }\quad$
(c) The work done by friction

$=-\mu mgl\quad$
The total work done

${ W }_{ s }+{ W }_{ f }=\Delta KE\Rightarrow -\mu mgl-\left( 1/2 \right) k{ l }^{ 2 }\Rightarrow \cfrac { 1 }{ 2 } k{ l }^{ 2 }+\mu mgl=\left( \cfrac { 1 }{ 2 } \right) m{ v }_{ 0 }^{ 2 }$

$\Rightarrow l=\cfrac { \mu mg }{ k } \left[ \sqrt { 1+\cfrac { k }{ m } { \left( \cfrac { { v }_{ 0 } }{ \mu g } \right) }^{ 2 } } -1 \right]$
1027956_1018333_ans_4dca0ecc74f94da19c786f9bb57c6f54.png

Figure shows the force-displacement graph for a moving body. What is the work done by the force in displacing body from $x=0$ to $x=35m$ ?

The area under the force-displacement graph gives the work done.

Let the area of the trapezium is ${A_1}$ and the area of the triangle from $35\;{\rm{m}}$ to $40\;{\rm{m}}$ is ${A_2}$ .

The area from $0\;{\rm{m}}$ to $35\;{\rm{m}}$ is given as,

$A = {A_1} - {A_2}$

$= \frac{1}{2} \times 10 \times \left( {20 + 40} \right) - \frac{1}{2} \times 5 \times 5$

$= 287.5$

Thus, the work done by the force is $287.5\;{\rm{J}}$ .

A spring of force constant k is cut in two parts at its one-third length. When both the parts are stretched by same amount. The work done in the two parts will be:
(Note - Spring constant of a spring is inversely proportional to length of spring.)

Spring constant is inversely proportional to length

Lengths are in ratio 1:3

Spring constant in first spring

$k_1=3k$

Spring constant in second spring

$\dfrac{3k}{2}$

Work done in first spring=

$3kx$

Work done in second spring=

$\dfrac{3kx}{2}$

Two springs $A$ and $B$ are identical but $A$ is harder than $B (k_A > k_B )$ . Let $W_A$ and $W_B$ represent the work done when the springs are stretched through the same of the distance and $W'_A$ and $W'_B$ are the work done when these are stretched by equal forces then which of the following is true:

$k_A>k_B$

When stretched by same distance-

$U=\dfrac{1}{2}kx^2=W$

$\implies W_A>W_B$ since

$k_A>k_B$

When stretched by same force-

$F=kx\implies x=\dfrac{F}{k}$

$W=\dfrac{1}{2}kx^2=\dfrac{F^2}{2k}$

$\implies W\propto \dfrac{1}{k}$

$\implies W'_A<W'_B$ since

$k_A>k_B$

Fig. shows a bead of mass $m$ moving with uniform speed $v$ through a $U-$ shaped smooth wire the wire has a semicircular bending between $A$ and $B$ . Calculate the average force exerted by the bead on the part $AB$ of the wire.

Change in linear momentum of the bead

$\Delta P = mv-(-mv) =2mv$

Time taken

$\Delta t = \dfrac{\pi d/2}{v} = \dfrac{\pi d}{2v}$

Average force

$F = \dfrac{\Delta P}{\Delta t} = \dfrac{2mv}{\pi d/2v} = \dfrac{4mv^2}{\pi d}$

Discuss the motion in vertical circle. Find an expression for the minimum velocities at the lowest point and top point. Also find tension at these points?

The motion of a Particle in Vertical Circle Whenever a body is released from a height, it travels vertically downward towards the surface of the earth.

This is due to the force of gravitational attraction exerted on the body by the earth.

The acceleration produced by this force is called acceleration due to gravity and is denoted by

$‘g’$ . Value of

$‘g’$ on the surface of the earth is taken to the 9.8

$m/s^2$ and it is same for all the bodies.

It means all bodies (whether an iron ball or a piece of paper), when dropped (u = 0) from same height should fall with the same rapidity and should take the same time to reach the earth, it is the minimum velocity given to the particle at the lowest point to complete the circle.

The tendency of the string to become slack is maximum when the particle is at the topmost point of the circle.

At the top, tension is given by T =

$\dfrac{mv_T^2}{R}$ - mg,

where

$v_T$ = speed of the particle at the top.

$\dfrac{ mv_T^2}{R}$ =

$T + mg$

For

$v_T$ to be minimum,

$T = 0$

$v_T = \sqrt{gR}$

$V_B$ be the critical velocity of the particle at the bottom,

then from conservation of energy:

$Mg(2R)$ +

$\dfrac{1}{2} mv_T^2$ = 0 +

$\dfrac{1}{2} mv_B^2$

$v_T = \sqrt{gR}$ >

$2 mgR$ +

$\dfrac{1}{2} mgR$

$\dfrac{1}{2 }mv_B^2$

$v_B = \sqrt{5 g R}$

Highest point

$H ( h = 2r )$

We have,

$v$ =

$\sqrt{u^2 - 2gh}$

$v$ =

$\sqrt{(\sqrt{5gr})^2 - 2g ( 2r )}$

$v$ =

$\sqrt{5gr - 4gr}$

$\sqrt{gr}$

Tension at the lowest point:

$T - mg$ =

$\dfrac{mv^2}{R}$ .........(since,v =

$\sqrt{5gr}$ )

$T = 6mg$

Tension at highest point:

$T + mg$ =

$\dfrac{mv^2}{R}$ .......(since, v =

$\sqrt{gr}$ )

$T = 0$

1131611_1033276_ans_211f73748be642b1afc870fde9658586.png

A locomotive of mass $m$ starts moving so that its velocity varies according to the law $v=\alpha \sqrt { s }$ , where $\alpha$ is constant and $s$ is the distance covered. Find the total work done by all the forces acting on the locomotive during the first $t$ seconds after the beginning of motion.

The velocity is given as,

$v = \alpha \sqrt s$

$\dfrac{{ds}}{{dt}} = \alpha \sqrt s$

$\int_0^x {\dfrac{{ds}}{{\alpha \sqrt s }}} = \int_0^t {dt}$

$\dfrac{{2\sqrt s }}{\alpha } = t$

$s = \dfrac{{{\alpha ^2}{t^2}}}{4}$

The acceleration is given as,

$a = v\dfrac{{dv}}{{ds}}$

$= \alpha \sqrt s \times \dfrac{\alpha }{{2\sqrt s }}$

$= \dfrac{{{\alpha ^2}}}{2}$

The total work done is given as,

$W = m \times a \times s$

$= m \times \dfrac{{{\alpha ^2}}}{2} \times \dfrac{{{\alpha ^2}{t^2}}}{4}$

$= \dfrac{{m{\alpha ^4}{t^2}}}{8}$

Thus, the total work done by all forces acting on the locomotive is $\dfrac{{m{\alpha ^4}{t^2}}}{8}$ .

The mass of the cyclist together with the bike is $90 \ kg$ . Calculate the increase in kinetic energy if the speed increases from $6 \ km/h$ to $12 \ km/h$ .

We know that,

$1km/h=\dfrac{5}{18}m/s$

Here, initial velocity is

$u=6km/h= 6 \times \dfrac{5}{18} = \dfrac{5}{3}m/s$

Final velocity is

$v=12km/h= 12 \times \dfrac{5}{18} = \dfrac{10}{3}m/s$

Change in Kinetic energy is given by

$\Delta K=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2=\dfrac{1}{2}m(v^2-u^2)$

$\Delta K=\dfrac{90}{2}\left(\dfrac{100}{9}-\dfrac{25}{9}\right)$

$\implies \Delta K=\dfrac{90}{2}\times \dfrac{75}{9}$

$\implies \Delta K=375J$

Find the work done by $F$ as it moves the particle from $A$ to $C$ (fig.) along each of the paths $ABC, ADC$ and $AC$ .

Work done along ABC

$= \int\limits_{0,0}^{1,0} {\left( {{x^2}{y^2}\hat i - {x^2}{y^2}\hat j} \right)} dx + \int\limits_{1,0}^{1,1} {\left( {{x^2}{y^2}\hat i - {x^2}{y^2}\hat j} \right)} dy$

$= 0 + 0$

Work done along ADC

$= \int\limits_{0,0}^{0,1} {\left( {{x^2}{y^2}\hat i - {x^2}{y^2}\hat j} \right)} dy + \int\limits_{0,1}^{1,1} {\left( {{x^2}{y^2}\hat i - {x^2}{y^2}\hat j} \right)} dx$

$= 0 + 0$

Work done along AC

$= \int\limits_{0,0}^{1,1} {\left( {{x^2}{y^2}\hat i - {x^2}{y^2}\hat j} \right)} dxdy$

$= 0$

A particle of mass $m$ moves on a straight line with its velocity varying with the distance travelled according to the equation $v = a \sqrt{x}$ , where $a$ is a constant. Find the total work done by all the forces during the displacement from $x = 0$ to $x = d$ .

The work done by all the forces is given as:

$W=\int F\cdot dx$

$W=\displaystyle\int_{0}^{d}{m\cdot \cfrac{dv}{dt}\cdot dx}$

$\cfrac{dv}{dt}=a\cfrac{1}{2\sqrt{x}}\cdot \cfrac{dx}{dt}$

$\Rightarrow \cfrac{dv}{dt}=\cfrac{a}{2\sqrt{x}}\cdot a\sqrt{x}=\cfrac{a^2}{2}$

$W=\displaystyle\int_{0}^{d}{m\cdot \cfrac{a^2}{2}\cdot dx}$

$\Rightarrow W=\cfrac{ma^2d}{2}$

$J$

Two point mass m are connected the light rod length i and is free to roatate in vertical plane as shown. Calculate the minimum horizontal velocity is given to mass so that it completes the circular motion in

when an object of mass m which is fixed to an end of an inextensible and light rod

and revolving around a circular path about the other end,

weight of the object and tension in the rod are getting adjusted to provide centripetal force as shown in figure,

Let us consider when the object is at B, as shown in figure.

Centripetal force $\frac { m{ v }^{ 2 } }{ l } =T-mgcosθ\Longrightarrow (1)$

where T is tension in the rod, m is mass of object and θ is the angular dispalcement when the vertical position is taken as reference.

When the object is at A, θ = 0. hence centripetal force $\frac { m{ v }^{ 2 } }{ l } =T-mg$

When the object is at the top most position C, θ = π, hence centripetal force $\frac { m{ v }^{ 2 } }{ l } =T+mg$

When the object is at top most position C, minimum required velocity ${ V }_{ min }$ is obtained,

by considering tension in the rod T is zero.

when we apply this condition in eqn.(3), we get ${ V }_{ min }=\sqrt { gl }$

Two point mass m are connected the light rod of length t and it is free to rotate in vertical plane as shown.Calculate the minimum horizontal velocity is given to mass so that it completes the circular motion in verticalfane

$v=0$

${ v }_{ 2 }-{ u }_{ 2 }=−2gh$

$0={ u }_{ 2 }-2gh\quad (h=l)$

$u=\sqrt { 2gl }$

A stone weighing $0.5\ kg$ tied to a rope of length $0.5\ m$ revolves along a circular path in a vertical. The tension of the rope at the bottom point of the circle is $415$ newton. To what height will the stone if the rope breaks at the moment when the velocity is directed upwards? $(g = 10 m/s^2)$

$m=0.5kg$

$v_1=$ speed at the bottom

$T=415N$

$r=$ length of the rope

$=0.5m$

$mg=0.5 \times 9.8=4.9N=5N$

$T-mag =mv^v_1/r$

$415-5=(0.5)V^v_1/0.5$

$\boxed{v_1=6.4m/s}$

$K.E$ at

$A = K.E$ at

$B + P.E$ at

$B$

$(0.5)^1mV^v_1=(0.5)mV^v_2+ mgr$

$\boxed{v_2=5.5m/s}$

$h$ = height risen by the stone above the point where velocity is upward

$h=\dfrac{V^v_2}{29}$

$h=(5.5)^v/(2 \times 9.8)$

$\boxed {h=1.54m}$

A running man has half the kinetic energy as that of a boy half his mass. The man speeds up by $1\ ms^{-1}$ . So that the kinetic energies of the boy and man are equal. Calculate the initial speeds of boy and man.

