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CBSE Questions for Class 9 Physics Work And Energy Quiz 15 - MCQExams.com
CBSE
Class 9 Physics
Work And Energy
Quiz 15
A man raises 1 kg wt. to a height of 100 cm and holds it there for 30 minutes. How much work has he performed?
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$$1\times 9.8J$$
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$$1\times 9.8\times 30\times 60J$$
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$$1\times 9.8\times 30J$$
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$$1\times 9.8\times 30erg$$
Explanation
Given
$$m=1\ kg\\h=100\ cm=1\ m \\t=30\ mit$$
Work dont $$W=$$ Force $$\cdot $$ Displacement
Force apply to lift the mass = Weight of the body= $$1\times 9.8$$ N
From above equation,
$$W=1\times 9.8\times 1\ J$$
$$W=1\times 9.8 \ J$$
Option A
A skater of weight $$30 \ kg$$ has initial speed $$32m/s$$ and second one of weight $$40 \ kg$$ has $$5m/s$$. After the collision, they stick together and have a speed $$5m/s$$. Then the loss in KE is
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$$48J$$
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$$96J$$
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Zero
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None of these
Explanation
Given:
Weight of the skater 1 is $$m_1=30 \ kg$$
His initial speed $$u_1=32 \ m/s$$
Weight of the skater 2 is $$m_2=40 \ kg$$
His initial speed $$u_2=5 \ m/s$$
After collision they stick together
Both of their speed $$v_1=v_2=v=5 \ m/s$$
We know that kinetic energy $$=\dfrac{1}{2}mv^2$$
Initial kinetic energy $$=KE_{skater 1}+KE_{skater 2}$$
$$=\dfrac{1}{2}m_1u_1^2+\dfrac{1}{2}m_2u_2^2$$
$$=\dfrac{1}{2}(30)(32)^2+\dfrac{1}{2}(40)(5)^2$$
$$=\dfrac{1}{2}(30)(1024)+\dfrac{1}{2}(40)(25)$$
$$=15360+500=15860 \ J$$
Final kinetic energy
$$=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2$$
$$=\dfrac{1}{2}m_1v^2+\dfrac{1}{2}m_2v^2=\dfrac{1}{2}(m_1+m_2)v^2$$
$$=\dfrac{1}{2}(30+40)(5)^2$$
$$=\dfrac{1}{2}(70)(25)=875 \ J$$
Loss in kinetic energy $$\Delta KE=15860-875$$
$$=14985 \ J$$
The answer is none of the given options. So option D is the answer.
If a body is released from a certain height, during its fall.
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Its kinetic energy increases and potential energy decreases.
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Its potential energy increases and kinetic energy decreases.
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Both potential and kinetic energy of that body increase
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Both potential and kinetic energy of the body decrease
When the force and the displacement are in the opposite directions, the work done by the force is ..........
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$$positive$$
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$$negative$$
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$$zero$$
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$$not form this$$
Explanation
Work done by the force,
$$W=\vec F.\vec r=Frcos\theta$$. . . . . . . .(1)
If the force and the displacement are in the opposite direction, then angle between the force and the displacement is $$180^0$$
$$cos\theta=cos180^0=-1$$
from equation (1),
$$W=-Fr$$
The work done by the force is negative.
The correct option is B.
A machine delivering constant power moves a body along straight line starting from rest. The distance moved by the body in time $$t$$ is proportional to
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$$t^{3/4}$$
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$$t^{3/2}$$
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$$\sqrt t$$
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$$t$$
Explanation
Let the force applied on the body be $$F$$ and the distance travelled by it be $$s$$.
We know that power $$P=\dfrac{work}{time}$$
$$P=\dfrac{Fs}{t}=F \times \dfrac{s}{t}$$
Speed $$v=\dfrac{Distance \ 's'}{Time \ 't'}$$
So power
$$P=Fv$$
The power delivered by the machine is constant.
