Work, Energy And Power - Class 11 Medical Physics - Extra Questions

If we rub our palms vigorously or rubbing a metal strip of an iron nail on a stone several time that is produced. there are the examples to illustrate that heat can be generated from the energy of motion

An electrolytic cell is an electrochemical cell which converts electrical energy to chemical energy by undergoing a redox reaction.

Moving Car, as it is the only one having kinetic energy. 

Various forms of energy are light, heat, sound, electrical, nuclear, chemical, etc. Although there are many specific types of energy, the two major forms are Kinetic Energy and Potential Energy.

When an electric current flows through the fan, the fan starts to rotate.
So electric energy gets converted into mechanical energy in a celing fan.

In an electric iron, electric current flowing through the wire heats up the wire.
So electric energy gets converted into heat energy in an electric iron.

Potential energy is converted completely into kinetic energy 

(i) In a loud speaker, electrical energy is converted to sound energy.

1.An electric fan converts electrical energy into Mechanical energy.
2.Coking gas converts chemical energy into heat energy.
3.The energy possessed by a spring is potential energy.
4.The energy possessed by a body due to its position is called potential energy.
5.The energy possessed by a body due to its motion is called kinetic energy.

compressed spring=potential energy

The energy stored on a capacitor can be expressed in terms of the work done by the battery. Voltage represents energy per unit charge, so the work to move a charge element dq from the negative plate to the positive plate is equal to$$Vdq$$ , where $$V$$ is the voltage on the capacitor. The voltage $$V$$ is proportional to the amount of charge which is already on the capacitor.

Element of energy stored: $$dE=dW=Vdq$$

Hence, relation is $$E=W$$ 

Let us consider both blocks and springs as the physical system. The center of mass of the system moves with an acceleration $$a=\cfrac { F }{ { m }_{ 1 }+{ m }_{ 2 } } $$ towards right. Let us work in the frame of center of mass. As this frame is a non-inertial frame (accelerated with respect to the ground) we have to apply a pseudo force $${ m }_{ 1 }a$$ on the block $${ m }_{ 1 }$$ and $${ m }_{ 2 }a$$ towards left on the block $${ m }_{ 2 }$$

Let us consider both blocks and spring as the physical system. The centre of mass of the system moves with acceleration $$a=\dfrac{F}{m_1+m_2}$$ towards right Let us work in the frame of centre of mass. As this frame is a non-inertial frame (accelerated with respect to the ground) we have to apply a pseudo force m1a towards left on the block $$m_1$$ and $$m_2$$ a towards left on the block $$m_2$$. 

$$L_{i} =L_{f} $$ (about bottommost point)
$$\displaystyle \therefore I\omega _{0} =2 [I \omega + mRv] $$ or
$$\displaystyle \left( \frac{1}{2} mR^{2} \right) \omega_{0} =2 \left[ \frac{1}{2} mR^{2} \omega +mR (\omega R) \right] $$
$$\displaystyle \therefore \omega =\frac{ \omega_{0}}{6}$$ and
$$\displaystyle v=\omega R =\frac{\omega_{0}R}{6}$$

The right matches are given below.
$$A\rightarrow 2$$,
$$B\rightarrow 1$$,
$$C\rightarrow 4$$,
$$D\rightarrow 3$$.

The different forms of energy other than mechanical energy are:

(a)Microphone converts sound energy into electrical energy.

Work done=Area enclosed from A to B

Given that,

Mass of liquid $$m=1.00g=.001\,kg$$

Height $$h=1km=1000\,m$$

Speed $$v=50\,m/s$$

We know that,

Work done by gravitational force is loss of potential energy is

  $$ P.E=mgh $$

 $$ P.E=0.001\times 10\times 1000 $$

 $$ P.E=10\,J $$

Now, gain in kinetic energy

  $$ K.E=\dfrac{1}{2}m{{v}^{2}} $$

 $$ K.E=\dfrac{1}{2}\times 0.001\times 50\times 50 $$

 $$ K.E=1.25\,J $$

Now,

Loss of potential energy = gain in kinetic energy +work done by resistive force

  $$ W=10-1.25 $$

 $$ W=8.75\,J $$

Hence, the work done by resistive force is $$8.75\ J$$

 

Lifiting a weight from the ground and putting it on a shelf is a good ex. Of work . the force is equal to the weight of the object, and the distance is equal to the weight of shelf ( W =Fxd).