1752871_1081352_ans_2602ab7e14b64087a9ce860c45004254.jpg

A particle of mass $12$ kg is acted upon by a force $f$ $= \left( {100 - 2{x^2}} \right)$ when $f$ is in newton and $x$ is in metre. Calculate the wrok done by this force in moving the particle $x = 0$ to $x = -10m$ . What will be the speed at $x = 10m$ if it starts from rest.

Given Force

$F=100-2{x}^{2}$

And we know work done

$\Delta{W}=Fdx$ where dx=displacement.

Now integrating the above equation from limits

$x=0$ to

$x=-10$ ,we get,

$W=\int _{ 0 }^{ -10 }{ (100-2{ x }^{ 2 })dx } \\ =|{ 100x| }_{ 0 }^{ -10 }-{ |\dfrac { 2 }{ 3 } ({ x }^{ 3 }| }_{ 0 }^{ -10 }\\ =\dfrac { -3000+2000 }{ 3 } =-333.33J$ negative sign implies work is done by the system.

Also we know

$W=Fs$ where s=displacement=

$vt$ where v=velocity and t= time.

And Force

$F=ma=m\dfrac{v}{t}$ given initial velocity u=0 m/s.

So work done

$W=m{v}^{2}$

$\sqrt{\dfrac{333.33}{12}}=v$

$v=5.27 m/s$

The work done by a force $\bar{F} = (-6x^3 i) N$ in displacing particle from $x = a m to x = -2 m$ is

$W=\displaystyle\int_{\vec{r_1}}^{\vec{r_2}}\vec{F}.\vec{dr}$

here,

$W=\displaystyle\int_{a}^{-2}F_x dx=-6\int_{a}^{-2}x^3dx=\dfrac{-6}{4}\left[x^4\right]_{a}^{-2}$

$\implies W=\dfrac{3}{2}(a^4-16 ) J$

A car and a truck are moving with the same velocity of $60 \ km/h$ . Which one has more kinetic energy ? $(M_T > M_C)$ .

We know,

$K.E = \dfrac{1}{2}MV^2$

$V_T = V_C = 60\ km/hr$

$\because M_T > M_C \implies K.E_T > K.E_C$

Hence, truck has more kinetic energy.

Suppose the rod with the balls $A$ and $B$ of the previous problem is clamped at the centre in such a way that it can rotate freely about a horizontal axis through the clapm. The system is kept at rest in the horizontal position. A particle $P$ of the same mass $m$ is dropped from a height $h$ on the ball $B$ . The particle collides with $B$ and sticks to it. (a) Find the angular momentum and and the angular speed of the system just after the collision. (b) What should be the minimum value of $h$ so that the system makes a full rotation after the collision.

$(a)$ Inertia of masses about centre

$O$ initially

$=m{(\cfrac{L}{2})}^{2}+m{(\cfrac{L}{2})}^{2}$

$\cfrac{m{L}^{2}}{2}$

So, angular momentum

$(Li)=\cfrac{m{L}^{2}}{2}\omega$

Now

Inertia of masses about

$O$ is finally

$=m{(\cfrac{L}{2})}^{2}+2m{(\cfrac{L}{2})}^{2}$

$=\cfrac{3}{4}m{L}^{2}$

So angular momentun

$({L}_{f)})=\cfrac{3}{4}m{L}^{2}\omega$

where,

${\omega}_{0}\rightarrow$ final angular momentum

Applying conservation of angular momentum

i.e.,

${L}_{i}={L}_{f}$

$\Rightarrow$

$\cfrac{m{L}^{2}}{2}\omega=\cfrac{3}{4}m{L}^{2}\omega$

$\Rightarrow$

${\omega}_{0}=\cfrac{2}{3}\omega$

$(b)$ When the mass

$2m$ will at the top most position and the mass

$m$ at the at the lowest point they will automatically rotate. In this position he total gain in potential energy

$=2mg\times (\frac{l}{2})-mg(\frac{l}{2})=mg(\frac{l}{2})$

$\Rightarrow mg\ \frac{l}{2}=\frac{l}{2}\ l\omega^{2}$

$\Rightarrow mg\ \frac{l}{2}=(\frac{1}{2}\times \frac{3ml^{2}}{4}\times (\frac{8gh}{9gl^{2}})$

$\Rightarrow h=\frac{3l}{2}$

1352188_1206910_ans_d28203f8116f41f0a49ff7ecf0600287.PNG

A copper wire of length $2.2\ m$ and a steel wire of length $1.6\ m$ , both of diameter $3.0\ mm$ , are connected end to end. When stretched by a load, the net elongation is found to be $0.70\ mm$ . Obtain the load applied.

1124790_1228940_ans_2e7dd2785df240f0bc6eb0e81eff2ccf.jpg

In the figure shown all the surfaces are frictionless, and mass of the block, m= $1$ kg. The block and wedge are held initially at rest. Now wedge is given a horizontal acceleration of $10 m/s^2$ by applying a force on the wedge, so that the block does not slip on the wedge. Then work done by the normal force in ground frame on the block in $\sqrt3$ seconds is

1560647_1241437_ans_c597d11bed17409db48598ae948541f8.jpg

Calculate the change that should be affected in the velocity of a body to maintain the same K.E, if mass of the body is increased to 4 times.

K.E = $\dfrac{1}{2}$ $mv ^{2} \implies$ $v=\sqrt {\dfrac{2 \times K.E}{m}}$ .

K.E is same, mass increased 4 times,

$New\ v=\sqrt {\dfrac{2 \times K.E}{4 \times m}}$ = $\dfrac{1}{2} \sqrt{\dfrac{2 \times K.E}{m}}$ = $\dfrac{v}{2}$

Hence, velocity becomes halfed.

A body of mass $15$ kg is raised to height of $10$ m. Calculate the potential energy at that height. What will be it's potential energy, if it is taken to double height.

Given,

$m= 15\ kg$ ; mass of the body

$h= 10\ m$ ; raised height

$g=9.8\ ms^{-2}$ ; acceleration due to gravity

Potential energy at that point-

$E = mgh$

$\Rightarrow E = 15 \times 9.8 \times 10$

$\Rightarrow E = 1470\ J$

Taken to double the height -

$h = 20\ m$

New potential energy-

$E' = 2mgh$

$\Rightarrow E' = 2 \times 1470\ J$

$\Rightarrow E' = 2940 J$

A body possess potential energy of $460\ J$ whose mass is $20\ kg$ and is raised to a certain height. What is the height when $g=10m/s^ {2}$

Given:-

$P.E=460 J$

$m=20 kg$

$g=10 m/s^2$

$h=?$

Potential energy

$P.E.=mgh$

$460=20 \times 10 \times h$

$h=\dfrac{460}{20 \times 10}$

$h=2.3 m$

The height at which the object is raised from the ground is 2.3 m.

The ball of mass 0.4 kg is dropped from some certain height .As it falls through 2 m. The kinetic energy acquired by it is.

$m=0.4kg$

$g=10m/s^2$

Let ball be dropped from some height

$h_1$

And it falls

$2m$ and new height is

$h_2$ so it loses height hence it loses Potential energy.This loss in potential energy goes into kinetic energy.

$So$

$loss$

$in$

$PE=mg( h_1-h_2)$ =

$mg(2)=0.4\times 10 \times 2=8Joules$

$KE$

$gained$ =

$8Joules$

A boy of mass $50\ kg$ climbs up a vertical height of $100\ m$ . Calculate the amount of potential energy he gains.

Given,

Mass of the boy,

$m=50\ kg$

Height,

$h=100\ m$

Acceleration due to gravity,

$a=g=9.8\ m/s^2$

Force,

$F=m\times a$

$F=m\times g$

$F=50\times 9.8$

$F=490\ N$

Work done,

$W=Force\times displacement$

$W=490\times 100$

$W=49000\ J$

A body of mass $10 k g$ is moving with a velocity $20 m s ^ { - 1 }$ . If the mass of the body is doubled and its velocity is halved, find the ratio of the initial kinetic energy to the final kinetic energy.

Given , Mass of object,

$m= 10\ kg$

Velocity of object,

$v=20 ms^{-1}$

Initial Kinetic energy,

$K_1=\dfrac12 \times 10\times 20^2 = 2000\ J$

Now, Final mass,

$m_2= 20\ kg$

Final Velocity,

$v_2 =10m/s$

Final Kinetic energy,

$K_2=\dfrac12 \times 20 \times 10^2 = 1000\ J$

Required Ratio =

$\dfrac{K_1}{K_2}=\dfrac{2000}{1000}=2:1$

Determine the kinetic energy possesed by a battle tank belt whose mass is m if the tank is moving with constant velocity v. Assume there is no slipping.

2049389_1398994_ans_365ba39cc5cc48c1a2af95bac4cd8a6a.jpg

Show that change in potential energy is given by $\frac { \frac { mgh }{ 1+h } }{ Re }$

2046885_1392878_ans_6282b76d56da45c5876a6cc59d87429c.jpg

Calculate the kinetic energy of a body of mass $20\, kg$ moving with velocity of $0.1\, m/s$ . How the kinetic energy will change if the velocity of the body be doubled ?

Given:

Mass of the body:

$m =20\ kg$

Velocity of the body:

$v = 0.1\ m/s$

$K.E = \dfrac{1}{2}mv^2 = \dfrac{1}{2}\times 20 \times 0.1^2 =0.1\ J$

When the velocity of the body is doubled -

$new \ v = 2 \times0.1\ m/s = 0.2\ m/s$

$new\ K.E = \dfrac{1}{2}mv^2 = \dfrac{1}{2}\times 20 \times 0.2^2 =0.4\ J$

New K.E becomes

$4\ times$ the old K.E.

a) Define the term 'energy' of a body. What is the SI unit of energy.
b) What are the various forms of energy?
c) Two bodies having equal masses are moving with uniform speeds of $v$ and $2v$ respectively. Find the ratio their kinetic energies.

a) Energy is defined as the ability to do work and joule is the SI unit of energy.

b) Following are the various forms of energy:
Kinetic energy, potential energy, nuclear energy, electrical energy, chemical energy, sound energy, light energy, and heat energy.

c) Let the mass of both the bodies be

$m$
Velocity of body

$1 , v1 = v$
Velocity of body

$2 , v2 = 2v$

$KE_1 = 1/2 \,mv^2 = 1/2 m(v_1)^2$

$KE_2 = 1/2 \,mv^2 = 1/2 m(v_2)^2$

$Ratio = KE_1 : KE_2 = 1/4$

The arrangement shown in Fig is at rest. An ideal spring of natural length $l_{0}$ having spring constant $k=220\ Nm^{-1}$ , is connected to block $A$ . Blocks $A$ and $B$ are connected by an ideal string passing through a friction less pulley. The mass of each block $A$ and $B$ is equal to $m=2\ kg$ when the spring was in natural length, the whole system is given an acceleration $\vec{a}$ as shown. If coefficient of friction of both surfaces is $\mu=0.25$ then find the maximum extension of the spring $(g=10\ ms^{-2})$

Apply work-energy theorem from the accelerating system's frame of reference

also, the pseudo force will act on both the blocks in opposite direction to the acceleration

$[W_{spring} + W_{friction} + W_{pseudo} ]_A + [W_{gravity} + W_{friction} + W_{pseudo}]_B = \Delta K.E. \\ \dfrac{-1}{2}kx^2 - \mu (mg+ma_y)x - ma_xx + mgx - \mu (ma_x)x+ ma_yx = 0 \\ \dfrac{-1}{2} * 220 *x^2 - 0.25 * (2 * 10+2*4)x - 2*4x + 2 * 10 x -0.25 * 2* 4*x +2 * 4x = 0 \\ -110 x^2 -7x - 8x + 20 x -2x + 8 x = 0 \\ -110x^2 - 11 x = 0 \\ x = \dfrac{11}{110} \\ x = \dfrac{1}{10}m = 10 cm$

In Figure the pulley shown is smooth. The spring and the string are light. Block $B$ slides down from the top along the fixed rough wedge of inclination $\theta$ . Assuming that the block reaches the end of the wedge, find the speed of the block at the end. Take the coefficient of friction between the block and the wedge to be $\mu$ and that the spring was relaxed when the block was released from the top of the wedge.

1728778_1734404_ans_0965d7ad25a64865ae64e36e7bb177a7.jpg

In Fig. $8-46$ , a spring with $k = 170\,N/m$ is at the top of a frictionless incline of angle $\theta = 37.0^\circ.$ The lower end of the incline is distance $D =1.00\, m$ from the end of the spring, which is at its relaxed length. A $2.00 \,kg$ canister is pushed against the spring until the spring is compressed $0.200\, m$ and released from rest. (a) What is the speed of the canister at the instant the spring returns to its relaxed length (which is when the canister loses contact with the spring)? (b) What is the speed of the canister when it reaches the lower end of the incline?