$$\therefore P=Fv=constant \ 'k'$$ . . . (i)
By Newton's second law $$Force \ 'F'=mass \ 'm' \times acceleration \ 'a'$$
$$\Rightarrow F=ma$$ . . . (ii)
By first equation of motion $$v=u+at$$
$$\Rightarrow v=at$$ . . . (iii)
($$\because u=0$$ as body starts from rest)
Using $$(ii)$$ and $$(iii)$$ in $$eq. (i)$$
$$Fv=k$$
$$(ma)(at)=k$$
$$a^2=\dfrac{k}{mt}$$
$$\Rightarrow a=\sqrt{\dfrac{k}{mt}}=\left( \dfrac{k}{mt}\right)^{\frac{1}{2}}$$ . . . (iv)
From second equation of motion,
$$s=ut+\dfrac{1}{2}at^2$$
$$\Rightarrow s=\dfrac{1}{2}at^2$$ . . . (v)
By substituting value of $$a$$ from $$(iv)$$ in $$(v)$$
$$s=\dfrac{1}{2} \left( \dfrac{k}{mt}\right)^{\frac{1}{2}}t^2 $$
$$s=\dfrac{1}{2} \left( \dfrac{k}{m}\right)^{\frac{1}{2}} \left( \dfrac{1}{t}\right)^\frac{1}{2} t^2 $$
$$s=\left[ \dfrac{1}{2} \left( \dfrac{k}{m}\right)^{\frac{1}{2}} \right] (t)^{-1/2} (t)^2 $$
The quantity in the square brackets $$[ \ ]$$ is constant so
$$s \propto (t)^{-1/2} (t)^2 $$
$$s \propto t^{\frac{-1}{2}+2}$$
$$s \propto t^{3/2}$$
So option $$(B)$$ is the answer.
Two bodies A and B have masses $$20\ kg$$ and $$5\ kg$$ respectively . Each one is acted upon by a force of $$4\ kg.-wt$$. If they acquire the same kinetic energy in times $$t_{ A }\ and\ t_{ B }$$, then the ratio $$\dfrac { t_{ A } }{ t_{ B } } $$:
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$$\dfrac { 1 }{ 2 } $$
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$$2$$
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$$\dfrac { 2 }{ 5 } $$
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$$\dfrac { 5 }{ 6 } $$
A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time $$t$$ is proportional to
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$$
t^{1 / 2}
$$
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$$
t^{3/ 4}
$$
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$$
t^{3/ 2}
$$
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$$
t^{2}
$$
If water is flowing in a pipe at a height 4m from the ground then its potential
energy per unit volume is (Reference is taken at ground, $$ g=10 \mathrm{m} / \mathrm{s}^{2} ) $$
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$$
20 \mathrm{kJ} / \mathrm{m}^{3}
$$
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$$
10 \mathrm{kJ} / \mathrm{m}^{3}
$$
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$$
40 \mathrm{kJ} / \mathrm{m}^{3}
$$
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$$
30 \mathrm{kJ} / \mathrm{m}^{3}
$$
A ball of mass 2 kg hits a floor with a speed of 4 m/s at an angle of $$ 60 ^o $$ with the normal.
If (e=1/2); then the change in the kinetic energy of the ball is
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-6J
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-12J
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-9J
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-3J
From an automatic gun a man fires 360 bullets per minute with a speed of 360 km/hr. If each bullet weighs 20g, the power of the gun is
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600 W
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300 W
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150 W
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75 W
Explanation
Power of gun = $$ \dfrac { \text{Total K.E. of fired bullet} }{ \text{time }} $$
Total KInetic Energy $$=n\times \dfrac{1}{2}mv^2$$
$$P = \dfrac {n \times \dfrac {1}{2}mv^2 }{t} = \dfrac {360}{60} \times \dfrac { 1}{2} \times 2 \times 10^{-2} \times ( 100)^2 = 600 W $$
A body of mass 200 g moving on a test has final K.E. of 50 J after travelling a distance of 10 cm. Assuming 90% loss of energy due to friction. The initial speed of the body is
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10 $$ms^{-1}$$
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$$10\sqrt{50}ms^{-1}$$
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$$5\sqrt{50}ms^{-1}$$
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$$50\sqrt{50}ms^{-1}$$
A block of mass $$20 kg$$ is being brought down by a chain. If block acquires a speed of $$2 m/s$$ in dropping down $$2 m$$. Find work done by the chain during the process. ($$g= 10m/s^{2}$$ )
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$$-360 J$$
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$$400 J$$
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$$360 J$$
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$$-280 J$$
Potential energy v/s displacement curve for one dimensional conservative field is shown. Force at A and B is respectively -
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Positive, Positive
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Positive, Negative
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Negative, Positive
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Negative, Negative
A lift weighing $$250 Kg$$ is to be lifted up at a constant velocity of $$0.20 m/s$$. What would be the minimum horsepower of the motor to be used?
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$$1.3 hp$$
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$$0.65 hp$$
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$$1.5 hp$$
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$$0.75 hp$$
The potential energy of a particle of mass 1 kg free to move along the x-axis is given by U($$x$$) =($$\frac{x^2}{2}$$-$$x$$) joule. If the total mechanical energy of the particle is 2J, then find the maximum speed of the particle. (Assuming only conservative force acts on the particle)
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$$\sqrt5$$ m/s
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$$\sqrt7$$ m/s
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$$\sqrt2$$ m/s
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$$\sqrt6$$ m/s
A ball is allowed to fall from a height of 10m. if there is $$40\%$$ loss of energy due is impact then after one impact ball will go up to
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10m
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8m
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4m
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6m
When a body falls freely towards the earth, then its total energy:
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Increases
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Decreases
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Remains constant
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First increases and then decreases
Explanation
When a body falls freely towards the earth, the sum of the potential energy and kinetic energy of the object would be the same at all points. That is,
$$potential\ energy + kinetic\ energy = constant$$
It obeys the law of conservation of energy. Due to this, its total energy remains constant.
