# Work and Energy : NCERT Exemplar Q.1 ¬† ¬† A rocket is moving up with a velocity v. If the velocity of this rocket is suddenly tripled, what will be the ratio of two kinetic energies?
Sol. ¬† ¬† ¬†
¬†Since K.E $\propto \,v^2$ , if velocity is tripled, the K.E will become 9 time. Ratio will be 1:9.

Q.2 ¬† ¬† Avinash can run with a speed of 8 $m{s^{ - 1}}$ against the frictional force of 10 N, and Kapil can move with a speed of 3 m${s^{ - 1}}$ against the frictional force of 25 N. Who is more powerful and why?
Sol. ¬† ¬† ¬† Power = F √ó v
¬† ¬† ¬† ¬† ¬† ¬† ¬†Power of Kapil = 10 √ó 8 = 80 W.
¬† ¬† ¬† ¬† ¬† ¬† ¬†Power of Avinash = 25 √ó 3 = 75 W.
¬† ¬† ¬† ¬† ¬† ¬† ¬† ¬†Kapil is more powerful.

Q.3 ¬† ¬† Can any object have mechanical energy even if its momentum is zero? Explain.
Sol.¬† ¬† ¬† ¬†Momentum of a body is zero,when its velocity is zero.The body may be having P.E at this position,Thus mechanical energy need not be zero,if momentum is zero.

Q.4 ¬† ¬† Can any object have momentum even if its mechanical energy is zero? Explain.
Sol.¬† ¬† ¬† ¬†No. If a body has momentum $\to$ velocity is not equal to zero. Therefore the body will have mechanical energy in the form of its kinetic energy.

Q.5 ¬† ¬† The power of a motor pump is 2 KW. How much water per minute the pump can raise to a height of 10 m?(Given g = 10 m${s^{ - 2}}$)
Sol. ¬† ¬† ¬† Work done = mgh = Power √ó Time
¬† ¬† ¬† ¬† ¬† ¬† ¬† Mass of water = ${{P \times t} \over {g \times h}} = {{2000 \times 60} \over {10 \times 10}} = 1200kg.$

Q.6 ¬† ¬† The weight of a person on a planet A is about half that on the earth. He can jump up to 0.4 m height on the surface of the earth. How high he can jump on the¬† planet A?
Sol. ¬† ¬†¬†¬†The weight of the person is half an object A.It means gravity of A planet = ${{g{\mkern 1mu} (earth)} \over 2}$. On earth the man can jump up to 0.4 m.It means he has energy = m √ó (0.4) √ó g
¬† ¬† ¬† ¬† ¬† ¬† ¬†Since, his energy is same; on the planet it will be¬†$m\left( {{g \over 2}} \right)h$
¬† ¬† ¬† ¬† ¬† ¬† ¬†Therefore,  mg √ó 0.4 = ${{mg} \over 2} \times h$
¬† ¬† ¬† ¬† ¬† ¬† ¬†Therefore,  h= 0.8 m on the planet.

Q.7 ¬† ¬† The velocity of a body moving in a straight line is increased by applying a constant force F, for some distance in the direction of the motion.
Prove that the increase in the kinetic energy of the body is equal to the work done by the force on the body.
Sol. ¬† ¬† ¬†¬†Let a constant force (F) be applied on a body of mass m , such that it moves a distance d.
¬† ¬† ¬† ¬† ¬† ¬† ¬†Than work done = F √ó d = mad...........(i)
¬† ¬† ¬† ¬† ¬† ¬† ¬†Now by Newton's 3rd eq. of motion.
¬† ¬† ¬† ¬† ¬† ¬† ¬†$v^2- u^2= 2ad$
¬† ¬† ¬† ¬† ¬† ¬† ¬†$\,ad = \,{{v^2- u^2 } \over 2}$ ..................(ii)
¬† ¬† ¬† ¬† ¬† ¬† ¬†Substituting the value of ad from (ii) in (i), we get ,
¬† ¬† ¬† ¬† ¬† ¬† ¬†Work done = $m = \,{{(v^2- u^2 )} \over 2} = {1 \over 2}mv^2- {1 \over 2}mu^2$
¬† ¬† ¬† ¬† ¬† ¬† ¬†= Final K.E - Initial K.E.
¬† ¬† ¬† ¬† ¬† ¬† ¬†Thus work done = change in kinetic energy

Q.8 ¬† ¬† Is it possible that an object is in the state of accelerated motion due to external force acting on it, but no work is being done by the force? Explain it with an example.
Sol. ¬† ¬† ¬†¬†A body in uniform circular motion is an example of accelerated motion. Consider the motion of earth around the sun.¬† The earth is constantly moving it a¬† direction perpendicular to the gravitational pull of sun,
hence work do by the gravitational force is zero. Thus work done can be zero for an accelerated body.

Q.9 ¬† ¬† A ball is dropped from a height of 10 in. If the energy of the ball reduces by 40% after striking the ground, how much high can the ball bounce back? (g = $10\,m/{s^{ - 2}}$)
Sol. ¬† ¬† ¬†¬†The ball has energy = mgh
¬† ¬† ¬† ¬† ¬† ¬† ¬† = mg 10.
¬† ¬† ¬† ¬† ¬† ¬† ¬†After striking the ground, energy left ${60 \over {100}}$ √ó mg 10. = 6 mg.
¬† ¬† ¬† ¬† ¬† ¬† ¬†If this energy is converted to P.E = mgh
¬† ¬† ¬† ¬† ¬† ¬† ¬† mgh = 6 mg
¬† ¬† ¬† ¬† ¬† ¬† ¬† h = 6 m

Q.10 ¬† ¬† If an electric iron of 1200 W is used for 30 minutes everyday, find electric energy consumed in the month of April.
Sol.¬† ¬† ¬† ¬† ¬†Energy consumed = P √ó t
¬† ¬† ¬† ¬† ¬† ¬† ¬† ¬† = 1200 √ó ${1 \over 2}hr$ √ó 30 days.
¬† ¬† ¬† ¬† ¬† ¬† ¬† ¬† = 1800 Wh = 18 KWh = 18 Units

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