All heights h are measured from the lower end of the incline (which is our reference position for computing gravitational potential energy

$mgh$ ). Our x axis is along the incline,with

$+x$ being uphill (so spring compression corresponds to

$x > 0$ ) and its origin being at the relaxed end of the spring. The height that corresponds to the canister's initial position(with spring compressed amount

$x = 0.200\, m$ ) is given by

$h_1 = (D + x) sin \,\theta,$ where

$\theta = 37^\circ.$

(a) Energy conservation leads to

$K_1 + U_1 = K_2 + U_2 \Rightarrow 0 + mg(D + x) sin \,\theta + \dfrac{1}{2} kx^2 = \dfrac{1}{2} mv_2^2 + mg \,D \,sin \,\theta$
which yields, using the data

$m = 2.00 \,kg$ and

$k = 170\,N/m,$

$v_2 =\sqrt{2gx \,sin\,\theta + kx^2/ m} = 2.40 \,m/s.$

(b) In this case, energy conservation leads to

$K_1 + U_1 = K_3 + U_3$

$0 + mg(D + x) sin\,\theta + \dfrac{1}{2} kx^2 = \dfrac{1}{2} mv_3^2 + 0$

which yields

$v_3 = \sqrt{2g(D + x) sin\,\theta + kx^2 / m} = 4.19 \,m/s.$

Figure 8-31 shows a ball with mass $m = 0.341\, kg$ attached to the end of a thin rod with length $L = 0.452\, m$ and negligible mass.The other end of the rod is pivoted so that the ball can move in a vertical circle. The rod is held horizontally as shown and then given enough of a downward push to cause the ball to swing down and around and just reach the vertically up position, with zero speed there. How much work is done on the ball by the gravitational force from the initial point to (a) the lowest point, (b) the highest point, and (c) the point on the right level with the initial point? If the gravitational potential energy of the ballEarth system is taken to be zero at the initial point, what is it when the ball reaches (d) the lowest point, (e) the highest point, and (f) the point on the right level with the initial point? (g) Suppose the rod were pushed harder so that the ball passed through the highest point with a nonzero speed.Would $\Delta \,U_g$ from the lowest point to the highest point then be greater than, less than, or the same as it was when the ball stopped at the highest point?

(a) The only force that does work on the ball is the force of gravity; the force of the rod is perpendicular to the path of the ball and so does no work. In going from its initial position to the lowest point on its path, the ball moves vertically through a distance equal to the length

$L$ of the rod, so the work done by the force of gravity is

$W = mgL = (0.341\,kg)(9.80\,m/s^2)(0.452\,m) = 1.51\,J.$

(b) In going from its initial position to the highest point on its path, the ball moves vertically through a distance equal to

$L,$ but this time the displacement is upward, opposite the direction of the force of gravity. The work done by the force of gravity is

$W = - mgL = - (0.341\,kg)(9.80\,m/s^2)(0.452\,m) = - 1.51\,J.$

(c) The final position of the ball is at the same height as its initial position. The displacement is horizontal, perpendicular to the force of gravity. The force of gravity does no work during this displacement.

(d) The force of gravity is conservative. The change in the gravitational potential energy of the ball-Earth system is the negative of the work done by gravity:

$\Delta U = -mgL = -(0.341\,kg)(9.80\,m/s^2)(0.452\,m) = -1.51\,J.$
as the ball goes to the lowest point.

(e) Continuing this line of reasoning, we find

$\Delta U = + mgL = (0.341\,kg)(9.80\,m/s^2)(0.452\,m) = 1.51\,J.$
as it goes to the highest point.

(f) Continuing this line of reasoning, we have

$\Delta U = 0$ as it goes to the point at the same height.
(g) The change in the gravitational potential energy depends only on the initial and final positions of the ball, not on its speed anywhere. The change in the potential energy is the same since the initial and final positions are the same.

We move a particle along an $x$ axis, first outward from $x = 1.0\, m$ to $x = 4.0\, m$ and then back to $x =1.0\, m,$ while an external force acts on it. That force is directed along the x axis, and its x component can have different values for the outward trip and for the return trip. Here are the values (in newtons) for four situations, where x is in meters:
Find the net work done on the particle by the external force for the round trip for each of the four situations. (e) For which, if any, is the external force conservative?

(a) The table shows that the force is

$+ (3.0\,N) \hat{i}$ while the displacement is in the

$+x$ direction

$( \vec{d} = + (3.0 \,m) \hat{i}),$ and it is

$-(3.0 \,N) \hat{i}$ while the displacement is in the

$-x$ direction. Using Eq.

$7-8$ for each part of the trip, and adding the results, we find the work done is

$18 \,J$ . This is not a conservative force field; if it had been, then the net work done would have been zero (since it returned to where it started).

(b) This, however, is a conservative force field, as can be easily verified by calculating that the net work done here is zero.

$\int \,2x \,dx$ so that we are adding two equivalent terms, where each equals

$x^2$ (evaluated at

$x = 4$ , minus its value at

$x = 1$ ). Thus, the work done is

$2 (4^2 - 1^2) = 30\,J.$

(d) This is another conservative force field, as can be easily verified by calculating that the net work done here is zero.

(e) The forces in (b) and (d) are conservative.

The cable of the $1800 \,kg$ elevator cab in Fig. $8-56$ snaps when the cab is at rest at the first floor, where the cab bottom is a distance $d = 3.7\, m$ above a spring of spring constant $k = 0.15 \,MN/m.$ A safety device clamps the cab against guide rails so that a constant frictional force of $4.4\, kN$ opposes the cabs motion. (a) Find the speed of the cab just before it hits the spring. (b) Find the maximum distance $x$ that the spring is compressed (the frictional force still acts during this compression). (c) Find the distance that the cab will bounce back up the shaft. (d) Using conservation of energy, find the approximate total distance that the cab will move before coming to rest. (Assume that the frictional force on the cab is negligible when the cab is stationary.)

1749417_1766653_ans_5564bede02b744a59c2e6f6207e40b49.jpg

In Fig. $8-53,$ a block of mass $m = 2.5\, kg$ slides head on into a spring of spring constant $k = 320\, N/m.$ When the block stops, it has compressed the spring by $7.5\, cm.$ The coefficient of kinetic friction between block and floor is $0.25.$ While the block is in contact with the spring and being brought to rest, what are (a)the work done by the spring force and (b) the increase in thermal energy of the blockfloor system? (c) What is the blocks speed just as it reaches the spring?

(a) With

$x = 0.075 \,m$ and

$k = 320 \,N/m,$ Eq.

$7-26$ yields

$W_s = -\dfrac{1}{2} kx^2 = -0.90 \,J.$ For later reference, this is equal to the negative of

$\Delta U.$

(b) Analyzing forces, we find

$F_{N} = mg,$ which means

$f_k = \mu_k F_{N} = \mu_k mg.$ With

$d = x,$ Eq.

$8-31$ yields

$\Delta E_{th} = f_kd = \mu_k mgx = (0.25)(2.5)(9.8) (0.075) = 0.46 \,J.$

$8-33$ ( with

$W = 0$ ) indicates that the initial kinetic energy is

$K_i = \Delta U + \Delta E_{th} = 0.90 + 0.46 = 1.36 \,J$

which yields to

$v_i = \sqrt{ 2K_i / m} = 1.0 \,m/s.$

The block lies on a horizontal frictionless surface, and the spring constant is $50 \ N/m$ . Initially, the spring is at its relaxed length and the block is stationary at position $x = 0$ . Then an applied force with a constant magnitude of $3.0 \ N$ pulls the block in the positive direction of the $x$ axis, stretching the spring until the block stops. When that stopping point is reached, what are (a) the position of the block, (b) the work that has been done on the block by the applied force, and (c) the work that has been done on the block by the spring force? During the block’s displacement, what are (d) the block’s position when its kinetic energy is maximum and (e) the value of that maximum kinetic energy?

(a) So we have

$Fx = \frac{1}{2} kx^2 \Rightarrow (3.0 N) x = \frac{1}{2}(50 N/m)x^2$

which (other than the trivial root) gives x = (3.0/25) m = 0.12 m.

(b) The work done by the applied force is

$W_a = Fx = (3.0 N) (0.12 m) = 0.36 J$ .

$W_s = -W_a = -0.36J$ .

(d) With

$k_f = K$ considered variable and

$K_i = 0$ , The equation gives

$K = Fx - \frac{1}{2}kx^2$ . We have the derivative of K with respect to x and set the resulting expression equal to zero, in order to find the position

$x_c$ that corresponds to a maximum value of K:

$x_c = \frac{F}{k} = (3.0 / 50) m = 0.060m$

We note that

$x_c$ is also the point where the applied and spring forces “balance.”

(e) At

$x_c$ we find

$K = K_{max} = 0.090 J$ .

A certain spring is found not to conform to Hookes law. The force (in newtons) it exerts when stretched a distance x (in meters) is found to have magnitude $52.8x + 38.4x^2$ in the direction opposing the stretch. (a) Compute the work required to stretch the spring from $x = 0.500\, m$ to $x = 1.00\, m.$ (b) With one end of the spring fixed, a particle of mass $2.17 \,kg$ is attached to the other end of the spring when it is stretched by an amount $x = 1.00\, m.$ If the particle is then released from rest, what is its speed at the instant the stretch in the spring is $x = 0.500\, m$ ? (c) Is the force exerted by the spring conservative or nonconservative? Explain.

(a) To stretch the spring an external force, equal in magnitude to the force of the spring but opposite to its direction, is applied. Since a spring stretched in the positive x direction exerts a force in the negative x direction, the applied force must be

$F = 52.8x + 38.4 x^2,$ in the

$+x$ direction. The work it does is

$\displaystyle W = \int_{0.50}^{1.00} (52.8 x + 38.4 x^2)dx =$

$\displaystyle \left ( \dfrac{52.8}{2}x^2 + \dfrac{38.4}{3}x^3 \right )\displaystyle \mid_{0.50}^{1.00} = 31.0 \,J.$

(b) The spring does

$31.0 \,J$ of work and this must be the increase in the kinetic energy of the particle. Its speed is then

$v = \sqrt{\dfrac{2K}{m}} = \sqrt{\dfrac{2(31.0 \,J)}{2.17 \,kg}} = 5.35 \,m/s.$

$x_1$ to any other point

$x_2$ depends only on

$x_1$ and

$x_2$ , not on details of the motion between

$x_1$ and

$x_2$ .

A force is applied of magnitude $80 \ N$ to hold the block stationary at $x = -2.0 \ cm$ . From that position, we then slowly move the block so that our force does $+4.0 \ J$ of work on the spring - block system; the block is then again stationary.What is the block’s position? (Hint: There are two answers.)

Since the block is stationary before and afterthe displacement. The work done by the applied force can be written as

$W_a = -W_s = \frac{1}{2}k(x_f^2 - x_i^2)$ .

The spring constant is k = (80 N) / (2.0 cm) =

$4.00 \times 10^3 N/m$ .

With

$W_a = 4.0 J$ , and

$x_i = -2.0 cm$ , we have

$x_f = \pm \sqrt{\frac{2W_a}{k} + x_f^2} = \pm \sqrt{\frac{2(4.0J)}{(4.0 \times 10^3 N/m)} + (-0.020 m)^2} = \pm 0.049m = \pm 4.0 cm$

A rope is used to pull a $3.57\, kg$ block at constant speed $4.06\, m$ along a horizontal floor. The force on the block from the rope is $7.68\, N$ and directed $15.0^\circ$ above the horizontal.What are (a) the work done by the ropes force, (b) the increase in thermal energy of the blockfloor system, and (c) the coefficient of kinetic friction between the block and floor?

(a) The work done on the block by the force in the rope is, using Eq.

$7-7,$

$W = Fd \,cos \,\theta = (7.68\,N)(4.06\,m) cos \,15.0^\circ = 30.1\,J.$

(b) Using f for the magnitude of the kinetic friction force, Eq.

