NCERT Solutions for Class 11 Physics Chapter 5: Work, Energy and Power (NCERT 2026–27)

These Class 11 Physics Chapter 5 solutions cover Work, Energy and Power with every NCERT Exercise question reproduced verbatim and solved step by step. All numericals are worked out in full with correct units and cross-checked against the official NCERT answer key, so you can revise and self-check with confidence for the 2026–27 session.

Class: 11 Subject: Physics Chapter: 5 Title: Work, Energy and Power Exercises: 23 (5.1–5.23) Session: 2026–27
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Class 11 Physics Chapter 5 Solutions – Overview

Chapter 5, Work, Energy and Power, gives precise physical meaning to three words we use loosely in daily life. It opens with the scalar (dot) product of vectors, then defines work as W = F·d = Fd cosθ and kinetic energy as K = ½mv². The central result is the work–energy theorem: the change in kinetic energy of a body equals the work done by the net force on it. The chapter then develops work done by a variable force (as the area under a force–displacement graph / a definite integral), the idea of potential energy, conservation of mechanical energy for conservative forces, the spring potential energy ½kx², the definition of power (P = W/t = F·v), and finally collisions — elastic and inelastic, in one and two dimensions. Mastering these ideas builds the foundation for thermodynamics, oscillations and modern physics later in the course.

Key Concepts & Definitions

Work: done by a constant force is the product of the component of the force along the displacement and the magnitude of the displacement: W = (F cosθ)d. Work is a scalar with dimensions [ML²T−²] and SI unit joule (J). It can be positive, negative or zero.

Kinetic energy (KE): the energy a body possesses by virtue of its motion, K = ½mv². A scalar, always positive.

Work–energy theorem: the change in kinetic energy of a body equals the net work done on it, Kf − Ki = Wnet. It holds for both constant and variable forces.

Potential energy (PE): energy stored by virtue of position or configuration. Gravitational PE near Earth = mgh; spring (elastic) PE = ½kx².

Conservative force: work done is path-independent (depends only on end points) and is zero over a closed path; a potential energy can be defined for it (e.g. gravity, spring). Friction is non-conservative.

Conservation of mechanical energy: if only conservative forces do work, K + V = constant.

Power: the time rate of doing work or transferring energy, average Pav = W/t, instantaneous P = F·v. SI unit watt (W); 1 hp = 746 W; 1 kWh = 3.6×106 J.

Collisions: linear momentum is conserved in all collisions. In an elastic collision KE is also conserved; in an inelastic collision some KE is lost (in a completely inelastic collision the bodies move together).

Important Formulas

Scalar product: A·B = AB cosθ = AxBx + AyBy + AzBz

Work (constant force): W = F·d = Fd cosθ

Work (variable force): W = ∫ F(x) dx (area under the F–x graph)

Kinetic energy: K = ½mv² = p²/2m

Work–energy theorem: Kf − Ki = Wnet

Gravitational PE: V = mgh  •  Spring PE: V = ½kx²

Power: Pav = W/t ,  Pinst = F·v

1-D elastic collision (m₂ at rest): v1f = [(m₁−m₂)/(m₁+m₂)]v1i ,  v2f = [2m₁/(m₁+m₂)]v1i

Completely inelastic (1-D): vf = m₁v1i/(m₁+m₂)

NCERT Solutions – Exercises (5.1 to 5.23)

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.

ANSWER (a) Positive. The applied force and the displacement of the bucket are both upward (same direction), so θ = 0° and W > 0. (b) Negative. Gravity acts downward while the bucket moves up, so θ = 180° and the work is negative. (c) Negative. Friction opposes the sliding motion (it acts up the incline while the body slides down), so θ = 180° and W < 0. (d) Positive. For uniform velocity the applied force must balance friction and acts in the direction of motion, so W > 0. (e) Negative. Air resistance always opposes the motion of the bob, so the work it does is negative (it removes energy and stops the pendulum).

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.

