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Work, Energy and Power NCERT Solutions, Complete PDF Content and Important Questions
NCERT Solutions for Class 11 Physics Chapter 5 Work, Energy and Power PDF Download (2026) + Important Questions Welcome to one of the most fascinating and high-scoring chapters in your Class 11 journey! If you've ever wondered how roller coasters work, why a stretched catapult packs so much punch, or how car crashes are analyzed, Class 11 Physics Chapter 5: Work, Energy and Power holds all the answers. For students preparing for the CBSE 2026 board exams, as well as competitive exams like JEE Main, JEE Advanced, and NEET, this chapter is an absolute game-changer. It bridges the concepts of kinematics and Newton's laws of motion, giving you a powerful new tool-the Work-Energy Theorem-to solve complex mechanics problems in seconds. In this comprehensive guide, we will break down the entire chapter with simple explanations, step-by-step NCERT solutions, and a handpicked list of board-style important questions.
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ChapterWork, Energy and Power
SubjectPhysics
Class11
BoardCBSE
DifficultyModerate to High
Exam Weightage7 - 9 Marks (Crucial for competitive exams)
Learning Objectives
- Understand the physics definition of work and.
- Calculate the work done by constant and variable forces.
- Apply the Work-Energy Theorem to.
- Solve complex motion problems without using kinematics.
- Differentiate between conservative and non-conservative forces.
- Apply the principle of Conservation of Mechanical Energy.
- Understand its real-world applications.
- Analyze 1D and 2D collisions, distinguishing between perfectly elastic and inelastic collisions.
Key Concepts, Definitions and Formulas
- Work-Energy Theorem: The net work done on an object by all forces equals its change in kinetic energy. $$W_{net} = K_f - K_i = \Delta K$$ Potential Energy ($U$): The energy possessed by a body due to its position or configuration
- Elastic Potential Energy of a Spring: $$U = \frac{1}{2}kx^2$$
- Conservation of Mechanical Energy: JSON { For conservative forces, the total mechanical energy (Kinetic + Potential) of an isolated system remains constant. $$K_i + U_i = K_f + U_f$$ * Power ($P$): The rate at which work is done or energy is transferred. $$P = \frac{dW}{dt} = \vec{F} \cdot \vec{v}$$ *
- Elastic Collision: Both momentum and"widgetSpec": { "height": "600px", "prompt": "
- Interactive Conservation of Energy simulator. Data State: initialValues = { mass: 2, height: 10 }
- Standard Layout. Inputs: Sliders for Mass (1 to 10 kg) and Initial Height kinetic energy are conserved. *
- Inelastic Collision: Momentum is conserved, but kinetic energy is not (some energy is lost as heat/sound)
- Standard Layout. Data State: initialValues = { height: 10, mass with g = 9.8." } }
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Full NCERT Solutions
NCERT Q5.1: 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. ####
Answer: * (a) Positive: The force exerted by the man (tension) is in the upward direction, and the displacement of the bucket is also upward. Since the angle $\theta = 0^\circ$, $\cos 0^\circ = 1$. Hence, $W > 0$. * (b) Negative: The gravitational force acts downwards, while the bucket moves upwards. The angle $\theta = 180^\circ$, and $\cos 180^\circ = -1$. Hence, $W < 0$. * (c) Negative: Frictional force always opposes relative motion. It acts exactly opposite to the direction of displacement ($\theta = 180^\circ$). * (d) Positive: Since the body moves with uniform velocity on a rough plane, the applied force must exactly balance the frictional force. The applied force and displacement are in the same direction ($\theta = 0^\circ$). * (e) Negative: The resistive force of air always acts in the direction opposite to the motion of the pendulum bob, meaning $\theta = 180^\circ$.