$8-29$ reveals that the increase in thermal energy is

$\Delta E_{th} = fd = (7.42\,N)(4.06\,m) = 30.1\,J.$

$\mu_k = f/F_{N}$ to obtain the coefficient of friction. Place the x-axis along the path of the block and the y-axis normal to the floor. The free-body diagram is shown below. The x and the y component of Newton's second law are

$x: F \,cos\,\theta - f = 0$

$y: F_{N} + F \,sin \,\theta - mg = 0,$

where m is the mass of the block, F is the force exerted by the rope, and

$\theta$ is the angle between that force and the horizontal.
The first equation gives

$f = F \,cos\,\theta = (7.68\,N)(cos \,15.0^\circ = 7.42\,N$

and the second gives

$F_{N} = mg - F \,sin\,\theta = (3.57\,kg)(9.8\,m/s^2) - (7.68\,N)sin \,15.0^\circ = 33.0\,N.$

Thus, the coefficient of kinetic friction is

$\mu_k = \dfrac{f}{F_{N}} = \dfrac{7.42\,N}{33.0\,N} = 0.225.$

1673726_1766596_ans_e50748d01c36447184a1c4a434f3f8cd.png

A CD case slides along a floor in the positive direction of an x axis while an applied force $\vec{F_a}$ acts on the case. The force is directed along the x axis and has the x component $F_{ax} = 9x - 3x^2$ , with x in meters and $F_{ax}$ in newtons. The case starts at rest at the position x = 0, and it moves until it is again at rest. (a) Plot the work $\vec{F_a}$ does on the case as a function of x. (b) At what position is the work maximum, and (c) what is that maximum value? (d) At what position has the work decreased to zero? (e) At what position is the case again at rest?

(a) Since we know that, the work becomes

$W = \frac{9}{2} x^2 – x^3$ . The plot is shown in the figure

(b) We see from the graph that its peak value occurs at

$x = 3.00 \ m$ .

This can be verified by taking the derivative of W and

setting equal to zero, or simply by noting that this is

where the force vanishes.

$W = \dfrac{9}{2} (3.00)^2 - (3.00)^3 = 13.50 J$ .

(d) We see from the graph (or from our analytic expression) that W = 0 at x = 4.50 m.

(e) The case is at rest when

$v = 0$ .

Since

$W = ΔK = mv^2 / 2$ , the condition implies

$W = 0$ .

This happens at

$x = 4.50 \ m$ .

1678176_1766762_ans_12228eeac02946209641269c4b2f5772.png

A spring with a spring constant of $3200 \,N/m$ is initially stretched until the elastic potential energy of the spring is $1.44\, J.$ ( $U = 0$ for the relaxed spring.) What is $\Delta U$ if the initial stretch is changed to (a) a stretch of $2.0 \,cm,$ (b) a compression of $2.0 \,cm,$ and(c) a compression of $4.0 \,cm$ ?

(a) In the initial situation, the elongation was (using Eq.

$8-11$ )

$x_i = \sqrt{2 (1.44) / 3200} = 0.030 \,m$ (or

$3.0 \,cm$ ).

In the next situation, the elongation is only

$2.0 \,cm$ (or

$0.020 \,m$ ), so we now have less stored energy (relative to what we had initially). Specifically,

$\Delta U = \dfrac{1}{2} (3200 \,N/m)(0.020 \,m)^2 - 1.44 \,J = -0.80 \,J.$

(b) The elastic stored energy for

$|x| = 0.020 \,m$ does not depend on whether this represents a stretch or a compression. The answer is the same as in part (a),

$\Delta U = -0.80 \,J.$

$|x| = 0.040 \,m,$ which is greater than

$x_i,$ so this represents an increase in the potential energy (relative to what we had initially). Specifically,

$\Delta U = \dfrac{1}{2} (3200 \,N/m)(0.040 \,m)^2 - 1.44 \,J = + 1.12 \,J \approx 1.1 \,J.$

A rocket moving up with a velocity $v$ . If the velocity of this rocket is suddenly tripled, what will be the ratio of two kinetic energies?

We know,

Kinetic energy,

$KE=\dfrac 12mv^2$

where,

$m=$ mass and

$v=$ velocity

If the velocity of this rocket is suddenly tripled.

$v'=3v$

Then, kinetic energy,

$KE'=\dfrac 12 mv'^2$

$KE'=\dfrac 12 m(3v)^2$

$KE'=\dfrac 92 mv^2$

The ratio of two kinetic energies,

$\dfrac{KE}{KE'}=\dfrac {\dfrac 12mv^2}{\dfrac 92 mv^2}$

$\dfrac{KE}{KE'}=\dfrac 19$

The spring in the muzzle of a childs spring gun has a spring constant of $700 \,N/m.$ To shoot a ball from the gun, first the spring is compressed and then the ball is placed on it. The guns trigger then releases the spring, which pushes the ball through the muzzle. The ball leaves the spring just as it leaves the outer end of the muzzle.When the gun is inclined upward by $30^\circ$ to the horizontal, a $57\, g$ ball is shot to a maximum height of $1.83\, m$ above the guns muzzle.Assume air drag on the ball is negligible. (a) At what speed does the spring launch the ball? (b) Assuming that friction on the ball within the gun can be neglected, find the springs initial compression distance.

(a) At the highest point, the velocity

$v = v_{x}$ is purely horizontal and is equal to the horizontal component of the launch velocity (see section

$4-6$ ):

$v_{ox} = v_o \,cos \theta$ , where

$\theta = 30^\circ$ in this problem. Equation

$8-17$ relates the kinetic energy at the highest point to the launch kinetic energy:

$K_o = mgy + \dfrac{1}{2} mv^2 = \dfrac{1}{2} mv^2_{ox} + \dfrac{1}{2} mv^2_{oy},$

with

$y = 1.83 \,m.$ Since the

$mv^2_{ox} / 2$ term on the left-hand side cancels the

$mv^2 / 2$ term on the right-hand side, this yields

$v_{oy} = \sqrt{ 2gy} \approx 6 \,m/s.$ With

$v_{oy} = v_o \,sin \theta,$ we obtain

$v_o = 11.98 \,m/s \approx 12 \,m/s.$

(b) Energy conservation (including now the energy stored elastically in the spring, Eq.

$8-11$ ) also applies to the motion along the muzzle (through a distance d that corresponds to a vertical height increase of

$d \,sin \theta$ ):

$\dfrac{1}{2} kd^2 = K_{o} + mg \,d \,sin \theta \Rightarrow d = 0.11 \,m.$

An adult weighing $600\ N$ ralses the center of gravity of his body by $0.25\ m$ while taking each step of $1\ m$ length in jogging. If he jogs for $6\ km$ , calculate the energy utilized by him in jogging assuming that there is no energy loss due to friction of ground and air. Assuming that the body of the adult is capable of converting $10\%$ of energy intake in the from of food, calculate the energy equivalent of food that would be required to compensate energy utilized for jogging.

Energy used up by raising the centre of gravity by

$0.25\ m$ by Jogger in one step

$=mgh$

$mg=600\ N$

$h=0.25m$

$\therefore$ Number of steps in

$6\ km=\dfrac {6000m}{1\ m}=6000$ steps
Energy utilised in

$6000\times 600\times 0.25\ J$
Since

$10\%$ of energy utilised in jogging

$\therefore$ Energy utilised in jogging

$=\dfrac {10}{100}\times 6000\times 600\times 0.25$

$=360000\times 0.25$

$=90000\ J=9\times 10^4\ J$

A $20\, kg$ object is acted on by a conservative force given by $F = -3.0\,x \,- 5.0 \,x^2,$ with $F$ in newtons and $x$ in meters. Take the potential energy associated with the force to be zero when the object is at $x = 0.$ (a) What is the potential energy of the system associated with the force when the object is at $x = 2.0 \,m$ ? (b) If the object has a velocity of $4.0\, m/s$ in the negative direction of the x axis when it is at $x = 5.0\, m,$ what is its speed when it passes through the origin? (c) What are the answers to (a) and (b) if the potential energy of the system is taken to be $-8.0\, J$ when the object is at $x = 0$ ?

From Eq.

$8-6$ , we find (with SI units understood)

$\displaystyle U(\xi) = -\int_{0}^{\xi} (-3x - 5x^2) \,dx = \dfrac{3}{2} \xi^2 + \dfrac{5}{3} \xi^3.$

(a) Using the above formula, we obtain

$U(2) \approx 19 \,J.$

(b) When its speed is

$v = 4 \,m/s$ , its mechanical energy is

$\dfrac{1}{2} mv^2 + U(5)$ . This must equal the energy at the origin:

$\dfrac{1}{2} mv^2 + U (5) = \dfrac{1}{2} mv^2_0 + U (0)$

so that the speed at the origin is

$v_0 = \sqrt{ v^2 + \dfrac{2}{m} ( U (5) - U (0))}.$

Thus, with

$U (5) = 246 \,J , U(0) = 0$ and

$m = 20 \,kg.$ we obtain

$v_0 = 6.4 \,m/s.$

$U(x) = -8 + \dfrac{3}{2} x^2 + \dfrac{5}{3} x^3.$

in this case. Therefore,

$U (2) = 11 \,J$ . But we still have

$v_0 = 6.4 \,m/s$ since that calculation only depended on the difference of potential energy values (specifically,

$U (5) - U (0)$ ).

A spring with spring constant $k = 200 \,N/m$ is suspended vertically with its upper end fixed to the ceiling and its lower end at position $y = 0.$ A block of weight $20\, N$ is attached to the lower end, held still for a moment, and then released. What are (a) the kinetic energy $K,$ (b) the change (from the initial value)in the gravitational potential energy $\Delta U_g,$ and (c) the change in the elastic potential energy $\Delta U_e$ of the spring - block system when the block is at $y = -5.0\,cm$ ? What are (d) $K$ , (e) $\Delta U_g,$ and (f) $\Delta U_e$ when $y = -10\,cm$ , (g) $K$ , (h) $\Delta U_g,$ and (i) $\Delta U_e$ when $y = -15\,cm,$ and (j) $K$ , (k) $\Delta U_g,$ and (l) $\Delta U_e$ when $y = -20\,cm$ ?

1747162_1767188_ans_235112d7ceba41868262f88ff4d1287c.jpg

Figure 9-55 shows a two-ended rocket that is initially stationary on a frictionless floor, with its center at the origin of an x axis.The rocket consists of a central block C (of mass $M = 6.00 \,kg$ )and blocks L and R (each of mass $m = 2.00 \,kg$ ) on the left and right sides. Small explosions can shoot either of the side blocks away from block C and along the x axis. Here is the sequence: (1) At time $t = 0$ , block L is shot to the left with a speed of $3.00 \,m/s$ relative to the velocity that the explosion gives the rest of the rocket. (2) Next, at time $t = 0.80 \,s,$ block R is shot to the right with a speed of $3.00 \,m/s$ relative to the velocity that block C then has. At $t = 2.80 \, s$ , what are (a) the velocity of block C and (b) the position of its center?

(a) With SI units understood, the velocity of block

$L$ (in the frame of reference indicated in the figure that goes with the problem) is

$( \nu_i - 3) \hat{i}$ . Thus, momentum conservation (for the explosion at

$t = 0$ ) gives

$m_L ( \nu_1 - 3) + ( m_C + m_g) \nu_1 = 0$

which leads to

$\nu_1 = \dfrac{3\, m_L}{m_L + m_C + m_g} = \dfrac{3(2\,kg)}{10\,kg} = 0.60 \, m/s$ .

Next at

$t = 0.80\,s$ , momentum conservation ( for the second explosion ) gives

$m_C \nu_2 + m_R ( \nu_2 + 3) = ( m_C + m_g)\nu_1 = (8\,kg) ( 0.60\, m/s) = 4.8 \, kg \cdot m/s$ .

This yields

$\nu_2 = -0.15$ . Thus , the velocity of block C after the second explosion is

$\nu_2 = - (0.15\, m/s) \hat{i}$ .

(b) Between

$t = 0$ and

$t = 0.80 \,s$ , the block moves

$\nu_1 \Delta t = (0.60\, m/s)(0.80\,s) = 0.48\,m$ .
Between

$t = 0.80 \,s$ and

$t = 2.80 \,s$ , it moves an additional

$\nu_2 \Delta t = (-0.15\,m/s )(2.00\,s) = -0.30\,m$ .

Its net displacement since

$t = 0$ is therefore

$0.48 \,m - 0.30 \,m = 0.18\,m$ .

Fasten one end of a vertical spring to a ceiling, attach a cabbage to the other end, and then slowly lower the cabbage until the upward force on it from the spring balances the gravitational force on it. Show that the loss of gravitational potential energy of the cabbageEarth system equals twice the gain in the springs potential energy.