ANSWER (step by step) Given m = 2 kg, F = 7 N, μ = 0.1, u = 0, t = 10 s, g = 9.8 m s−². Friction force f = μmg = 0.1 × 2 × 9.8 = 1.96 N (opposing motion). Net force = 7 − 1.96 = 5.04 N. Acceleration a = Fnet/m = 5.04/2 = 2.52 m s−². Displacement in 10 s: s = ½at² = ½ × 2.52 × 10² = 126 m. (a) Wapplied = F × s = 7 × 126 = 882 J. (b) Wfriction = −f × s = −1.96 × 126 = −246.96 J ≈ −247 J. (c) Wnet = Fnet × s = 5.04 × 126 = 635 J (also 882 − 247 = 635 J). (d) ΔK = Kf − Ki. Final speed v = u + at = 0 + 2.52 × 10 = 25.2 m s−¹; ΔK = ½ × 2 × 25.2² − 0 = 635 J. Interpretation: the change in kinetic energy (635 J) equals the work done by the net force (635 J), confirming the work–energy theorem.

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.

ANSWER A particle can exist only where its total energy E is not less than the potential energy, i.e. where kinetic energy K = E − V(x) ≥ 0. Wherever V(x) > E, the particle cannot be found. (a) The particle cannot exist where V(x) > E, i.e. for x > a. Minimum total energy required = 0. (Context: a particle near a step/wall potential.) (b) Here V(x) > E for the whole range, so the particle cannot exist anywhere (−∞ < x < ∞). The minimum total energy needed = V₁. (c) The particle cannot exist where V(x) > E, i.e. for x < a and x > b (it is trapped in the well between a and b). Minimum total energy = −V₁. (Context: a bound particle in a potential well.) (d) The forbidden regions are −b/2 < x < −a/2 and a/2 < x < b/2 (the two barriers where V > E). Minimum total energy = −V₁. (Context: a periodic / double-well type potential.)

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.

ANSWER (step by step) Total energy E = K + V. At a turning point the particle momentarily stops, so K = 0 and all the energy is potential: E = V(x). So ½kx² = E ⇒ x = ±√(2E/k). Substituting E = 1 J and k = 0.5 N m−¹: x = ±√(2 × 1 / 0.5) = ±√4 = ±2 m. For |x| > 2 m, V(x) > 1 J would make K negative, which is impossible. Hence the particle cannot move beyond x = ±2 m and must turn back there. (Verified.)

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?

ANSWER (a) The rocket. The heat is generated at the expense of the rocket’s own (kinetic + chemical) energy, so its mass and energy decrease as the casing burns. (b) Gravity is a conservative force, so the work it does equals the negative of the change in potential energy. Over one complete orbit the comet returns to the same point, so the change in potential energy is zero; hence the total work done over a full orbit is zero. (c) As the satellite descends its gravitational potential energy decreases. Although a little total (mechanical) energy is lost to friction, the fall in PE is much larger, so the kinetic energy — and therefore the speed — actually increases as it spirals closer to Earth. (d) The work done is greater in the second case. In case (i) the man supports the load on his hands; the force is vertical while the displacement is horizontal (θ = 90°), so the work done on the load is zero. In case (ii) he must pull the rope along the direction he walks to keep the hanging mass moving, so he does positive work.

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.

ANSWER (a) decreases. For a conservative force ΔV = −W, so positive work means the potential energy falls. (b) kinetic energy. Work done against friction draws on the body’s kinetic energy. (c) external force. Internal forces cancel in pairs (Newton’s third law), so only the net external force changes the total momentum. (d) total linear momentum, and (for an isolated system) total energy. Kinetic energy is not conserved in an inelastic collision.

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.

ANSWER (a) False. In an elastic collision it is the total momentum and total KE of the system that are conserved, not those of each body individually. (b) False. Total energy is conserved only for an isolated system. If external forces transfer energy in or out, the system’s total energy can change. (c) False. This is true only for conservative forces. For non-conservative forces such as friction, the work over a closed loop is not zero. (d) False (usually true but not always). Generally the final KE is less, but in some cases involving internal stored energy (e.g. an explosive collision) the final KE can be greater. So it is not always less.

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).

