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NCERT Solutions · Class 11 Physics · NCERT Physics Part I and II Part I · Chapter 5

Chapter 5: Work, Energy and Power (Work, Energy and Power)

Q: The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = kx²/2, 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 Fig. 5.12. Show that a particle of total energy 1 J moving under this potential must 'turn back' when it reaches x = ± 2 m.

K = 0 at x = ±2 m, so the particle turns back there.

Q: 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? (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? (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? (d) In Fig. 5.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 5.13(ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater?

(a) Rocket (b) Gravity is conservative; closed orbit (c) PE falls faster than total energy, so KE rises (d) Case (ii)

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

(a) decreases (b) kinetic (c) external force (d) total linear momentum (and total energy)

Q: 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)? (b) Is the total linear momentum conserved during the short time of an elastic collision of two balls? (c) What are the answers to (a) and (b) for an inelastic collision? (d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).

(a) No (b) Yes (c) Momentum yes, KE no (d) Elastic

Step-by-step solutions to all 23 exercises of Chapter 5, Work, Energy and Power (NCERT Class 11 Physics Part I, 2026-27): sign of work, the work-energy theorem with friction, potential energy curves, conservative forces, elastic and inelastic collisions, power, the raindrop, pump and windmill problems, and the pendulum collision. All 23 questions are answered, with the key answer highlighted.

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Work $W = F \cdot d = F d cos θ$ . Work–energy theorem: $W_{net} = Δ K$ . For a conservative force, $W = - Δ U$ . Power $P = \frac{d W}{d t} = F \cdot v$ . In every collision total momentum is conserved; in an elastic collision kinetic energy is also conserved. Equal masses in a 1-D elastic collision exchange velocities.

Exercises

5.1

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: (a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket. (b) work done by gravitational force in the above case, (c) work done by friction on a body sliding down an inclined plane, (d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity, (e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.

Solution

Work is positive when the force has a component along the displacement and negative when it opposes it.

(a) Man's force up, bucket moves up: positive. (b) Gravity down, bucket moves up: negative. (c) Friction acts up the incline, body moves down: negative. (d) Applied force along the motion (balancing friction): positive. (e) Air resistance opposes the motion: negative.

(a) + (b) − (c) − (d) + (e) −

5.2

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, (b) work done by friction in 10 s, (c) work done by the net force on the body in 10 s, (d) change in kinetic energy of the body in 10 s, and interpret your results.

Solution

Friction $f = μ m g = 0.1 \times 2 \times 9.8 = 1.96$ N. Net force $= 7 - 1.96 = 5.04$ N, so $a = 2.52 m s^{- 2}$ .

In 10 s: $s = \frac{1}{2} a t^{2} = \frac{1}{2} (2.52) (100) = 126$ m; $v = a t = 25.2 m s^{- 1}$ .

(a) $W_{F} = 7 \times 126 = 882$ J (b) $W_{f} = - 1.96 \times 126 = - 247$ J (c) $W_{net} = 5.04 \times 126 = 635$ J (d) $Δ K = \frac{1}{2} (2) (25.2)^{2} - 0 = 635$ J.

The work done by the net force equals the change in kinetic energy: the work-energy theorem.

(a) 882 J (b) −247 J (c) 635 J (d) 635 J; W_net = ΔK

5.3

Given in Fig. 5.11 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. [(a) V = 0 for x < a, V = V₀ for x > a, with 0 < E < V₀. (b) V(x) is a staircase that stays above E everywhere. (c) V = V₀ for x < a and x > b, V = −V₁ for a < x < b, with 0 < E < V₀. (d) a well of depth −V₁ for |x| < a/2, with V rising above E for a/2 < |x| < b/2 and falling to zero beyond.]

Solution

Since $K = E - V (x) \geq 0$ , the particle can only be where $V (x) \leq E$ . The minimum total energy is the least value of V(x).