1747215_1767189_ans_39e0180782ea429d8901e66b9cba64b9.jpg

A raindrop of mass $1.00\ g$ falling from a height of $1\ km$ on the ground with a speed of $50\ m/s$ . Calculate:
The loss of $P.E.$ of the drop.

Drop mass,

$m=0.001\ kg, h=1\ km=1000\ m$
Speed of

$v=50\ m/s$

$PE$ at highest point of drop =

$mgh=0.001\times 10\times 1000=10\ J$ so loss of

$PE=10\ J$

A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are : [Choose the correct alternative]
Lowest Point Highest Point
$mg - T_1$ $mg + T_2$
$mg + T_1$ $mg - T_2$
$mg + T_1 -(mv_1^2)/R$ $mg - T_2 + (mv_1^2)/R$
$mg - T_1 -(mv_1^2)/R$ $mg + T_2 + (mv_1^2)/R$
$T_1$ and $v_1$ denote the tension and speed at the lowest point. $T_2$ and $v_2$ denote corresponding values at the highest point.

The net force acting on stone at the lowest point directed vertically downward =

$mg - T_1$ & and the highest point =

$mg + T_2$ . Hence option (a) is correct answer
1769867_1844917_ans_9ec8b38a00234ab98c7dadb8617ff416.png

Consider a body of mass 'm' attached to a string of length 'L'. If the ring is forming a vertical circle, derive an expression for velocity and tension at any point. Also, find the velocity that is required for mass to just reach the peak point of the circle.

Ref. image 1

Let

$'theta'$ be the angle the string makes with the vertical at any instant of time. Let

$v_x$ be the velocity of body at the lowermost point x. The distance traveled by the body from the given point 'p' to 'X' in 'y' direction is given by.

$XY = L - L \cos \theta = L(1 - \cos \theta)$

Ref. image 2

$T - mg \cos \theta = \frac{mv^2}{L}$
Using the equation

$v^2 = u^2 + 2as$ , we can write

$vx^2 = v^2 + 2g (XY)$
i.e.

$c^2 = vx^2 - 2 g L (1 - \cos \theta)$ .........(1)
Tension,

$T = mg \cos \theta + \frac{mv^2}{L}$ using (1)

$T = mg \cos \theta + \frac{m}{L}(Vx^2 - 2gL)(1 - \cos \theta)$

$= mg \cos \theta + \frac{mv_x^2 }{L} - 2 mg ( 1 - \cos \theta)$

$T = \frac{mv_x^2}{L} + mg (3 \cos \theta - 2)$
For the vertical circle to be reached v = 0, &

$v_x = ?$ &

$\theta = 180^0$

$\Rightarrow v_x^2 = v^2 + 2gL (1 - \cos\theta)$

$v_x^2 = 0 + 2g L(1 - (-1))$

$v_x = \sqrt{4gL}$

$v_x = 2\sqrt{gL}$
1770099_1845371_ans_983b3c95d75f42bfa4d9424b51c7da54.png

Let the mass of the flower pot 15 kg and hight of the sunshade 4m. What is the kinetic energy of the flower pot just before it touches the ground ?

$K = \dfrac{1}{2}mv^{2}$

$v^{2}=u^{2}+2as$

$v^2=0+2\times 10\times 4=80$
Kinetic energy

$K=\dfrac{1}{2}mv^{2}$

$K=1/2\times 15\times 80=600J$

A ball is placed on a compressed spring. What form of energy does the spring possess? On releasing the spring, the ball flies away. Give a reason.

The compressed spring has elastic potential energy due to its compressed state. Which it is released, the potential energy of the spring changes into kinetic energy which does work on the ball if placed on it and changes into kinetic energy of the ball due to which it flies away.
1713047_1806787_ans_8b2e5aac731442a7bcaf32c21c330aec.png

1713047_1806787_ans_8b2e5aac731442a7bcaf32c21c330aec.png

A body of mass $10\, kg$ is moving with a velocity $20\, m \,s^{-1}$ . If the mass of the body is doubled and velocity is halved. Find the ratio of the initial kinetic energy to the final kinetic energy.

Let initial Mass,

$m_1= 10\,kg$

and velocity,

$v_1 = 20\ m/s$

Final mass,

$m_2= 2 \times 10 = 20\, kg$

and velocity,

$v_2= 20/2 = 10\,m/s$

Initial kinetic energy,

$KE_1= \frac {1}{2} \times mass \times (velocity)^2$

$\Rightarrow KE_1= \frac {1}{2} \times 10 \times (20)^2$

$\Rightarrow KE_1 = \frac {1}{2} \times 10 \times 20 \times 20$

$\Rightarrow KE_1 = 2000\, J$

Final kinetic energy,

$KE_2 = \dfrac{1}{2} \times mass \times (velocity)^2$

$\Rightarrow KE_2= \frac {1}{2} \times 20 \times (10)^2$

$\Rightarrow KE_2 = \frac {1}{2} \times 20 \times 10 \times 10$

$\Rightarrow KE_2= 1000\, J$

$\Rightarrow \dfrac{KE_1}{KE_2}= \dfrac{2000}{1000}$

$\Rightarrow \dfrac{KE_1}{KE_2}= \dfrac{2}{1}$

$\Rightarrow \dfrac{KE_1}{KE_2}= 2:1$

Give one example of a body that has potential energy. In each of the following:
(a) due to its position at a height,
(b) due to its elongated stretched state.

(a)

$P.E.$ due to its position at a height :

Water at a height has

$P.E.$ stored in it. Falling water from a height can be used to do work like turning a wheel,

(b)

$P.E.$ Due to its elongated stretched state :

A stretched rubber band (elongated state) has potential energy. it does work in restoring itself to its original state. A pebble placed on the stretched rubber catapujt , is thrown away when it is released to restore its originals state.

1694940_1802750_ans_9ece8d338978445fa0e189105e8bf075.png

Give the two names which store kinetic energy and potential energy in it on the basis of daily life observation.

The energy of an object by virtue of its motion is known as kinetic energy.

The energy of an object due to a change in position is known as potential energy.

Water in the overhead tanks and striking hammer have potential as well as kinetic energies. Flying birds and running fans have both potentials as well as kinetic energies.

A ball of mass $0.5\, kg$ slows down from a speed of $5 \,m \,s^{-1}$ to that of $3 \,m\, s^{-1}$ . Calculate the change in kinetic energy of the ball.

Mass of ball

$= 0.5\,kg$

Initial velocity

$= 5\, m/s$

Initial kinetic energy

$= \dfrac{1}{2} \times mass \times (velocity)^2$

$= \dfrac{1}{2} \times 0.5 \times(5)^2$

$= \dfrac{1}{2} \times \times 0.5 \times 6.25\, J = 6.25\,J$

Final velocity of the ball

$= 3 \, m/s$

Final kinetic energy of the ball

$= \dfrac{1}{2} \times mass \times (velocity)^2$

$= \dfrac{1}{2} \times 0.5 \times(3)^2$

$= \dfrac{1}{2} \times 0.5 \times 9 = 2.25\, J$

Change in the kinetic energy of the ball

$= 2.25\, J- 6.25\, J= -4\, J$

There is a decrease in the kinetic energy of the ball.

Why is following situation impossible? A new high-speed roller coaster is claimed to be a safe that the passengers fo not need to wear seat belts or any other restraining device. The coster is design with a vertical circular section over which the coaster travels on the inside of the circle so that the passenger are upside down section i s $12.0m$ and the coster enters the bottom of the circular section at a speed of $22.0 m/s$ . Assume the coster moves without friction on the track and model teh coaster as a particle.

1900278_1864948_ans_020071e6c3974c6d8c09436b363ace13.jpg

A box of mass 5kg was raised from the ground to the second floor of height 7m. Calculate the potential energy of the box with respect to the ground floor?

Mass m = 5g
Height from ground to the second floor

$h = 7m$

$g = 10 m/s^{2}$

Potential energy with respect to the ground -

$U = mgh = 5kg \times 10m/s^{2}\times 7=350\ J$

An object of mass $m_{1} = 4.00 kg$ is tied to an object of mass $m_{2} = 3.00 kg$ with String $1$ of length $l= 0.500 m$ . The combination is swung in a vertical circular path on a second string, String $2$ , of length $l= 0.500 m$ . During the motion, the two strings are collinear at all times as shown in below figure. At the top of its motion, $m_{2}$ is traveling at $v = 4.00 m/s$ . (a) What is the tension in String $1$ at this instant? (b) What is the tension in String $2$ at this instant? (c) Which string will break first if the combination is rotated faster and faster?

The radius of the path of object

$1$ is twice that of object

$2$ . Because the strings are always “collinear,” both objects take the same time interval to travel around their respective circles; therefore, the speed of object

$1$ is twice that of object

$2$ .
The free-body diagrams are shown below. We are given

$m_{1}= 4.00 kg, m_{2} = 3.00 kg, v = 4.00 m/s$ , and

$l = 0.500m$
Taking down as the positive direction, we have
Object

$1$ :

$T_{1}+m_{1}g=\dfrac{m_{1}v_{1}^{2}}{r_{1}}$ , where

$v_{1}=2v,r_{1}=2l$
Object

$2$ :

$T_{2}-T_{1}+m_{2}g=\dfrac{m_{2}V_{2}^{2}}{r_{2}}$ , where

$v_{2}=v,r_{2}=2l$
(a) From above:

$T_{1}=\dfrac{m_{1}v_{1}^{2}}{r_{1}}-m_{1}g=m_{1}\left ( \dfrac{v_{1}^{2}}{r_{1}}-g \right )$

$T_{1}=(4.00kg)\left [ \dfrac{[2(4.00m/s)]^{2}}{2(0.500m)}-9.80m/s^{2} \right ]$

$T_{1}=216.8N=217N$
(b) From above:

$T_{2}=T_{1}+\dfrac{m_{2}v_{2}^{2}}{r_{2}}-m_{2}g$

$T_{2}=T_{1}+m_{2}\left ( \dfrac{v_{2}^{2}}{r_{2}}-g \right )$

$T_{2}=T_{1}+(3.00kg)\left [ \dfrac{(4.00m/s)^{2}}{0.500m}-9.80m/s^{2} \right ]$

$T_{2}=216.8N+66.6N=283.4N=283N$
(c) From above,

$T_{2} > T_{1}$ always, so string

$2$ will break first.
1802342_1860757_ans_1d1480bedaad43d8b345d0d8a6c6d72e.png

A ball of mass $m = 0.275 kg$ swings in a vertical circular path on a string $L = 0.850 m$ long as in above figure. (a) What are the forces acting on the ball at any point on the path? (b) Draw force diagrams for the ball when it is at the bottom of the circle and when it is at the top. (c) If its speed is $5.20 m/s$ at the top of the circle, what is the tension in the string there? (d) If the string breaks when its tension exceeds $22.5 N$ , what is the maximum speed the ball can have at the bottom before that happens?

(a) At each point on the vertical circular path, two forces are acting on the ball (refer to below figure):
(1) The downward gravitational force with constant magnitude

$F_{g} = mg$ .
(2) The tension force in the string, always directed toward the center of the path.
(b) Below figure shows the forces acting on the ball when it is at the highest point on the path (left-hand diagram) and when it is at the bottom of the circular path (right-hand diagram). Note that the gravitational force has the same magnitude and direction at each point on the circular path. The tension force varies in magnitude at different points and is always directed toward the
center of the path.
(c) At the top of the circle,

$F_{c}=mv^{2}/r=T+F_{g}$ or,

$T=\dfrac{mv^{2}}{r}-F_{g}=\dfrac{mv^{2}}{r}-mg=m\left ( \dfrac{v^{2}}{r}-g \right )$

$=(0.275kg)\left [ \dfrac{(5.20m/s)^{2}}{0.850m}-9.80m/s^{2} \right ]=6.05N$
(d) At the bottom of the circle,

$F_{c}=mv^{2}/r=T-F_{g}=T-mg$ and solving for the speed gives

$v^{2}=\dfrac{r}{m}(T-mg)=r\left ( \dfrac{T}{m}-g \right )$ and

$v=\sqrt{r\left ( \dfrac{T}{m}-g \right )}$
If the string is at the breaking point at the bottom of the circle, then

$T = 22.5 N$ , and the speed of the object at this point must be

$v=\sqrt{(0.850m)\left ( \dfrac{22.5N}{0.275kg}-9.80m/s^{2} \right )}=7.82m/s$
1802348_1860766_ans_8e368170616e428abc879daa7e013999.png

the angle $\theta$ between the thread and the vertical at the moment when the total acceleration vector of the sphere is directed horizontally.