ANSWER (a) No. During contact the balls deform, so part of the kinetic energy is temporarily stored as elastic potential energy. KE is conserved only after the collision is over, not at every instant during it. (b) Yes. Linear momentum is conserved at every instant of the collision because the mutual forces are internal and equal-and-opposite (Newton’s third law). (c) For an inelastic collision: linear momentum is still conserved during (and after) the collision; kinetic energy is not conserved — some of it is permanently lost to heat, sound and deformation. (d) Elastic. If the potential energy depends only on the separation of centres, the force is conservative; the energy stored during deformation is fully recovered, so the collision is 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) t1/2   (ii) t   (iii) t3/2   (iv) t²

ANSWER (step by step) For constant acceleration a from rest, v = at and F = ma = constant. Power P = Fv = (ma)(at) = ma²t. Since m and a are constant, P ∝ t. Correct option: (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) t1/2   (ii) t   (iii) t3/2   (iv) t²

ANSWER (step by step) Constant power P means work W = Pt, and W = ½mv² (from rest), so ½mv² = Pt ⇒ v ∝ t1/2. Then displacement s = ∫v dt ∝ ∫t1/2 dt ∝ t3/2. Correct option: (iii) t3/2.

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?

ANSWER (step by step) Displacement along z-axis: d = 4k̂ m. Work W = F·d = (−î + 2ĵ + 3k̂)·(4k̂). Only the z-components contribute: W = 3 × 4 = 12 J. (Verified.)

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).

ANSWER (step by step) From K = ½mv², speed v = √(2K/m). Ratio ve/vp = √[(2Ke/me) / (2Kp/mp)] = √[(Ke/Kp) × (mp/me)]. Ke/Kp = 10/100 = 0.1, and mp/me = 1.67×10−²⁷ / 9.11×10−³¹ = 1833. ve/vp = √(0.1 × 1833) = √183.3 = 13.5. Since the ratio > 1, the electron is faster, with ve/vp ≈ 13.5. (Verified.)

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?

ANSWER (step by step) Radius r = 2 mm = 2×10−³ m; density of water ρ = 1000 kg m−³; g = 9.8 m s−². Mass m = ρ × (4/3)πr³ = 1000 × (4/3)π(2×10−³)³ = 3.35×10−⁵ kg (about 33.5 mg). Each half of the fall = 250 m. Work by gravity over a 250 m drop: Wg = mgh = 3.35×10−⁵ × 9.8 × 250 = 0.082 J. So the gravitational force does 0.082 J in the first half and 0.082 J in the second half (it depends only on height, not on speed). Total work by gravity over 500 m = 2 × 0.082 = 0.164 J. Final KE at ground (v = 10 m s−¹): Kf = ½mv² = ½ × 3.35×10−⁵ × 10² = 1.675×10−³ J. Initial KE = 0. By the work–energy theorem, Wg + Wresistive = ΔK, so Wresistive = ΔK − Wg = 0.00168 − 0.164 = −0.163 J. The resistive force does −0.163 J over the whole journey (negative, as it opposes the fall). (Verified.)

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?

ANSWER Momentum is conserved. Considering the molecule + wall as one system, the wall gains a tiny recoil momentum equal and opposite to the change in the molecule’s momentum, so total momentum is conserved. (Because the wall is so massive, its recoil speed is negligible, and the wall is treated as stationary.) The collision is elastic. The molecule rebounds with the same speed, so its kinetic energy is unchanged; since kinetic energy 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?

ANSWER (step by step) Volume V = 30 m³, so mass of water m = ρV = 1000 × 30 = 3×10⁴ kg. Time t = 15 min = 900 s, height h = 40 m, g = 9.8 m s−². Useful (output) work = mgh = 3×10⁴ × 9.8 × 40 = 1.176×10⁷ J. Output power = mgh/t = 1.176×10⁷ / 900 = 1.307×10⁴ W = 13.07 kW. Efficiency = 30% = 0.30, so input (electric) power = output power / efficiency = 13.07 / 0.30 = 43.6 kW. (Verified.)

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?

ANSWER (step by step) Let each ball have mass m. Before: one ball moves with speed V, the other two are at rest. Total momentum = mV; total KE = ½mV². For an elastic collision both momentum and KE must be conserved. Check the option where, after collision, one ball moves off with speed V and the other two remain at rest: Momentum after = mV (conserved ✓); KE after = ½mV² (conserved ✓). The alternative (two balls each moving with V/2) gives momentum 2m(V/2) = mV (ok) but KE = 2×½m(V/2)² = ¼mV² ≠ ½mV² (KE not conserved), so it is not possible. Correct result: option (b) — only the last ball moves off with speed V while the first two stay at rest. (Verified.)