(a) For x > a, V₀ > E: the particle cannot be found at x > a. Minimum energy 0. (Context: a ball on a floor meeting a step it cannot climb.)

(b) V(x) > E at every x: the particle cannot be found anywhere. Its total energy must be at least the lowest level of V(x) (marked V₁ in the NCERT answer).

(c) For x < a and x > b, V₀ > E: it cannot be found at x < a or x > b; it is trapped in the well. Minimum energy −V₁. (Context: a particle in a box / an electron bound in a well.)

(d) In the regions −b/2 < x < −a/2 and a/2 < x < b/2, V > E: forbidden. Minimum energy −V₁. (Context: an alpha particle inside a nucleus, held by a barrier.)

(a) x > a; 0 (b) everywhere; the lowest level of V(x) (V₁) (c) x < a and x > b; −V₁ (d) −b/2 < x < −a/2 and a/2 < x < b/2; −V₁

5.4

The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = kx²/2, 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 Fig. 5.12. Show that a particle of total energy 1 J moving under this potential must 'turn back' when it reaches x = ± 2 m.

Solution

$E = K + V$ , and $K \geq 0$ . At the turning points $K = 0$ , so $V = E$ :

$\frac{1}{2} (0.5) x^{2} = 1 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2$ m.

Beyond |x| = 2 m, V would exceed 1 J and K would be negative, which is impossible. So the particle turns back at x = ±2 m.

K = 0 at x = ±2 m, so the particle turns back there.

5.5

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? (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? (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? (d) In Fig. 5.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 5.13(ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater?

Solution

(a) The rocket. The heat comes from the rocket's own mechanical energy (its kinetic energy decreases).

(b) Gravity is conservative: the work done over a closed path is zero. After a full orbit the comet is back where it started, so the change in potential energy, and hence the work, is zero.

(c) As the satellite comes closer, its potential energy decreases more than its total energy is lost to friction. Total energy $E = - \frac{GM m}{2 r}$ becomes more negative, and $K = \frac{GM m}{2 r}$ increases. So the speed increases while the total energy decreases.

(d) (i) He holds the mass with an upward force while walking horizontally: force ⊥ displacement, so the work by his force on the mass is zero. (ii) He pulls the rope with a force of $15 \times 9.8 = 147$ N along his 2 m displacement: $W = 147 \times 2 = 294$ J. More work is done in case (ii).

(a) Rocket (b) Gravity is conservative; closed orbit (c) PE falls faster than total energy, so KE rises (d) Case (ii)

5.6

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

Solution

(a) $W = - Δ U$ : positive work means U decreases.

(b) Friction removes kinetic energy.

(d) Total linear momentum is unchanged (and the total energy too, if the system is isolated); kinetic energy is not conserved.

(a) decreases (b) kinetic (c) external force (d) total linear momentum (and total energy)

5.7

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. (b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present. (c) Work done in the motion of a body over a closed loop is zero for every force in nature. (d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.

Solution

(a) False. The total momentum and total energy of the system are conserved, not those of each body: the bodies exchange momentum and energy.

(b) False. External forces can do work on the system and change its energy. (Only the energy of an isolated system is conserved.)

(c) False. This is true only for conservative forces. Friction does non-zero (negative) work over a closed loop.

(d) False (usually true, but not always). If energy is released during the collision, e.g. when an explosive or a compressed spring is triggered, the final kinetic energy can exceed the initial.

(a) F (b) F (c) F (d) F

5.8

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)? (b) Is the total linear momentum conserved during the short time of an elastic collision of two balls? (c) What are the answers to (a) and (b) for an inelastic collision? (d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).

Solution

(a) No. During contact the balls are deformed; part of the kinetic energy is stored as elastic potential energy. It is fully returned only after they separate.

(b) Yes. No external force acts, so total momentum is conserved at every instant.

(c) Momentum is conserved during and after the collision; kinetic energy is not conserved, even after it is over.