We have

$\vec{w}=w_{i}\hat{u_{i}}+w_{n}\hat{u_{n}}$ thus

$w_{y}=w_{t(y)}+w_{n(y)}$

But in accordance with the problem

$w_{y}=0$

So,

$w_{t(y)}+w_{n(y)}=0$

or,

$g\sin \theta\sin \theta+2g\cos^{2}\theta(-\cos \theta)=0$

or,

$\cos\theta=\dfrac{1}{\sqrt{3}}$ or,

$\theta=54.7^{o}$

1792905_1851909_ans_d3d00f43d6084b52a08497563b92d4ad.png

Find the work to be performed in order to blow a soap bubble of radius $R$ if the outside air pressure is equal to $p_0$ and the surface tension of the soap water solution is equal to $\alpha$ .

1901161_1852488_ans_a61e76c5e68b4b23aaec9ede3e2c63c7.jpg

One end of a cord is fixed and a small $0.500kg$ object is attached to the other end, where it swings in a section of a vertical circle of radius $2.00 m$ as shown in above figure. When $\theta = 20.0^{0}$ , the speed of the object is $8.00 m/s$ . At this instant, find (a) the tension in the string, (b) the tangential and radial components of acceleration, and (c) the total acceleration. (d) Is your answer changed if the object is swinging down toward its lowest point instead of swinging up? (e) Explain your answer to part (d).

(a) Consider radial forces on the object, taking inward as positive

$\sum F_{r}=ma_{r}:T-mg\cos\theta=\dfrac{mv^{2}}{r}$
solving for the tension gives

$T=mg\cos\theta+\dfrac{mv^{2}}{r}$

$=(0.500kg)(9.80m/s^{2})\cos20.0^{0}+(0.500kg)(8.00m/s^{2})/2.00m$

$=4.60N+16.0N=20.6N$
(b) We already found the radial component of acceleration

$a_{r}=\frac{v^{2}}{r}=\frac{(8.00m/s)^{2}}{2.00m}=32.0m/s^{2}$ inward.
Consider the tangential forces on the object:

$\sum F_{t}=ma_{t}:mg\sin\theta=ma_{t}$
Solving for the tangential component of acceleration gives

$a_{t} = g \sinθ = (9.80 m/s^{2} )\sin20.0^{0}$

$= 3.35 m/s^{2}$ downward tangent to the circle.
(c) The magnitude of the acceleration is

$a=\sqrt{a_{r}^{2}+a_{t}^{2}}=\sqrt{(32.0m/s^{2})^{2}+(3.35m/s^{2})^{2}}=32.2m/s^{2}$
at an angle of

$\tan^{-1}\left ( \dfrac{3.35m/s^{2}}{32.0m/s^{2}} \right )=5.98^{0}$
Thus, the acceleration is

$32.2 m/s^{2}$ inward and below the cord at

$5.98^{0}$
(d) No change.
(e) If the object is swinging down it is gaining speed, and if the object is swinging up it is losing speed, but the forces are the same; therefore, its acceleration is regardless of the direction of swing.

A stone of mass 2kg its thrown upward from the ground with a velocity of 3m/s. When it reaches maximum height, calculate its potential energy.

Given,

$u = 3m/s \\v = 0$

$a= g = 10 m/s^{2}$ (velocity of a body moving upwqard is decreasing so acceleration is negative)

$v^{2}=u^{2}+2as$

$0=3^2+2\times -10\times s$

$0=9-20s$

$20s = 9,\ \ \ s=\dfrac{9}{20}$

$U = mgh = 20 \times 10\times \dfrac{9}{20}=90J$

A ball whirls around in a vertical circle art the end of a string The other end of the string is fixed at the center of the circle. Assuming the total energy of the ball-Rath system remain constant, show that the tension in the string at the bottom is greater than the tension at the top by six times the ball's weight.

1902394_1865203_ans_91e8e4c08f7240eba4478bb4388e2eb7.jpg

A ball of mass $m=300 g$ is connected by a strong string of length $L=80.0\ cm$ to a pivot and held in place with the string vertical. A wind exerts constant force $F$ to the right on the ball as shown Figure The ball is released from rest. The wind makes it swing up to attain maximum height $J$ above its starting point before it swings down again as $F$ approaches infinity?

1979002_1865369_ans_1c749c35c04f4a25b400220ed34cf034.jpg

While Lost-a-Lot ponders his next move in the situation described in problem $19$ and illustrated in figure $P12.19$ , the enemy attacks! An incoming projectile breaks off the stone ledge so that the end of the drawbridge can be lowered past the wall where it usually rests. In addition, a fragment of the projectile bounces up and cuts the drawbridge cable! The hinge between the castle wall and the bridge is frictionless, and the bridge swings down freely until it is vertical and smacks into the vertical castle wall below the castle entrance.
Immediately before it strikes the castle wall.

Data according to problem mentioned in ques which is missing:

$\theta_o=20^o$

$l=8.00\,m$

$m=2000\,kg$

Force exerted by the hinge on the bridge before the cable breaks:

According to FBD as shown in the figure we can write:

$R_{x}=0$ there is no force in horizontal direction before the cable breaks.

and we know that:

$a=\omega^{2}\left(\dfrac{1}{2} \ell\right)=(1.56 \mathrm{rad} / \mathrm{s})^{2}(4.0 \mathrm{m})=9.67 \mathrm{m} / \mathrm{s}^{2}$

$R_{y}-m g=m a$

solving and putting values according to the question:

$\therefore R_{y}=(2000 \mathrm{kg})\left(9.8 \mathrm{m} / \mathrm{s}^{2}+9.67 \mathrm{m} / \mathrm{s}^{2}\right)=38.9 \mathrm{kN}$

Thus:

$\quad R_{y}=38.9 \hat{\mathbf{j}} \mathrm{kN}$

1893118_1865798_ans_fb8de78460254594b83f2792950ef722.png

Why is the following situation impossible? An athlete test her hand strength by having an assistance hang weight from her belt as she hangs onto a horizontal bar with her hands. When the weights hanging on her belt have increased to $80\%$ on her body weight, her hands can no longer supports her and she drops to the floor. Frustrated at not meeting her hand-strength goal, she decides to swing on a trapeze. The trapeze consists of a bar suspended by two parallel ropes, each of length $l$ , allowing performers to swing in a vertical circular arc Fig The athlete hold the bar and steps off an elevated platform, starting from rest with the ropes at an angle $\theta_i=60.0^o$ with respect to the vertical. As she swings several times back and forth in a circular arc, she forgets her frustrations related to the hand-strength test. Assume the size of the performer's body is small compared to the length $l$ and air resistance is negligible.

1978448_1865132_ans_7d6bb3b265bc47a4abe6bbcd49e3b7ac.jpg

Two hypothetical planets of masses $m_1$ and $m_2$ and radii $r_1$ and $r_2$ respectively, are nearly at rest when they are an infinite distance apart. Because of their gravitational attraction, they head towards each other on a collision course.
When their center to center separation is $d$ , find expressions for the speed to each planet and for their relative speed.
Note : Both the energy and momentum of the isolated two planet system are constant.

1980770_1867288_ans_feb48a899a444fbe98e64b731aa77c46.jpeg

While Lost-a-Lot ponders his next move in the situation described in problem $19$ and illustrated in figure $P12.19$ , the enemy attacks! An incoming projectile breaks off the stone ledge so that the end of the drawbridge can be lowered past the wall where it usually rests. In addition, a fragment of the projectile bounces up and cuts the drawbridge cable! The hinge between the castle wall and the bridge is frictionless, and the bridge swings down freely until it is vertical and smacks into the vertical castle wall below the castle entrance.
Immediately after the cable breaks

Data according to problem mentioned in ques which is missing:

$\theta_o=20^o$

$l=8.00\,m$

$m=2000\,kg$

Force exerted by the hinge on the bridge immediately after the cable breaks:

The linear acceleration of the bridge is:

$a=\dfrac{1}{2} \ell \alpha=\dfrac{1}{2}(8.0 \mathrm{m})\left(1.73 \mathrm{rad} / \mathrm{s}^{2}\right)=6.907 \mathrm{m} / \mathrm{s}^{2}$

According to FBD s shown in diagram:

The force at the hinge + the force of gravity produce the acceleration:

$a=6.907 \mathrm{m} / \mathrm{s}^{2}$ at right angles to the bridge.

$R_{x}=m a_{x}=(2000 \mathrm{kg})\left(6.907 \mathrm{m} / \mathrm{s}^{2}\right) \cos 250^{\circ}=-4.72 \mathrm{kN}$

$R_{y}-m g=m a_{y}$

Solving and putting values according to the question:

$\therefore R_{y}=m\left(g+a_{y}\right)=(2000 \mathrm{kg})\left[9.80 \mathrm{m} / \mathrm{s}^{2}+\left(6.907 \mathrm{m} / \mathrm{s}^{2}\right) \sin 250^{\circ}\right]=6.62 \mathrm{kN}$

Thus:

$\mathrm{R}=(-4.72 \hat{\mathbf{i}}+6.62 \hat{\mathbf{j}}) \mathrm{kN}$

1893112_1865794_ans_adbdcbfc0a6c4ff7adca05a35be20644.png

	Lower Point	Highest Point
a.	mg - $T_1$	mg + $T_2$
b.	mg + $T_1$	mg - $T_2$
c.	mg + $T_1 - (mv^2_1) / R$	mg - $T + (mv^2_1) / R$
d.	mg - $T_1 - (mv^2_1) / R$	mg + $T_2 + (mv^2_1) / R$

Lowest Point	Highest Point
$mg - T_1$	$mg + T_2$
$mg + T_1$	$mg - T_2$
$mg + T_1 -(mv_1^2)/R$	$mg - T_2 + (mv_1^2)/R$
$mg - T_1 -(mv_1^2)/R$	$mg + T_2 + (mv_1^2)/R$

Work,Energy And Power - Class 11 Engineering Physics - Extra Questions

Force ,F= 6x2+4x+76{x^2} + 4x + 7. If x=0x = 0 to x=2x=2 ,calculate W.D.

Define the following and give one example.Hybrid rocket propellants.

Spring of force content 100N/m is extended to 240 cm and released. Find out the work done when the extension has reduced to 10 cm.

A particle moves along X -axis from x=0x=0 to x=1mx=1m under the influence of a force given by F=3x2+2x−10F=3x^2+2x-10. Work done in the process is

A string can bear a maximum tension of 3.16kg−wt3.16kg-wt. What maximum number of revolutions per second can be made by a stone of mass 49g49g tied to one end of a 1m1m long piece of this string so that the string may not break? (π=22/7\pi =22/7)

A single force acts on a 3.0 kg particle-like object whose position is given by x=3.0t−4.0t2+1.0t3x = 3.0 t - 4.0 t^2 + 1.0 t^3, with x in meters and t in seconds. Find the work done by the force from t = 0 to t = 4.0 s.

Fill in the blanksWork done when a body of mass 100 kg lifted up to a height of 1 mater is ..........

Show that a force that does no work must be a velocity deependent force.

An object of mass 1000 kg is traveling with a velocity 72 km/h. Calculate the work done to bring it rest.

Estimate the work done on an object of mass 80 kg to change its velocity from 5m /s to 10 m /s.

A 20 kg object is being lifted through a height of ______ m when 484 J of work is done on it.

What should be the change in velocity of a body, if its mass is increased by four times for the same K.E of the body?

Calculate the work done to raise a body of the mass 30 kg30 kg to a height of 50 m 50 m (g = 10 m s^{-2})(g = 10 m s^{-2})

Two bodies A and B of equal masses are kept at heights of h and 2h respectively. What will be the ratio of their potential energy?

Compute the work which must be performed (in Kgf-mKgf-m) to slowly pump water out of a hemispherical reservoir of radius R=0.6 mR=0.6 m

A spring does not obey Hooke's law. Rather it follows, F = kx - bx^2 + cx^3F = kx - bx^2 + cx^3. If the spring has natural length ll and compressed length l'l' then find the work done.