$$W - E$$ principle$$ \Rightarrow $$

The change is K.E equal to the work dons on the object

Initial velocity and final velocity is  $${{V}_{initial}}=1000\,{{s}^{-1}}\,\,and\,{{V}_{final}}=500\,m{{s}^{-1}}$$

Loss in kinetic energy $$\Delta K.E=\dfrac{1}{2}M\left( V_{final}^{2}-V_{initial}^{2} \right)=\dfrac{1}{2}0.01\times \left( {{500}^{2}}-{{1000}^{2}} \right)=-3.75\,kJ$$

Work by resistive force, $$W={{F}_{r}}d\,\,\,\,\,(where,\,F=average\,resistive\,force,\,\,\,d=displacement\,)$$

From conservation of energy

Decrease in kinetic Energy = Work done by resistive force

$$ {{F}_{r}}d=-3.75\,kJ $$

$$ {{F}_{r}}=\dfrac{-3.750}{d}=\dfrac{-3.75}{0.025}=-150\,kN $$

Resistive force is $$150\,kN$$ , opposite to the direction of motion.

The total mechanical energy is the total potential energy it possesses before its fall. 
$$E = mgh = 10 \times 10 \times 5 = 500 J$$ 
It is equal to the maximum potential energy $$E = mgh =10 \times 10 \times 5 = 500 J.$$

We know that $$v^2=u^ 2+2as$$ 

$$1.$$Collision means two objects coming into contact with each other for a very short period. In other words, collision is a reciprocative interaction between two masses for a very short interval wherein the momentum and energy of the colliding masses changes. While playing carroms, you might have noticed the effect of a striker on coins when they both collide.

Let $$h$$ ne the minimum height using conservation of mechanical energy
$$U_i + K_i = U_f + K_f$$
$$U_i = mgh$$
$$K_j = Oj$$ [starts from rest]
$$U_f = mg(5m) \,\,K_f = \dfrac{1}{2} mV_T^2 \,\,V_T=$$ speed of the top.

So $$mgh \,to = mg(6m) + \dfrac{1}{2}mV_T^2$$

$$V_T^2 = 2g(n - 6)$$   ....(1)
Now using Newton's $$2^{nd}$$ law at the top of the loop.

$$\Sigma f_y = \dfrac{mV_T^2}{R} \,\,R = 3m$$ (radius of the path)

$$\Sigma f_y = Mg + N = \dfrac{mV_T^2}{R}$$

So for minimum $$h \Longrightarrow N = 0$$
$$Mg = \dfrac{mV_T^2}{R} \Longrightarrow V_T^2 = Rg$$  ...(2)

So from (1) and (2) $$R = 3m$$
$$Rg = 2g (h - 6) \Longrightarrow 3g = 2g (h - 6)$$
$$n = 6 + 1.5 = 7.5m$$

Elastic Collision: In the elastic collision total momentum, the total energy and the total kinetic energy are conserved. However, the total mechanical energy is not converted into any other energy form as the forces involved in the short interaction are conserved in nature. Consider from the above graph two masses, m₁ and  m₂ moving with speed u₁ and u₂. The speed after the collision of these masses is v₁ and v₂

m1u1 + m2u2 = m1v1 + m2v2

Inelastic Collision: In the inelastic collision, the objects stick to each other or move in the same direction. The total kinetic energy in this form of collision is not conserved but the total momentum and energy are conserved. During this kind of collision, the energy is transformed into other energy forms like heat and light. Since during the phenomenon the two masses follow the law of conservation of momentum and move in the same direction 

m1u1 + m2u2 = (m1+ m2)v

From the equation,

 According to wording of the problem suggests that it is only necessary to examine the contribution from the rope which would be the $$“W_a”$$

Yes, mechanical energy comprises both potential energy and kinetic energy. Momentum is zero which means velocity is zero. Hence their energy but the object may possess potential energy.

No, since mechanical energy is zero, there is no potential energy and no kinetic energy. Kinetic energy being zero, velocity is zero. Hence, there will be no momentum.

According to work-energy theorem,
Change in KE is equal to work done by all the forces acting on the body. Let us assume that only one force (retarding force) is acting on the body, therefore,
KE of the body = Work done by retarding force KE of the body = Retarding force x Displacement
As KE of the bodies and retarding forces applied on them are same, therefore, both bodies will travel equal distances before coming to rest.