NCERT Q5.2: 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. ####
Answer: Given: Mass $m = 2$ kg, $u = 0$, $F = 7$ N, $\mu_k = 0.1$, $t = 10$ s, $g = 9.8$ m/s2 Frictional force $f = \mu_k N = \mu_k mg = 0.1 \times 2 \times 9.8 = 1.96$ N. Net force $F_{net} = F - f = 7 - 1.96 = 5.04$ N. Acceleration $a = \frac{F_{net}}{m} = \frac{5.04}{2} = 2.52$ m/s2. Displacement in 10 s: $$s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(2.52)(10)^2 = 126 \text{ m}$$ (a) Work done by applied force: $$W_a = F \times s = 7 \times 126 = 882 \text{ J}$$ (b) Work done by friction: $$W_f = -f \times s = -1.96 \times 126 = -246.96 \text{ J}$$ (c) Work done by net force: $$W_{net} = F_{net} \times s = 5.04 \times 126 = 635.04 \text{ J}$$ (d) Change in kinetic energy: Final velocity $v = u + at = 0 + 2.52(10) = 25.2$ m/s. Change in KE $\Delta K = \frac{1}{2}mv^2 - 0 = \frac{1}{2}(2)(25.2)^2 = 635.04 \text{ J}$. Interpretation: The work done by the net force exactly equals the change in kinetic energy, which beautifully verifies the Work-Energy Theorem.
NCERT Q5.3: Given in Figure 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. *(Refer to NCERT textbook for figures).* ####
Answer: Core Concept: Total Energy ($E$) = Kinetic Energy ($K$) + Potential Energy ($V$). Therefore, $K = E - V$. Since Kinetic Energy can never be negative ($K \ge 0$), a particle can never be found in a region where Potential Energy is greater than Total Energy ($V > E$). * (a) The particle cannot be found in the region $x > a$ because $V(x) > E$. Minimum total energy must be zero. * (b) The particle cannot be found in regions $x < a$ and $x > b$ because $V(x) > E$. Minimum total energy must be $-V_1$. * (c) The particle cannot be found in regions $x < a$ and $x > b$ because $V(x) > E$. Minimum total energy must be $-V_1$. * (d) The particle cannot be found in regions where $V(x) > E$. Minimum total energy must be $-V_1$.
NCERT Q5.4: The potential energy function for a particle executing linear simple harmonic motion is given by $V(x) = \frac{kx^2}{2}$, where $k$ is the force constant of the oscillator. For $k = 0.5$ N/m, the graph of $V(x)$ versus $x$ is drawn in Fig. Show that a particle of total energy 1 J moving under this potential must 'turn back' when it reaches $x = \pm 2$ m. ####
Answer: Given: Total Energy $E = 1$ J, Force constant $k = 0.5$ N/m. At the extreme positions (where the particle "turns back"), its velocity becomes zero. Hence, Kinetic Energy $K = 0$. By conservation of energy: $E = K + V(x)$ $$1 = 0 + \frac{1}{2}kx^2$$ $$1 = \frac{1}{2}(0.5)x^2$$ $$1 = 0.25x^2 \implies x^2 = \frac{1}{0.25} = 4$$ $$x = \pm 2 \text{ m}$$ Therefore, the particle turns back when it reaches $x = \pm 2$ m.
NCERT Q5.5: 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? ####
Answer: * (a) The heat energy required for burning is obtained at the expense of the rocket's mechanical energy (specifically, its kinetic and potential energy). As it overcomes air friction, its mechanical energy continuously decreases, transforming into heat. * (b) Gravitational force is a conservative force. A core property of any conservative force is that the total work done by it over a closed path (like a complete orbit) is exactly zero, regardless of the shape of the path.
Extra Important Questions for 2026 Exams
Important Q1: A light and a heavy body have the same kinetic energy. Which one has greater momentum? A) The light body B) The heavy body C) Both have equal momentum D) Cannot be determined
Answer: B) The heavy body Explanation: $K = \frac{P^2}{2m} \implies P = \sqrt{2mK}$. Since $K$ is constant, $P \propto \sqrt{m}$. The heavier body has a larger mass, hence greater momentum.
Important Q2: The work done by the centripetal force on a body moving in a circular path is: A) Positive B) Negative C) Zero D) Dependent on radius
Answer: C) Zero Explanation: Centripetal force acts towards the center, while displacement is tangential. The angle $\theta = 90^\circ$. $W = Fd \cos 90^\circ = 0$.