Find work done by a force \vec{F} = x \hat{i} + xy \hat{j}\vec{F} = x \hat{i} + xy \hat{j} acting on a particle to displace it from point O(0, 0) to C(2, 2).

A molecule in a gas container hits a horizontal wall with speed 200 m s^{-1}s^{-1} and angle 30^030^0 with the normal, and rebounds with the same speed. Is momentum conserved in the collision ? Is the collision elastic or inelastic ?

An object of mass, m is moving with a constant velocity, vv. How much work should be done on the object in order to bring the object to rest?

Certain force acting on a 20\ kg20\ kg mass changes its velocity from 5\ ms^{-1}5\ ms^{-1} to 2\ ms^{-1}2\ ms^{-1}. Calculate the work done by the force.

Calculate the work required to be done to stop a car of 1500\ kg1500\ kg moving at the velocity of 60\ km/h60\ km/h?

A force is applied on a body of mass 20 kg20 kg moving with a velocity of 40 m{s}^{-1}40 m{s}^{-1}. The body attains a velocity of 50 m{s}^{-1}50 m{s}^{-1} in 22 seconds. Calculate the work done to attain this velocity.

A moving body weighing 400N400N possesses 500 J500 J of kinetic energy. Calculate the velocity with which the body is moving. (g= 10 ms^{-1})(g= 10 ms^{-1})

Define kinetic energy.

You did 150 \ joules150 \ joules of work by lifting a 120 \ Newton120 \ Newton backpack.a) How high did you lift the backpack?b) How much did the backpack weigh in pounds?

Look at the two diagrams given below. Calculate the work done in compressing each spring completely, assuming that the force applied remains constant.

A particle is moving along a vertical circle of radius R. The velocity of particle at P will be (assume critical condition at C).

A particle of mass mm moving with a speed vv hits elastically another stationary particle of mass 2m2m in a fixed smooth horizontal circular tube of radius rr. Find the time when the next collision will take place?

Two springs have force constants k_{1}k_{1} and k_{2}(k_{1} > k_{2})k_{2}(k_{1} > k_{2}). On which spring is more work done, if (i) they are stretched by the same force and (ii) they are stretched by same distance?

The bob of a simple pendulum of length 1m1m has mass 100g100g and a speed of 1.4m/s1.4m/s at the lowest point in its path. Find the tension in the string at this instant.

A particle of mass mm is fixed to one end of a light rigid rod of length ll and rotated in a vertical circular path about its other end. The minimum speed of the particle at its highest point must be:

Suppose the amplitude of a simple pendulum having a bob of mass mm is { \theta }_{ 0 }{ \theta }_{ 0 }. Find the tension in the string when the bob is at its extreme position.

Two capacitors { C }_{ 1 }{ C }_{ 1 } and { C }_{ 2 }{ C }_{ 2 } have equal amount of energy stored in them. What is the ratio of potential differences across their plates.

A stone tied to an inextensible string of length l = 1 \,ml = 1 \,m is kept horizontal. If it is released, find the angular speed of the stone when the string makes an angle \theta =30^o\theta =30^o with horizontal.

The displacement xx of a body of mass 1 kg on smooth horizontal surface as a function of time t is given by x=\dfrac { { t }^{ 3 } }{ 3 } x=\dfrac { { t }^{ 3 } }{ 3 } (where x is in metres and t is in seconds). Find the work done by the external agent for the first one second.

A bicycle chain of length 1.6 m and of mass 1 kg is lying on a horizontal floor. What is the work done in lifting it with one end touching the floor and the other end 1.6 m above the floor?(take g={ 10 }ms^{ -2 }{ 10 }ms^{ -2 })

An object of mass 0.50.5 kg attached to a rod of length 0.50.5 m is whirled in a vertical circle at constant angular speed. If the maximum tension in the string is 55 kg wt., Calculate (i) speed of stone (ii) maximum number of revolutions it can complete in a minute.

What is work done in lifting a body of mass 2020 kg at a height of 22m above the ground? (g=10m/s^2)(g=10m/s^2)

Two springs AA and BB are identical but AA is harder than B\ (k_{A} > k_{B})B\ (k_{A} > k_{B}). On which spring more work will be done if: (a)They are stretched through the same distance (b)They are stretched by same force?

A stone of mass 1kg1kg is tied to a light string of length l=10ml=10m is whirling in a circular path in the vertical plane. If the ratio of maximum to minimum tensions in the string is 33, find the speeds of the stone at the lowest and highest points.

Position-time variation of a particle of mass 1\ kg1\ kg is x=(2t^{3}+3)mx=(2t^{3}+3)m. Calculate the work done by the force in first 2\ seconds2\ seconds .

A particle moves along the x-axis from x=0x=0 to x=5mx=5m under the influence of a force FF (in N) given by F=3{x}^{2}-2x+7F=3{x}^{2}-2x+7. Calculate the work done by this force

Calculate the kinetic energy of an object weighing 600 \ kg600 \ kg moving at a velocity of 30 \ m/s30 \ m/s

A 4\ kg4\ kg swing in a vertical circle at the end of the chord 1\ m1\ m long. Find out the maximum speed at which it can swing if chord can sustain maximum tension of 140\ N140\ N.

A ball moving with a momentum of 5kgm/s5kgm/s strikes against a wall at angle of {45^ \circ }{45^ \circ } and is deflected at the same angle. Calculate the change in momentum.

A force F={x}^{2}+2x+1F={x}^{2}+2x+1 acts on a particle in the x direction, where FF is in newton and xx in metre. Find the work done by this force during a displacement from x=-1mx=-1m to x=2.0mx=2.0m

An object is displaced from position vector r_{1} = (2\hat{i} + 3\hat{j})\ mr_{1} = (2\hat{i} + 3\hat{j})\ m to r_{2} = (4\hat{i} + 6\hat{j})\ mr_{2} = (4\hat{i} + 6\hat{j})\ m under a force F = (3x^2 \hat{i} + 2y\hat{j})\ NF = (3x^2 \hat{i} + 2y\hat{j})\ N. Find the work done by this force.

A particle is moving in a circular path of radius a under the action of an attractive potential U=-\dfrac{K}{2r^2}U=-\dfrac{K}{2r^2}. Its total energy is

A position-dependent force f=7-2x+3x^2f=7-2x+3x^2 newton acts on a small body of mass 22kg and displaces from x=0x=0 to x=5x=5. Calculate the work done by the force.

Find the ratio of the weights, as measured by a spring balance, of a 1\,kg1\,kg block of iron and a 1\,kg1\,kg block of wood.Density of iron = 7800\,kg/m^37800\,kg/m^3, density of wood = 800\,kg/m^3800\,kg/m^3 and density of air= 1.293\,kg\,m^31.293\,kg\,m^3

When the bucket filled with water is resolved fast in a vertical circle, the water does not fall even when the bucket is inverted.

The relation between displacement xx and time tt for a body of mass 2 Kg2 Kg moving under the action of a force is given by x=\dfrac{t^3}{3}x=\dfrac{t^3}{3}, where xx is in meter and tt in second, calculate the work done by the body in first 22 seconds.

Calculate the work done in moving the object from x=2x=2 to x=3x=3 m from the given graph.

A 2 kg block slides on a horizontal floor with a speed of 4 m/s. It strikes an uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is 15 N and spring constant is 10,000 N/m. The spring compresses by :-

The work done by an applied variable force F=x+x^3F=x+x^3 from x=0mx=0m to x=2mx=2m,where xx is displacement ,is

If force on a block is F=8/x^2F=8/x^2 where xx is displacement ,then find total work done in moving block from from x=2mx=2m to x=6mx=6m

Explain why water stored in a dam has potential energy.

A body of mass 22 kg starts with an initial velocity 55 m/s. If the body is acted upon by a time dependent force (F) as shown in figure, then work done on the body in 2020 s is?

A car moves on a flat circular track of 200200 m radius with a constant speed of 2020 if the car weighs 400400 kg the centripetal force acting on the car is:

A stone of mass 1 Kg, tied to the end of a string of length 30cm, is whirled in a vertical circle such that its angular velocity when the string is horizontal is 10 rad/s. If the tension in the string at that instant is 5T, find the value of T

Find the velocity of a body of mass 100g100g having a kinetic energy of 20J20J.

Seen the displacement X and time t for a body of mass 2 Kg moving x=\dfrac { t^ 3 }{ 3 } x=\dfrac { t^ 3 }{ 3 } , where x is in meter and t in sec. Calculate the work done by body in first 2 second.

A stone of mass 1 kg is tied up with an inextensible string of length l = 10/3 m. It is rotated in a vertical circle of l radius. If a ratio of maximum and minimum tension in the string is 4 and the value of g is 10 m/s^2 m/s^2 then calculate a speed of stone at the highest point of a circle.

A porter lift a luggage of 15\ kg15\ kg from the ground and puts it on her head 2\ m2\ m above the ground. Calculate the work done by her on the luggage? Take g=10\ ms^{-2}g=10\ ms^{-2}

A pendulum bob has a speed of 3ms^{-1}3ms^{-1} at its lowest position. The pendulum is 0.5m0.5m long. What is the speed of the bob, when the length makes an angle of 60^\circ60^\circ to the vertical? (g = 10ms^{-2})(g = 10ms^{-2})

A position dependent force F = 7-2x + 3x^2 F = 7-2x + 3x^2 newton (where x is in meter) on a small body of mass 2 Kg & displace it from x = 0 x = 0 to x = 5 x = 5 m. Calculate the work done.

A force that object A exerts on object B is observed over a 10 second interval,as shown on the graph.How much work did object A do during that 10 s?

The car AA of mass 1500kg1500kg travelling at 40m/s40m/s collides with another car 'B''B' of mass 1250kg1250kg travelling at 25m/s25m/s in the same direction. After collision the velocity of car 'A''A' becomes 30m/s30m/s. calculate the velocity of car 'B''B' after the collision.

A sphere is suspended by a thread of length L. What minimum horizontal velocity to has to be imparted to it so that if completes a full vertical revolution?

Two objects having equal masses are moving with uniform velocities of 2 m/s and 6m/s respectively Calculate the ratio of their kinetic energies.

The bob of a simple pendulum at rest position is given a velocity V in horizontal direction so that bob describes verticle circle of radius equal to length of pendulum L . If the tension in string is 4 times weight of of bob when the string is horizontal , the velocity of bob :

A ball of mass 100 g is suspended a string 40 cm long. Keeping the string taut , the ball describes a horizontal circle of radius 10 cm Find the angular speed.

If the velocity of the body increases 3 times how does its K.E changes?

A boy weighing 50\, kg50\, kg climbs up a vertical height of 100\,m100\,m. Calculate the amount of work done by him. How much potential energy does he gain (g=9.8\, m/{s}^{2}g=9.8\, m/{s}^{2})

If the speed of a body is halved, what will be the change in its kinetic energy?

Certain force acting on a 20\ kg20\ kg mass changes its velocity from 5 m/{s}5 m/{s} to 2 m/{s}2 m/{s}. Calculate the work done by the force.

A man weighing 70 kg carries a weight of 10 kg to top of a tower 100 m high. Calculate the work done by the man

What is Kinetic Energy?

A car of mass 2000\ kg2000\ kg changes its speed from 18\ km/h18\ km/h to 90\ km/hr90\ km/hr. Find the work done by the engine.

Force ,F= $6{x^2} + 4x + 7$ . If $x = 0$ to $x=2$ ,calculate W.D.

Define the following and give one example.
Hybrid rocket propellants.

A particle moves along X -axis from $x=0$ to $x=1m$ under the influence of a force given by $F=3x^2+2x-10$ . Work done in the process is

A string can bear a maximum tension of $3.16kg-wt$ . What maximum number of revolutions per second can be made by a stone of mass $49g$ tied to one end of a $1m$ long piece of this string so that the string may not break? ( $\pi =22/7$ )

A single force acts on a 3.0 kg particle-like object whose position is given by $x = 3.0 t - 4.0 t^2 + 1.0 t^3$ , with x in meters and t in seconds. Find the work done by the force from t = 0 to t = 4.0 s.

Fill in the blanks
Work done when a body of mass 100 kg lifted up to a height of 1 mater is ..........

Calculate the work done to raise a body of the mass $30 kg$ to a height of $50 m$ $(g = 10 m s^{-2})$

Compute the work which must be performed (in $Kgf-m$ ) to slowly pump water out of a hemispherical reservoir of radius $R=0.6 m$

A spring does not obey Hooke's law. Rather it follows, $F = kx - bx^2 + cx^3$ . If the spring has natural length $l$ and compressed length $l'$ then find the work done.