The various forms include potential energy, kinetic energy, heat energy, chemical energy, electrical energy and light energy.

Yes, when force is normal to displacement, no transfer of energy lakes place. 

(A) Energy conversation	(B) Instruments
(i) Electrical energy to sound energy	(a) Speaker
(ii) Electrical energy to heat energy	(b) Electric heater
(iii) Mechanical energy to electrical energy	(c) Dynamo (generators)
(iv) Light energy to electrical energy	(d) Solar cell

Energy is the capacity of a body to do work. Different forms of energy are: Potential energy, kinetic energy, heat energy, chemical energy, electrical energy, 

When a wire is stretched, the internal reactive force comes in to play which opposes this action. Hence to overcome this opposing force, work is done. This work is stored in the form of elastic energy in the wire.

Consider path $$(a)$$
$$a = g \sin \theta$$
velocity on reaching ground is
$$V^{2} = u^{2} + zal$$
$$V^{2} = 2 g\sin \theta l$$
where $$l =\frac{h}{\sin\theta}$$
$$v = \sqrt{2g\sin\theta \frac{h}{\sin\theta}}=\sqrt{2gh}$$
work done = change in KE
$$= \frac{1}{2}mv^{2} – 0 = \frac{1}{2}* 2gh * m = mgh$$
Consider Path $$(b)$$
$$a = g$$
velocity on reaching ground is
$$v^{2} = u^{2} + 2gh$$
$$v^{2} = 2gh$$
$$v = \sqrt{2gh}$$
work done = change in KE.
$$\frac{1}{2}mv^{2} = mgh$$.

Mechanical energy 
Heat energy 
Electrical energy 
Chemical energy 
Light energy 
Nuclear energy 

Generator : Mechanical energy $$\rightarrow $$ electrical energy
Cell (while discharging) : Chemical energy $$\rightarrow $$ electrical energy
Battery (While discharging) : Chemical energy $$\rightarrow $$ electrical energy
Solar cell : Solar energy $$\rightarrow $$electrical energy

Potential energy will decrease and kinetic energy will increase. 

In given situation only potential energy is available.

Obviously the elongation in the cord, $$\Delta l=l_0(\sec{\theta}-1)$$, at the moment the sliding first starts and at the moment horizontal projection of spring force equals the limiting friction.
So, $$\kappa_1\Delta l\sin{\theta}=kN$$ (1)
(where $$\kappa_1$$ is the elastic constant).
From Newton's law in projection form along vertical direction
$$\kappa_1\Delta l\cos{\theta}+N=mg$$
or, $$N=mg-\kappa_1\Delta l\cos{\theta}$$ (2)
From (1) and (2),
$$\kappa_1\Delta l\sin{\theta}=k(mg-\kappa_1\Delta l\cos{\theta})$$

or, $$\displaystyle\kappa_1=\frac{kmg}{\Delta l\sin{\theta}+k\Delta l\cos{\theta}}$$

From the equation of the increment of mechanical energy, $$\Delta U+\Delta T=A_{fr}$$

or, $$\displaystyle\left(\frac{1}{2}\kappa_1\Delta l^2\right)=A_{fr}$$

or, $$\displaystyle\frac{kmg\Delta l^2}{2\Delta l(\sin{\theta}+k\cos{\theta})}=A_{fr}$$

Thus $$\displaystyle A_{fr}=\frac{kmgl_0(\sec{\theta}-1)}{2(\sin{\theta}-k\cos{\theta})}=0.09\:J$$ (on substitution)

Since, the collision is head on, the particle 1 will continue moving along the same line as before the collision, but there will be a change in the magnitude of its velocity vector. Let it starts moving with velocity $$v_1$$ and particle 2 with $$v_2$$ after collision, then from the conservation of momentum
$$mu=mv_1+mv_2$$ or, $$u=v_1+v_2$$.....(1)
And from the condition, given,

$$\displaystyle\eta=\frac{\displaystyle\frac{1}{2}mu^2-\left(\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\right)}{\displaystyle\frac{1}{2}mu^2}=1-\frac{v_1^2+v_2^2}{u^2}$$

or, $$v_1^2+v_2^2=(1-\eta)u^2$$.... (2)
From (1) and (2),
$$v_1^2+{(u-v_1)}^2=(1-\eta)u^2$$
or, $$v_1^2+u^2-2uv_1+v_1^2=(1-\eta)u^2$$
or, $$2v_1^2-2v_1u+\eta u^2=0$$