Important Q3: In a perfectly inelastic collision, which of the following is conserved? A) Kinetic Energy only B) Momentum only C) Both Kinetic Energy and Momentum D) Neither
Answer: B) Momentum only Explanation: In any collision without external forces, linear momentum is conserved. However, in an inelastic collision, kinetic energy is lost as heat or deformation.
Important Q4: Define a conservative force. Give one example.
Answer: A force is conservative if the work done by it in moving a particle from one point to another depends only on the initial and final positions, and is independent of the path taken. Example: Gravitational force, Electrostatic force.
Important Q5: State the Work-Energy Theorem.
Answer: The Work-Energy Theorem states that the net work done by all the forces acting on a body is exactly equal to the change in its kinetic energy. ($W_{net} = \Delta K$).
Important Q6: Two springs A and B ($k_A > k_B$) are stretched by the same force. On which spring is more work done?
Answer: Work done $W = \frac{F^2}{2k}$. Since the force $F$ is the same, $W \propto \frac{1}{k}$. Because $k_A > k_B$, the work done on spring B will be greater.
Important Q7: Derive the expression for the potential energy of an elastic stretched spring.
Answer: Let a spring of force constant $k$ be stretched by distance $x$. The restoring force is $F = -kx$. The external force applied is $F_{ext} = kx$. The small work done $dW$ to stretch it by an infinitesimally small distance $dx$ is $dW = F_{ext} dx = kx \, dx$. Total work done $W = \int_{0}^{x} kx \, dx = k \left[ \frac{x^2}{2} \right]_{0}^{x} = \frac{1}{2}kx^2$. This work done is stored in the spring as its elastic potential energy ($U$). Thus, $U = \frac{1}{2}kx^2$.
Important Q8: Show that in a perfectly elastic one-dimensional collision between two bodies, the relative velocity of approach is equal to the relative velocity of separation.
Answer: Let bodies of masses $m_1$ and $m_2$ have initial velocities $u_1$ and $u_2$, and final velocities $v_1$ and $v_2$. Conservation of Momentum: $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \implies m_1(u_1 - v_1) = m_2(v_2 - u_2)$ --- (Eq 1) Conservation of Kinetic Energy: $\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$ $m_1(u_1^2 - v_1^2) = m_2(v_2^2 - u_2^2)$ $m_1(u_1 - v_1)(u_1 + v_1) = m_2(v_2 - u_2)(v_2 + u_2)$ --- (Eq 2) Dividing Eq 2 by Eq 1: $u_1 + v_1 = v_2 + u_2$ $u_1 - u_2 = v_2 - v_1$ Here, $(u_1 - u_2)$ is the velocity of approach and $(v_2 - v_1)$ is the velocity of separation. Hence proved.
Important Q9: 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?
Answer: Volume $V = 30$ m3. Mass of water $m = V \times \text{density} = 30 \times 1000 = 30000$ kg. Height $h = 40$ m. Time $t = 15 \times 60 = 900$ s. $g = 9.8$ m/s2. Useful work done by pump $W = mgh = 30000 \times 9.8 \times 40 = 11760000$ J. Output Power $P_{out} = \frac{W}{t} = \frac{11760000}{900} = 13066.67$ W = 13.067 kW. Efficiency $\eta = \frac{P_{out}}{P_{in}} = 30\% = 0.3$. Input Power $P_{in} = \frac{P_{out}}{0.3} = \frac{13.067}{0.3} = 43.56$ kW. The pump consumes 43.56 kW of electric power.
Important Q10: What is the total mechanical energy of the car at Point A?
Answer: At A, it is at rest, so $K = 0$. $E = U = mgh = 500 \times 9.8 \times 50 = 245000 \text{ J}$ or 245 kJ.
Important Q11: Find the speed of the car when it reaches the bottom of the track (Point B).
Answer: By conservation of energy, $E_B = E_A = 245000$ J. At B ($h = 0$), $U = 0$, so $K = 245000$ J. $\frac{1}{2}mv^2 = 245000 \implies \frac{1}{2}(500)v^2 = 245000 \implies v^2 = \frac{245000}{250} = 980$. $v = \sqrt{980} \approx 31.3$ m/s.