(d) Elastic. A potential energy depending only on separation means a conservative force; when the balls separate to their original separation, all the stored energy returns as kinetic energy.

(a) No (b) Yes (c) Momentum yes, KE no (d) Elastic

5.9

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 (i) t^(1/2) (ii) t (iii) t^(3/2) (iv) t²

Solution

$v = a t$ , $F = ma$ (constant), so $P = F v = ma \cdot a t = m a^{2} t \propto t$ .

(ii) t

5.10

A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to (i) t^(1/2) (ii) t (iii) t^(3/2) (iv) t²

Solution

$P = F v = m v \frac{d v}{d t} = const \Rightarrow \frac{1}{2} m v^{2} = P t \Rightarrow v \propto t^{1/2}$ .

$x = \int v d t \propto t^{3/2}$ .

(iii) t^(3/2)

5.11

A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by F = –î + 2ĵ + 3k̂ N where î, ĵ, k̂ are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis?

Solution

$d = 4 \hat{k}$ m. $W = F \cdot d = (- 1) (0) + 2 (0) + 3 (4)$ .

12 J

5.12

An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton? Obtain the ratio of their speeds. (electron mass = 9.11 × 10⁻³¹ kg, proton mass = 1.67 × 10⁻²⁷ kg, 1 eV = 1.60 × 10⁻¹⁹ J).

Solution

$K = \frac{1}{2} m v^{2} \Rightarrow v = \frac{2 K}{m}$ , so

$\frac{v _{e}}{v _{p}} = \frac{K _{e}}{K _{p}} \cdot \frac{m _{p}}{m _{e}} = \frac{10}{100} \times \frac{1.67 \times 1 0 ^{- 27}}{9.11 \times 1 0 ^{- 31}} = 0.1 \times 1833 = 183.3 = 13.5$

The electron is faster; v_e/v_p ≈ 13.5

5.13

A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 m s⁻¹?

Solution

Mass $m = ρ \cdot \frac{4}{3} π r^{3} = 1000 \times \frac{4}{3} \times 3.14 \times (2 \times 1 0^{- 3})^{3} = 3.35 \times 1 0^{- 5}$ kg.

Work by gravity in each half (250 m): $m g h = 3.35 \times 1 0^{- 5} \times 9.8 \times 250 = 0.082$ J. Total by gravity $= 0.164$ J.

Final KE $= \frac{1}{2} (3.35 \times 1 0^{- 5}) (10)^{2} = 1.68 \times 1 0^{- 3}$ J.

Work-energy theorem: $W_{g} + W_{r} = Δ K \Rightarrow W_{r} = 0.00168 - 0.164 = - 0.163$ J.

0.082 J in each half; work by resistance = −0.163 J

5.14

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

Solution

The molecule's normal momentum component reverses, so its own momentum changes. But the momentum of the molecule + wall system is conserved: the wall takes a recoil momentum equal to the change, which gives it a negligible velocity because of its huge mass.

The molecule rebounds with the same speed, so kinetic energy is conserved: the collision is elastic.

Momentum of molecule + wall is conserved; the collision is elastic.

5.15

A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m³ in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?

Solution

Mass of water $= 30 \times 1000 = 3 \times 1 0^{4}$ kg. Useful work $= m g h = 3 \times 1 0^{4} \times 9.8 \times 40 = 1.176 \times 1 0^{7}$ J in 900 s.

Output power $= \frac{1.176 \times 1 0 ^{7}}{900} = 1.307 \times 1 0^{4}$ W. Input $= \frac{1.307 \times 1 0 ^{4}}{0.30} = 4.36 \times 1 0^{4}$ W.

43.6 kW

5.16

Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following (Fig. 5.14) is a possible result after collision? [(i) ball 1 at rest, balls 2 and 3 each moving with V/2 (ii) balls 1 and 2 at rest, ball 3 moving with V (iii) all three moving with V/3]

Solution

Both momentum (mV) and kinetic energy ( $\frac{1}{2} m V^{2}$ ) must be the same after the collision.