So, $$\displaystyle v_1=2u\pm\frac{\sqrt{4u^2-8\eta u^2}}{4}$$

$$\displaystyle\frac{1}{2}\left[u\pm\sqrt{u^2-2\eta u^2}\right]=\frac{1}{2}u(1\pm\sqrt{1-2\eta})$$

Positive sign gives the velocity of the 2nd particle which lies ahead. The negative sign is correct for $$v_1$$.
So, $$\displaystyle v_1=\frac{1}{2}u(1-\sqrt{1-2\eta})=5\:m/s$$ will continue moving in the same direction.
Note that $$v_1=0$$ if $$\eta=0$$ as it must.

Since the collision is head on elastic between two identical bodies, so the speeds of the two bodies are interchanged. Thus, the second body will start moving with velocity $${ V }_{ 1 }$$ after the collision.
$$Applying\quad conservation\quad ofangular\quad momentum:\\ m{ V }_{ 1 }R=mVR+I\omega =mVR+\frac { 2 }{ 5 } m{ R }^{ 2 }\omega \\ and\quad since\quad the\quad ball\quad is\quad rolling\quad after\quad long\quad time\quad \omega =\frac { V }{ R } \\ \Rightarrow m{ V }_{ 1 }R=mVR+\frac { 2 }{ 5 } mV{ R }\Rightarrow V=\frac { 5 }{ 7 } { V }_{ 1 }$$

$$T=40N={ F }_{ s }\\ kr\quad =\quad { F }_{ s }\\ kx\quad =\quad 40\quad \Longrightarrow k\quad =\quad \frac { 40 }{ \left( 0.25-0.2 \right)  } =\frac { 40 }{ 0.05 } =\quad 800N/m\\ s=\quad ut+1/2a{ t }^{ 2 }\\ 5=\quad 1/2\times 10{ t }^{ 2 }\\ =t\quad =\quad 1\quad sec\\ \\ \triangle \left( \frac { 1 }{ 2 } k{ x }^{ 2 } \right) =\quad \triangle \left( \frac { 1 }{ 2 } m{ v }^{ 2 } \right) \\ =\quad 800\times { 0.25 }^{ 2 }-800\times { .2 }^{ 2 }\quad =\quad 8{ v }^{ 2 }\\ v=1.5m/s$$

(a )In the instant right after the impulse, the particle can be safely assumed to be in circular motion. Hence, it will require centripetal force, which will be provided by the resultant of $$mg$$ and $$N$$.

Noting that the vertical displacement is $$10.0\, m – 1.50\, m = 8.50\, m$$ 

Yes. Here, energy will transfer from foot of the player to football. So, we can say work is done on the ball. 

Figure shows the top view of the trajectory of the ball. Since the collision of the ball with the wall and the bottom of the well are elastic, the magnitude of the horizontal component of the ball velocity remains unchanged and equal to $$\nu$$ The horizontal distances between points of two successive collisions are $$AA_1 =A_1 A_2= A_2 A_3=.... 2 r \cos \alpha$$. The time between two successive collisions of the ball with the wall of the well is $$t_1=2 r \cos \alpha/ \nu$$
 The vertical component of the ball velocity does not change upon a collision with the wall and reverse its sign upon a collision with the bottom.
The magnitude of the vertical velocity component for the first impact against the bottom is $$\sqrt{2 g H}$$ and the time of motion from the top to the bottom of the well is $$t_2=\sqrt{2H / g}$$
Figure shows vertical plane development of the polyhedron$$A_1 A_2 A_3 ...$$ ON this development the segments of the trajectory of the ball inside the well are parabolas (complete parabola are the segments of the trajectory between successive impact against the bottom). The ball can "get out" of the well in the moment of the maximum ascent along the parabola coincides with the moment of an impact against the wall (i.e. at the moment of maximum ascent, the ball is at point $$A_n$$ of the well edge). The time $$t_1$$ is connected to the time $$t_2$$ through the following relation :
where $$n$$ and $$k$$ are integral and mutually prime numbers. Substituting the values of $$t_1$$ and $$t_2$$ we obtain the relation between $$\nu, H, r$$ and $$\alpha$$ for which the ball can "get out" of the well:
$$\dfrac{nr \cos \alpha}{\nu}= k \sqrt{\dfrac{2H}{g}}$$