Important Q12: What is the kinetic energy of the car at Point C?
Answer: Total energy at C must equal 245000 J. Potential energy at C, $U_C = mgh_C = 500 \times 9.8 \times 20 = 98000$ J. $K_C = E - U_C = 245000 - 98000 = 147000 \text{ J}$.
Important Q13: Assertion (A): The work done by friction is always negative. Reason (R): Frictional force acts in the direction opposite to the displacement.
Answer: A. Since friction opposes motion, $\theta = 180^\circ$, making $\cos 180^\circ = -1$, resulting in negative work.
Important Q14: Assertion (A): A spring has potential energy only when it is compressed, not when stretched. Reason (R): The restoring force always tries to bring the spring back to its mean position.
Answer: D. Assertion is false because a spring stores elastic potential energy ($U = \frac{1}{2}kx^2$) whether it is compressed or stretched. Reason is true.
Important Q15: Assertion (A): In an elastic collision, both momentum and kinetic energy are conserved. Reason (R): Internal forces do not change the total momentum of the system.
Answer: B. Both are true, but R does not explain why kinetic energy is conserved in an elastic collision.
FAQ Section
**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.** ####
* **(a) Positive:** The force exerted by the man (tension) is in the upward direction, and the displacement of the bucket is also upward. Since the angle $\theta = 0^\circ$, $\cos 0^\circ = 1$. Hence, $W > 0$. * **(b) Negative:** The gravitational force acts downwards, while the bucket moves upwards. The angle $\theta = 180^\circ$, and $\cos 180^\circ = -1$. Hence, $W < 0$.
**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.** ####
**Given:** Mass $m = 2$ kg, $u = 0$, $F = 7$ N, $\mu_k = 0.1$, $t = 10$ s, $g = 9.8$ m/s2 Frictional force $f = \mu_k N = \mu_k mg = 0.1 \times 2 \times 9.8 = 1.96$ N. Net force $F_{net} = F - f = 7 - 1.96 = 5.04$ N. Acceleration $a = \frac{F_{net}}{m} = \frac{5.04}{2} = 2.52$ m/s2. Displacement in 10 s: $$s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(2.
**Given in Figure are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case.** *(Refer to NCERT textbook for figures).* ####
**Core Concept:** Total Energy ($E$) = Kinetic Energy ($K$) + Potential Energy ($V$). Therefore, $K = E - V$. Since Kinetic Energy can never be negative ($K \ge 0$), a particle can never be found in a region where Potential Energy is greater than Total Energy ($V > E$). * **(a)** The particle cannot be found in the region $x > a$ because $V(x) > E$. Minimum total energy must be zero.
**The potential energy function for a particle executing linear simple harmonic motion is given by $V(x) = \frac{kx^2}{2}$, where $k$ is the force constant of the oscillator. For $k = 0.5$ N/m, the graph of $V(x)$ versus $x$ is drawn in Fig. Show that a particle of total energy 1 J moving under this potential must 'turn back' when it reaches $x = \pm 2$ m.** ####
**Given:** Total Energy $E = 1$ J, Force constant $k = 0.5$ N/m. At the extreme positions (where the particle "turns back"), its velocity becomes zero. Hence, Kinetic Energy $K = 0$. By conservation of energy: $E = K + V(x)$ $$1 = 0 + \frac{1}{2}kx^2$$ $$1 = \frac{1}{2}(0.5)x^2$$ $$1 = 0.25x^2 \implies x^2 = \frac{1}{0.25} = 4$$ $$x = \pm 2 \text{ m}$$ Therefore, the particle turns back when it reaches $x = \pm 2$ m.
More Class 11 Physics Chapters
Ch 1: Units and MeasurementsCh 2: Motion in a Straight LineCh 3: Motion in a PlaneCh 4: Laws of MotionCh 5: Work, Energy and PowerCh 6: System of Particles and Rotational MotionCh 7: GravitationCh 8: Mechanical Properties of SolidsCh 9: Mechanical Properties of FluidsCh 10: Thermal Properties of MatterCh 11: ThermodynamicsCh 12: Kinetic TheoryCh 13: OscillationsCh 14: Waves