Find work done by a force $\vec{F} = x \hat{i} + xy \hat{j}$ acting on a particle to displace it from point O(0, 0) to C(2, 2).

A molecule in a gas container hits a horizontal wall with speed 200 m $s^{-1}$ and angle $30^0$ with the normal, and rebounds with the same speed. Is momentum conserved in the collision ? Is the collision elastic or inelastic ?

An object of mass, m is moving with a constant velocity, $v$ . How much work should be done on the object in order to bring the object to rest?

Certain force acting on a $20\ kg$ mass changes its velocity from $5\ ms^{-1}$ to $2\ ms^{-1}$ . Calculate the work done by the force.

Calculate the work required to be done to stop a car of $1500\ kg$ moving at the velocity of $60\ km/h$ ?

A force is applied on a body of mass $20 kg$ moving with a velocity of $40 m{s}^{-1}$ . The body attains a velocity of $50 m{s}^{-1}$ in $2$ seconds. Calculate the work done to attain this velocity.

A moving body weighing $400N$ possesses $500 J$ of kinetic energy. Calculate the velocity with which the body is moving. $(g= 10 ms^{-1})$

You did $150 \ joules$ of work by lifting a $120 \ Newton$ backpack.

a) How high did you lift the backpack?
b) How much did the backpack weigh in pounds?

A particle of mass $m$ moving with a speed $v$ hits elastically another stationary particle of mass $2m$ in a fixed smooth horizontal circular tube of radius $r$ . Find the time when the next collision will take place?

Two springs have force constants $k_{1}$ and $k_{2}(k_{1} > k_{2})$ . On which spring is more work done, if (i) they are stretched by the same force and (ii) they are stretched by same distance?

The bob of a simple pendulum of length $1m$ has mass $100g$ and a speed of $1.4m/s$ at the lowest point in its path. Find the tension in the string at this instant.

A particle of mass $m$ is fixed to one end of a light rigid rod of length $l$ and rotated in a vertical circular path about its other end. The minimum speed of the particle at its highest point must be:

Suppose the amplitude of a simple pendulum having a bob of mass $m$ is ${ \theta }_{ 0 }$ . Find the tension in the string when the bob is at its extreme position.

Two capacitors ${ C }_{ 1 }$ and ${ C }_{ 2 }$ have equal amount of energy stored in them. What is the ratio of potential differences across their plates.

A stone tied to an inextensible string of length $l = 1 \,m$ is kept horizontal. If it is released, find the angular speed of the stone when the string makes an angle $\theta =30^o$ with horizontal.

The displacement $x$ of a body of mass 1 kg on smooth horizontal surface as a function of time t is given by $x=\dfrac { { t }^{ 3 } }{ 3 }$ (where x is in metres and t is in seconds). Find the work done by the external agent for the first one second.

A bicycle chain of length 1.6 m and of mass 1 kg is lying on a horizontal floor. What is the work done in lifting it with one end touching the floor and the other end 1.6 m above the floor?
(take g= ${ 10 }ms^{ -2 }$ )

An object of mass $0.5$ kg attached to a rod of length $0.5$ m is whirled in a vertical circle at constant angular speed. If the maximum tension in the string is $5$ kg wt., Calculate
(i) speed of stone
(ii) maximum number of revolutions it can complete in a minute.

What is work done in lifting a body of mass $20$ kg at a height of $2$ m above the ground? $(g=10m/s^2)$

Two springs $A$ and $B$ are identical but $A$ is harder than $B\ (k_{A} > k_{B})$ . On which spring more work will be done if: (a)They are stretched through the same distance (b)They are stretched by same force?

A stone of mass $1kg$ is tied to a light string of length $l=10m$ is whirling in a circular path in the vertical plane. If the ratio of maximum to minimum tensions in the string is $3$ , find the speeds of the stone at the lowest and highest points.

Position-time variation of a particle of mass $1\ kg$ is $x=(2t^{3}+3)m$ . Calculate the work done by the force in first $2\ seconds$ .

A particle moves along the x-axis from $x=0$ to $x=5m$ under the influence of a force $F$ (in N) given by $F=3{x}^{2}-2x+7$ . Calculate the work done by this force

Calculate the kinetic energy of an object weighing $600 \ kg$ moving at a velocity of $30 \ m/s$

A $4\ kg$ swing in a vertical circle at the end of the chord $1\ m$ long. Find out the maximum speed at which it can swing if chord can sustain maximum tension of $140\ N$ .

A ball moving with a momentum of $5kgm/s$ strikes against a wall at angle of ${45^ \circ }$ and is deflected at the same angle. Calculate the change in momentum.

A force $F={x}^{2}+2x+1$ acts on a particle in the x direction, where $F$ is in newton and $x$ in metre. Find the work done by this force during a displacement from $x=-1m$ to $x=2.0m$

An object is displaced from position vector $r_{1} = (2\hat{i} + 3\hat{j})\ m$ to $r_{2} = (4\hat{i} + 6\hat{j})\ m$ under a force $F = (3x^2 \hat{i} + 2y\hat{j})\ N$ . Find the work done by this force.

A particle is moving in a circular path of radius a under the action of an attractive potential $U=-\dfrac{K}{2r^2}$ . Its total energy is

A position-dependent force $f=7-2x+3x^2$ newton acts on a small body of mass $2$ kg and displaces from $x=0$ to $x=5$ . Calculate the work done by the force.

Find the ratio of the weights, as measured by a spring balance, of a $1\,kg$ block of iron and a $1\,kg$ block of wood.
Density of iron = $7800\,kg/m^3$ , density of wood = $800\,kg/m^3$ and density of air= $1.293\,kg\,m^3$

The relation between displacement $x$ and time $t$ for a body of mass $2 Kg$ moving under the action of a force is given by $x=\dfrac{t^3}{3}$ , where $x$ is in meter and $t$ in second, calculate the work done by the body in first $2$ seconds.

Calculate the work done in moving the object from $x=2$ to $x=3$ m from the given graph.

The work done by an applied variable force $F=x+x^3$ from $x=0m$ to $x=2m$ ,where $x$ is displacement ,is

If force on a block is $F=8/x^2$ where $x$ is displacement ,then find total work done in moving block from from $x=2m$ to $x=6m$

A body of mass $2$ kg starts with an initial velocity $5$ m/s. If the body is acted upon by a time dependent force (F) as shown in figure, then work done on the body in $20$ s is?

A car moves on a flat circular track of $200$ m radius with a constant speed of $20$ if the car weighs $400$ kg the centripetal force acting on the car is:

Find the velocity of a body of mass $100g$ having a kinetic energy of $20J$ .

Seen the displacement X and time t for a body of mass 2 Kg moving $x=\dfrac { t^ 3 }{ 3 }$ , where x is in meter and t in sec. Calculate the work done by body in first 2 second.

A stone of mass 1 kg is tied up with an inextensible string of length l = 10/3 m. It is rotated in a vertical circle of l radius. If a ratio of maximum and minimum tension in the string is 4 and the value of g is 10 $m/s^2$ then calculate a speed of stone at the highest point of a circle.

A porter lift a luggage of $15\ kg$ from the ground and puts it on her head $2\ m$ above the ground. Calculate the work done by her on the luggage? Take $g=10\ ms^{-2}$

A pendulum bob has a speed of $3ms^{-1}$ at its lowest position. The pendulum is $0.5m$ long. What is the speed of the bob, when the length makes an angle of $60^\circ$ to the vertical? $(g = 10ms^{-2})$

A position dependent force $F = 7-2x + 3x^2$ newton (where x is in meter) on a small body of mass 2 Kg & displace it from $x = 0$ to $x = 5$ m. Calculate the work done.

The car $A$ of mass $1500kg$ travelling at $40m/s$ collides with another car $'B'$ of mass $1250kg$ travelling at $25m/s$ in the same direction. After collision the velocity of car $'A'$ becomes $30m/s$ . calculate the velocity of car $'B'$ after the collision.

A boy weighing $50\, kg$ climbs up a vertical height of $100\,m$ . Calculate the amount of work done by him. How much potential energy does he gain ( $g=9.8\, m/{s}^{2}$ )

Certain force acting on a $20\ kg$ mass changes its velocity from $5 m/{s}$ to $2 m/{s}$ . Calculate the work done by the force.

A car of mass $2000\ kg$ changes its speed from $18\ km/h$ to $90\ km/hr$ . Find the work done by the engine.

What is the work done to increase the velocity of a car from $30 \ km { h }^{ -1 }$ to $60 \ km { h }^{ -1 }$ , if the mass of the car is $1500 \ kg$ ?

To what height should a box of mass 150 kg be lifted, so that its potential energy may become 7350 joules? $\left( {g = 9.8\;m/{s^2}} \right).$

A ball of mass $100g$ is thrown with a speed of $15$ m/s. Calculate its kinetic energy.

In a conical pendulum, a string of length 120 cm is fixed at rigid support and carries a mass of 150g at its free end. If the mass is revolved in a horizontal circle of radius 0.2m around a vertical axis, calculate tension in the string $\left ( g=9.8m/s^{2} \right )$

Two masses $m$ and $2m$ are dropped from height $h$ and $2h$ . On reaching the ground, which will have greater kinetic energy and why?

In loading a trunk, a man lifts boxes of 100 N each through a height f 1.5 m,
(a) How much work does he do in lifting oe box?
(b) How much energy is transferred when one box is lifted ?
(c) if the man lifts 4 boxes per minutes, at what power is he working ? $(g = 10 m s^{-2})$

Find the mass of a body which has $10\, J$ of kinetic energy while moving at a speed of $3\, m/s$ .

What is the kinetic energy of a body of mass $1\,kg$ moving with a speed of $2\,m/s$ ?

State whether the following statement is true or false :
The potential energy of a body of mass $1\,kg$ kept at a height of $1\,m$ is $1\,J.$

How does the kinetic energy of a moving body depends on its
a) speed
b) mass

A weightlifter is lifting weights of mass $200\,kg$ up to a height of $2$ meters. If $g = 9.8\,m/s^2 ,$ calculate :
a) potential energy acquired by the weight
b) Work done by the weightlifter

How fast should a man of $50\,kg$ run so that his kinetic energy will be $625\,J$ ?

If a car of mass $1200 \ kg$ is moving along a highway at $120 \ km/h$ , what is the car’s kinetic energy as determined by someone standing alongside the highway?

What is the spring constant of a spring that stores $25\, J$ of elastic potential energy when compressed by $7.5 \,cm$ ?

A can of sardines is made to move along the x-axis from x = 0.25 m to x = 1.25 m by a force with a magnitude given by $F = exp(-4x^2)$ , with x in meters and F in newtons. (Here exp is the exponential function.). How much work is done on the can by the force?

An automobile with passengers has weight $16 \,400 \,N$ and is moving at $113\, km/h$ when the driver brakes, sliding to a stop. The frictional force on the wheels from the road has a magnitude of $8230\,N$ . Find the stopping distance.

A $91 \,kg$ man lying on a surface of negligible friction shoves a $68 \,g$ stone away from himself, giving it a speed of $4.0\, m/s.$ What speed does the man acquire as a result?

Calculate the kinetic energy of a body of mass $50 \,kg$ moving with a velocity of $50 \,cms^{-1}$

A $4.0 \,kg$ mess kit sliding on a frictionless surface explodes into two $2.0\, kg$ parts: $3.0\, m/s,$ due north, and $5.0\,m/s, 30^{\circ}$ north of east .What is the original speed of the mess kit?

Numerical integration. A breadbox is made to move along an x axis from x = 0.15 m to x = 1.20 m by a force with a magnitude given by $F = exp(-2x^2)$ , with x in meters and F in newtons. (Here exp is the exponential function.) How much work is done on the breadbox by the force?

A raindrop of mass $1.00\ g$ falling from a height of $1\ km$ hits the ground with a speed of $50\ m/s$ . Calculate the gain in $K.E.$ of the drop.

Find the energy possessed by an object of mass $10 \,kg$ when it is at a height of $6m$ above the ground.
Take, $g = 9.8 \,ms^{-1}$

To what height should a body of mass 5 kg be raised so that its potential energy is 490 J?
Take, $g = 9.8 \,m/s^2$

A ball is drop from a height of $10\ m$ . If the energy of the ball reduced by $40\%$ after striking the ground, how much height can the ball bounce back?