A single conservative force acts on a $$1kg$$ particle that moves along x-axis.Potential energy $$U(x)$$ is given by $$U(x)=20+{(x - 3)^2},$$ where x is in m.At $$x=0$$.particle has kinetic energy of $$20j$$ .
Find value of $$x$$ at which body is in equilibrium
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$$0\,m$$
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$$-2\,m$$
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$$3\,m$$
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$$1\,m$$
When a man increases his speed by $$2 m/s$$, he finds that his kinetic energy is doubled, the original speed of the man is
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$$2(\sqrt {2} - 1) m/s$$
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$$2(\sqrt {2} + 1) m/s$$
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$$4.5\ m/s$$
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None of these
Explanation
According to question,
Initial kinetic energy, $$ E_1 = \dfrac{1}{2} mv^2 $$ ....(i)
Then, he increases his speed by $$2 \ m/s$$
2m/s
and K.E. is doubled.
Final kinetic energy $$ E_2=2E_1= \dfrac{1}{2}m(v+2)^2 $$ ....(ii)
Let multiply both sides of (i) by 2,
$$2E_1=mv^2$$ . . . (iii)
Equating (ii) and (iii),
$$mv^2=\dfrac{1}{2}m(v+2)^2$$
$$2v^2=v^2+4v+4$$
Solving this using quadratic formula,
$$ v=(2+2\sqrt{2})\ m/s $$
$$= 2(1+\sqrt{2})\ m/s$$
We reject the other value of $$v$$ as it must be positive because we considered $$v$$ to be speed which is a non-negative quantity or in this case, positive.
Option B is correct.
A constant force of $$5N$$ is applied on a block of mass $$20\ kg$$ for a distance of $$2.0\ m$$, the kinetic energy acquired by the block is
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$$20\ J$$
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$$15\ J$$
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$$10\ J$$
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$$5\ J$$
Explanation
Acceleration $$a = \dfrac {F}{m} =\dfrac {5}{20} = \dfrac {1}{4} m/s^{2}$$
$$ u =0 $$
$$v = \sqrt {(2as)} = \sqrt {\left (\dfrac {2\times 2}{4}\right )} = 1\ m/s$$
$$KE = \dfrac {1}{2} mv^{2} = \dfrac {1}{2} \times 20\times (1)^{2} = 10\ J$$
(or)
Work done, $$W = Fs$$ $$= 5\times 2 = 10\ J$$.
A body at rest can have
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Kinetic Energy
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momentum
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both momentum and kinetic Energy
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potential Energy
Explanation
A body at rest can have potential energy because PE is due to position of object but it does not have kinetic energy because KE is due to motion but object is at rest so no KE will be present.Momentum is also not present because momentum is due to velocity and velocity is zero in this case.
For example an object at a roof at rest have no KE but it has gravitational potential energy due to it's height above ground given by $$mgh$$.
Choose whether true or false:-
As the object reaches the ground, its kinetic energy is the highest.
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True
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False
Explanation
When the object is about to reach the ground, $$h = 0$$ and $$v$$ will be the highest.Because it accelerates due to gravity.
Since $$KE=\dfrac{mv^2}{2}$$ and $$PE=mgh$$
Therefore, the kinetic energy would be the largest and potential energy the least at ground. Thus, the given statement is true.
The graph below show the force acting on a particle moves along the positive $$x-$$axis from the origin to $$x=x_1$$. The force if parallel to the $$x-$$axis and is conservative. The maximum magnitude $$F_1$$ has the same values for all graphs. Rank the solutions according to the change in the potential energy associated with the force. least (or most negative) to greatest (or most positive)
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$$2, 1, 3$$
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$$1, 3, 2$$
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$$2, 3, 1$$
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$$3, 2, 1$$
A ball rolling on the ground possesses
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kinetic energy
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potential energy
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no energy
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heat energy
Explanation
A moving car
possesses
mechanical energy due to its motion (kinetic energy). A moving baseball
possesses
mechanical energy due to both its high speed (kinetic energy) and its vertical position above the
ground
(gravitational potential energy).