(i) Momentum $= m \frac{V}{2} + m \frac{V}{2} = mV$ ✔; KE $= 2 \cdot \frac{1}{2} m \frac{V ^{2}}{4} = \frac{1}{4} m V^{2}$ ✘.

(ii) Momentum $mV$ ✔; KE $\frac{1}{2} m V^{2}$ ✔.

(iii) Momentum $3 m \frac{V}{3} = mV$ ✔; KE $= \frac{3}{2} m \frac{V ^{2}}{9} = \frac{1}{6} m V^{2}$ ✘.

Case (ii): balls 1 and 2 at rest, ball 3 moves with V.

5.17

The bob A of a pendulum released from 30° to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 5.15. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.

Solution

In a head-on elastic collision between equal masses, the velocities are exchanged. A comes to rest and B moves off with A's speed.

A does not rise at all; it transfers its entire momentum to B.

5.18

The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?

Solution

The bob falls through $h = l = 1.5$ m. 95% of $m g h$ becomes kinetic energy:

$\frac{1}{2} m v^{2} = 0.95 m g h \Rightarrow v = 2 \times 9.8 \times 1.5 \times 0.95 = 27.9$

5.3 m s⁻¹

5.19

A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg s⁻¹. What is the speed of the trolley after the entire sand bag is empty?

Solution

The sand leaks out vertically and leaves with the trolley's own horizontal velocity, so it exerts no horizontal force on the trolley. With no external horizontal force, the trolley's speed does not change.

27 km h⁻¹ (unchanged)

5.20

A body of mass 0.5 kg travels in a straight line with velocity v = a x^(3/2) where a = 5 m^(–1/2) s⁻¹. What is the work done by the net force during its displacement from x = 0 to x = 2 m?

Solution

$v (0) = 0$ ; $v (2) = 5 \times 2^{3/2} = 5 \times 22 = 102 m s^{- 1}$ .

$W = Δ K = \frac{1}{2} (0.5) (102)^{2} - 0 = 0.25 \times 200$ .

50 J

5.21

The blades of a windmill sweep out a circle of area A. (a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t? (b) What is the kinetic energy of the air? (c) Assume that the windmill converts 25% of the wind's energy into electrical energy, and that A = 30 m², v = 36 km/h and the density of air is 1.2 kg m⁻³. What is the electrical power produced?

Solution

(a) Volume through the circle in time t $= A v t$ , so $m = ρ A v t$ .

(b) $K = \frac{1}{2} m v^{2} = \frac{1}{2} ρ A v^{3} t$ .

(c) $v = 10 m s^{- 1}$ . Wind power $= \frac{K}{t} = \frac{1}{2} \times 1.2 \times 30 \times 1 0^{3} = 18000$ W. Electrical power $= 0.25 \times 18000 = 4500$ W.

(a) ρAvt (b) ½ρAv³t (c) 4.5 kW

5.22

A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated. (a) How much work does she do against the gravitational force? (b) Fat supplies 3.8 × 10⁷ J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?

Solution

(a) $W = n m g h = 1000 \times 10 \times 9.8 \times 0.5 = 49000$ J.

(b) Useful energy per kg of fat $= 0.2 \times 3.8 \times 1 0^{7} = 7.6 \times 1 0^{6}$ J. Fat used $= \frac{49000}{7.6 \times 1 0 ^{6}} = 6.45 \times 1 0^{- 3}$ kg.

(a) 49,000 J (b) 6.45 × 10⁻³ kg

5.23

A family uses 8 kW of power. (a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW? (b) Compare this area to that of the roof of a typical house.

Solution

(a) Useful power per m² $= 0.2 \times 200 = 40$ W. Area $= \frac{8000}{40} = 200 m^{2}$ .

(b) 200 m² is about 14 m × 14 m: comparable to the roof of a large house.

(a) 200 m² (b) about the roof of a large house (14 m × 14 m)

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