Class 11 Physics Chapter 5 Work Energy and Power is a foundational chapter that explains how energy is transferred and transformed through work and motion. In this chapter, you’ll learn key concepts such as the work-energy theorem, kinetic and potential energy, conservation of mechanical energy, and types of collisions.

Class 11 Physics Chapter 5 Work Energy and Power

This page provides all NCERT exercise solutions in plain text format with clearly explained steps to help students understand the logic behind each answer. Whether you’re preparing for exams or revising the basics, this is your complete guide to mastering Chapter 5.

Class 11 Physics Chapter 5 Work Energy and Power Textbook Answers

5.1 – Class 11 Physics Chapter 5 Work Energy and Power

Q: 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.

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

5.2 – Class 11 Physics Chapter 5 Work Energy and Power

Q: 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.

Answer:
Mass, m = 2 kg
Applied force, F = 7 N
Frictional force, f = μmg = 0.1 × 2 × 9.8 = 1.96 N
Net force = 7 – 1.96 = 5.04 N
Acceleration = F_net / m = 5.04 / 2 = 2.52 m/s²
Displacement in 10 s = (1/2) × a × t² = 0.5 × 2.52 × 100 = 126 m

(a) Work by applied force = 7 × 126 = 882 J
(b) Work by friction = –1.96 × 126 = –246.96 J
(c) Work by net force = 5.04 × 126 = 635.04 J
(d) Kinetic energy = (1/2) × 2 × (2.52 × 10)² = 635.04 J

Interpretation: Work done by net force equals the change in kinetic energy (work-energy theorem is verified).

5.3 – Class 11 Physics Chapter 5 Work Energy and Power

Q: 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.

Answer:

  • A particle cannot exist where potential energy > total energy.
  • Turning points are where total energy = potential energy.
  • Minimum energy is the minimum value of the potential curve.
  • Graphs represent SHM (spring), potential wells, and step potentials.
  • Particle is confined to regions where total energy ≥ potential energy.

5.4 – Class 11 Physics Chapter 5 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 = 0.5 N/m. The graph of V(x) vs 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.

Answer:
V(x) = (1/2)kx² = 0.25x²
Set V(x) = 1 J → 0.25x² = 1 → x² = 4 → x = ±2 m
So, particle turns back at x = ±2 m when kinetic energy becomes zero.

5.5 – Class 11 Physics Chapter 5 Work Energy and Power

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?

Answer:
(a) Rocket — its kinetic energy is converted into heat.
(b) Because gravitational force is conservative, total work over closed path is zero.
(c) It loses potential energy which is converted into kinetic energy, so speed increases.
(d) Work is greater in (ii) because lifting involves vertical displacement against gravity.

5.6 – Class 11 Physics Chapter 5 Work Energy and Power

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 / linear momentum / energy of the system of two bodies.

Answer:
(a) decreases
(b) kinetic
(c) external force
(d) linear momentum

5.7 – Class 11 Physics Chapter 5 Work Energy and Power

Q: 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.

Answer:
(a) False – Only total momentum and total kinetic energy of system is conserved.
(b) False – Only in isolated systems without non-conservative forces.
(c) False – True only for conservative forces.
(d) True – Kinetic energy is lost as heat or sound.

5.8

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?

Answer:
(a) No – It may convert to deformation energy during contact.
(b) Yes – Momentum is conserved in absence of external forces.
(c) (a) No – Kinetic energy not conserved in inelastic collisions.
  (b) Yes – Momentum still conserved.
(d) Elastic – If force depends only on separation, it is conservative.

5.9

Q: 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) t¹ᐟ²  (ii) t  (iii) t³ᐟ²  (iv) t²

Answer:
Velocity, v = at
Kinetic energy, KE = (1/2)mv² = (1/2)m a²t²
Power, P = d(KE)/dt = ma²t → Power ∝ t
Correct option: (ii) t

5.10

Q: A body is moving unidirectionally under the influence of a source of constant power. Its displacement is proportional to
(i) t¹ᐟ²  (ii) t  (iii) t³ᐟ²  (iv) t²

Answer:
P = constant = d(KE)/dt = mv dv/dt
⇒ v dv = (P/m) dt → integrate → v ∝ √t
Displacement = ∫v dt ∝ ∫√t dt ∝ t³ᐟ²
Correct option: (iii) t³ᐟ²

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