A body is falling a height $$h$$. After it has fallen a height $$\dfrac{h}{2}$$. it will possess
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only potential energy
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only kinetic energy
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half potential energy and half kinetic energy
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more kinetic and less potential energy
Explanation
$$(c)$$ half potential and half kinetic energy
Explanation: When the body is height $$h$$; its potential energy is at maximum and kinetic energy is zero. When the body hits the ground, its potential energy becomes zero and kinetic energy is at maximum. At mid-way i.e., half the height; its potential energy becomes half of the maximum potential energy and same happens to the kinetic energy.
The height of the water dam in the hydroelectric power station is $$20$$ m. How much water. in $$1$$ second, should fall on turbine, so that $$1$$ MW power is generated ?
($$g = 10m/s^2$$)
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$$5000\,kg$$
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$$10000\,kg$$
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$$500\,kg$$
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$$7,500\,kg$$
Explanation
We know that,
$$Power(P)=\Bigg(\dfrac{\text{Work Done}}{\text{Time taken}}\Bigg)$$
$$\text{work done} = mgh$$
$$\therefore P=\dfrac{mgh}{t}$$
$$g = 10\ ms^{-2}$$
$$h = 20\ m$$
$$time = 1\ s$$
$$\Rightarrow 1 \times 10^6=\dfrac{m(10)(20)}{1}$$
$$\Rightarrow m=5000\,kg$$
The engine of a car of mass $$1500$$ kg. keeps the car moving with constant velocity $$v=5\ m/s$$
. If frictional force is $$1000$$ N, the power of the engine is...
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$$5\,kW$$
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$$7.5\,kW$$
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$$15\,kW$$
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$$75\,kW$$
Explanation
Given,
Frictional force, $$F=1000\,N$$
To keep the car running at constant speed the engine must apply same amount of force as friction.
$$Power=\dfrac{\text{Work done}}{\text{Time taken}}=\dfrac{F \times S}{t}$$
$$\Rightarrow P=F \times \Bigg(\dfrac{S}{t}\Bigg)=Fv$$
$$\Rightarrow P=1000 \times 5=5000\,W=5\,kW$$
Four alternatives are given to each of the following incomplete statements/questions. Choose the right answer:
Which of the following object has higher potential energy?
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mass $$ = 10\,kg$$ , $$ g = 9.8\, ms^{-2} , h = 10\,m $$
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mass $$ = 5\,kg , g = 9.8\, ms^{-2} , h = 12\,m $$
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mass $$ = 8\,kg , g = 9.8\,ms^{-2} , h = 100\,m $$
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mass $$ = 6\,kg , g = 9.8\,ms^{-2} , h = 20\,m $$
Explanation
The potential energy is given as-
$$PE = mgH$$
$$m$$; mass of the object
$$g = 9.8/ ms^{-2}$$
$$h$$; height gained from ground
For option A:
$$m = 10\ kg,\ h = 10\ m$$
$$PE = 10 \times 9.8 \times 10$$
$$\Rightarrow PE = 980\ J$$
For option B:
$$m = 5\ kg,\ h = 12\ m$$
$$PE = 5 \times 9.8 \times 12$$
$$\Rightarrow PE = 588\ J$$
For option C:
$$m = 8\ kg,\ h = 100\ m$$
$$PE = 8 \times 9.8 \times 100$$
$$\Rightarrow PE = 7840\ J$$
For option D:
$$m = 6\ kg,\ h = 20\ m$$
$$PE = 6 \times 9.8 \times 20$$
$$\Rightarrow PE = 1176\ J$$
For option C potential energy is highest so option C is the correct answer.
Work done is equal to?
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Change in kinetic energy
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Kinetic energy
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Acceleration
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None of these
Explanation
The work done is equal to the change in the kinetic energy of an object, hence option $$A$$ is correct.
Is the work required to be done by an external force on an object on a frictionless, horizontal surface to accelerate it from a speed $$v$$ to a speed $$2v$$.
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equal to the work required to accelerate the object from $$v = 0$$ to $$v$$,
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twice the work required to accelerate the object from $$v = 0$$ to $$v$$
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three times the work required to accelerate the object from $$v = 0$$ to $$v$$,
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four times the work required to accelerate the object from $$0$$ to $$v$$
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not known without knowledge of the acceleration
Explanation
Answer (c). The net work needed to accelerate the object from $$v = 0$$ to $$v$$ is
$$W_{1}=KE_{1f}-KE_{1i}=\dfrac{1}{2}mv^{2}-\dfrac{1}{2}m(0)^{2}=\dfrac{1}{2}mv^{2}$$
The work required to accelerate the object from speed $$v$$ to speed $$2v$$ is
$$W_{2}=KE_{2f}-KE_{2i}=\dfrac{1}{2}m(2v)^{2}-\dfrac{1}{2}mv^{2}$$
$$=\dfrac{1}{2}m(4v^{2}-v^{2})=3\left ( \dfrac{1}{2}mv^{2} \right )=3W_{1}$$
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