Class 11 Physics

Work Energy Power

NCERT Solution

Part 1

Question 1: The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:

  1. Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
  2. Work done by gravitational force in the above case.
  3. Work done by friction on a body sliding down an inclined plane.
  4. Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.
  5. Work done by the resistive force of air on a vibrating pendulum in bringing it to rest.

Answer: (a) Positive, (b) Negative, (c) Negative, (d) Positive, (e) Negative

Question 2: A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the

(a) Work done by the applied force in 10 s

Answer: Given, m = 2 kg, μ = 0.1, F = 7 N, u = 0, t = 10 s

We know F = ma

So, `a=F/m=7/2=3.5` m s-2

Frictional force can be calculated as follows:


= 0.1 × 2 &imes; -9.8 = -1.96 N

Acceleration produced by frictional force can be calculated as follows:

`a_f=-(1.96)/2=-0.98` m s-2

Total acceleration of the body = a + af

= 3.5 – 0.98 = 2.52 m s-2

Now, distance can be calculated as follows:


`=1/2xx2.52xx10^2=1.26xx100=126` m

So, work done by applied force = F × s

= 7 × 126 = 882 J

(b) Work done by friction in 10 s

Answer: Wf = -1.96 × 126 = -246.96 J

(c) Work done by the net force on the body in 10 s

Answer: Net force = F + Ff

= 7 – 1.96 = 5.04 N

Work done by net force

Wnet = 5.04 × 126 = 635 J

(d) Change in kinetic energy of the body in 10 s

Answer: We need to find final speed which can be calculated as follows:


= 2.52 × 10 = 25.2 m s-1

Change in kinetic energy


`=1/2xx2xx25.2^2=635.04` J

Question 3: Given figures are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.

Answer: Total energy of a system is given by following equation:

E = PE + KE

So, KE = E – PE

This means that kinetic energy of a system can never be negative. So, particle will not exist when KE becomes negative. This also means that potential energy should always be less than total energy.

potential energy function graph

In the 1st figure, if x > a then PE > E

So, particle cannot exist in this region.

potential energy function graph

In 2nd figure, value of x is always above the level of E. In other words, for all values of x, PE > E.

So, particle cannot exist in this region.

potential energy function graph

In 3rd figure, for x > a and x < b, potential energy = -V1. For other values of x, potential energy is more than total energy. Particle can exist between a and b and the minimum energy required is –V1.

potential energy function graph

In 4th figure, potential energy is less than total energy between `-a/2` and `a/2`. Particle can exist in this region, and minimum energy is –V1

Question 4: The potential energy function for a particle executing linear simple harmonic motion is given by `V(x)=(kx^2)/2` where k is the force constant of the oscillator. For k = 0.5 N m-1, the graph of V(x) versus x is shown in given figure. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ±2 m.

potential energy function graph

Answer: Given, k = 0.5 N m-1, total energy E = 1 J

We know KE = `1/2mv^2`

E = V + K

Or, `1=1/2kx^2+1/2mv^2`

Velocity v = 0, at the point of turn back, and hence KE also becomes zero.

So, total energy at the point of turn back can be given as follows:


Or, `1=1/2xx0.5x^2`

Or, `0.25x^2=1`

Or, `x^2=1/(0.25)=4`

Or, `x=±2` m

This shows that the particle must turn back from a distance of ±2 m.

Question 5: Answer the following:

(a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?

Answer: We know:

Total Energy = Potential Energy + Kinetic Energy

Or, `E=mg\h+1/2mv^2`

The potential energy of the rocket increases with increase in height. It means that kinetic energy must decrease in order to keep total energy at constant level. Kinetic energy varies directly as mass and as square of velocity. With increase in velocity of the rocket, the mass must reduce in order to keep the total energy at constant level. This shows that the heat energy is obtained from loss of mass. So, the energy is coming at expense of the rocket.

(b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?

Answer: Gravitational force is a conservative force, and work done by a conservative force on a closed path is zero. Hence, work done by the gravitational force over every complete orbit of the comet is zero.

(c) An artificial satellite, orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth?

Answer: We know that total energy of an object is constant. When the satellite comes closer to the earth its potential energy decreases due to reduction in height. There must be commensurate increase in potential energy to keep the total energy at constant level. This is facilitated by increase in speed of the satellite.

(d) In this figure, the man walks 2 m carrying a mass of 15 kg on his hands. In the second figure, he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hands at its other end. In which case is the work done greater?

work done by man

Answer: Work done is zero in the first case because direction of displacement is perpendicular to direction of applied force. The man is working positive work in the second case because direction of displacement is in direction of force being applied.

Work done in second case = F × s

= 15 × 9.8 × 2 = 294 J

Question 6: Underline the correct alternative:

(a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered.

Answer: Decreases: When a force does positive work on a body, it causes displacement in the direction of force. As a result, the body moves towards the center of force, i.e. separation between two decreases. Due to this, potential energy decreases when a force does positive work on a body.

(b) Work done by a body against friction always results in a loss of its kinetic/potential energy.

Answer: Kinetic Energy: Work done by a body against friction reduces speed. Hence, it results in reduction of kinetic energy.

(c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system.

Answer: External force: Internal forces have no impact on change in momentum.

(d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.

Answer: Total linear momentum: It remains constant no matter if collision is elastic or inelastic.

Question 7: State if each of the following statements is true or false. Give reasons for your answer.

(a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.

Answer: False: The total momentum and total energy of a system of bodies are conserved. It is not about individual bodies.

(b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.

Answer: False: Total energy of the system is influenced by external energy.

(c) Work done in the motion of a body over a closed loop is zero for every force in nature.

Answer: False: It is true only for conservative forces.

(d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.

Answer: True: In case of inelastic collision, some energy is always dissipated in the form of heat, sound or light. So, final KE is always less than initial KE.

Question 8: Answer carefully, with reasons:

(a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact)?

Answer: NO: During collision, KE of the ball gets converted into PE. So, KE is not conserved during the short time of collision.

(b) Is the total linear momentum conserved during the short time of an elastic collision of two balls?

Answer: YES: Total linear momentum is always conserved.

(c) What are the answers to (a) and (b) for an inelastic collision?

Answer: The answers remain the same.

(d) If the potential energy of two billiard balls depends only on the separation distance between their centers, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).

Answer: Elastic, because potential energy depends only on separation distance between centers of the balls.

Question 9: A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to

  1. `t^(1/2)`
  2. `t`
  3. `t^(3/2)`
  4. `t^2`

Answer: (b) t

We know: `v = u + at`

Or, `v=at`

Moreover, `P=F×v`

And `F=ma`

Substituting values of v and F in this equation, we get:


This equation shows power is directly proportional to time.

Question 10: A body is moving uni-directionally under the influence of a source of constant power. Its displacement in time t is proportional to

  1. `t^(1/2)`
  2. `t`
  3. `t^(3/2)`
  4. `t^2`

Answer: (c) `t^(3/2)`

Power = Force × velocity

Dimensions can be written as follows:

P = [MLT][LT-2]

= ML2T-3]

Since P is constant, hence

L2T-3 = constant

Or, `(L^2)/(T^(-3))` = constant

Or, `L^2∝T^3`

Or, `L'T^(3/2)`