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.

ANSWER In a one-dimensional elastic collision between two equal masses, the moving body comes to rest and the stationary body moves off with the first body’s velocity (Case I: v1f = 0, v2f = v1i). So bob A transfers its entire momentum and kinetic energy to bob B and stops at the lowest point. Therefore bob A does not rise at all after the collision (it rises to a height of zero).

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?

ANSWER (step by step) Released from the horizontal position, the bob falls a height equal to the length L = 1.5 m. Initial PE = mgL. 5% is lost to air resistance, so 95% becomes kinetic energy at the bottom: ½mv² = 0.95 × mgL. v = √(2 × 0.95 × gL) = √(2 × 0.95 × 9.8 × 1.5) = √27.93 = 5.28 m s−¹ ≈ 5.3 m s−¹. (Verified.)

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?

ANSWER The sand leaks out vertically and carries away the same horizontal velocity as the trolley, exerting no horizontal force on it. The track is frictionless, so there is no external horizontal force on the system. Hence the horizontal momentum (and therefore the speed) of the trolley does not change. The trolley continues at its original speed = 27 km/h (no change). The leak rate of 0.05 kg s−¹ is irrelevant to the final speed.

5.20 A body of mass 0.5 kg travels in a straight line with velocity v = a x3/2 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?

ANSWER (step by step) By the work–energy theorem, W = ΔK = ½m(vf² − vi²). At x = 0: vi = a(0)3/2 = 0. At x = 2 m: vf = a(2)3/2 = 5 × 2√2 = 5 × 2.828 = 14.14 m s−¹. W = ½ × 0.5 × (14.14² − 0) = 0.25 × 200 = 50 J. (Verified.)

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?

ANSWER (step by step) (a) In time t the air travels a length vt, sweeping a cylinder of volume A·vt. Mass m = ρ × (A v t) = ρAvt. (b) Kinetic energy K = ½mv² = ½(ρAvt)v² = ½ρAv³t. (c) Convert v = 36 km/h = 10 m s−¹. KE per second (power in the wind) = ½ρAv³ = ½ × 1.2 × 30 × 10³ = 18000 W. Electrical power = 25% of this = 0.25 × 18000 = 4500 W = 4.5 kW. (Verified.)

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?

ANSWER (step by step) Given m = 10 kg, h = 0.5 m, number of lifts n = 1000, g = 9.8 m s−². (a) Work against gravity = n × mgh = 1000 × 10 × 9.8 × 0.5 = 49000 J. (b) Only 20% of the fat energy becomes mechanical work, so energy drawn from fat = 49000 / 0.20 = 2.45×10⁵ J. Fat used = energy from fat / energy per kg = 2.45×10⁵ / 3.8×10⁷ = 6.45×10−³ kg (about 6.45 g). (Verified.)

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.

ANSWER (step by step) Required power P = 8 kW = 8000 W. (a) Useful power per square metre = 20% of 200 W = 0.20 × 200 = 40 W m−². Area needed = P / (useful power per m²) = 8000 / 40 = 200 m². (b) 200 m² is comparable to the roof of a large house of dimensions about 14 m × 14 m. (Verified.)

Extra Practice Questions

Short Answer Type Questions

Q1. Define the work–energy theorem and state one situation where it is more convenient than Newton’s laws.

ANSWERThe work–energy theorem states that the net work done on a body equals the change in its kinetic energy, Kf − Ki = Wnet. It is more convenient when the force is variable or unknown but the change in speed (or work) is known — for example, finding the resistive force on a bullet from its loss of kinetic energy.

Q2. A 1 kg block moving at 4 m s−¹ compresses a spring of constant 100 N m−¹ on a frictionless surface. Find the maximum compression.

ANSWERAt maximum compression all KE becomes spring PE: ½mv² = ½kx². So x = v√(m/k) = 4 × √(1/100) = 4 × 0.1 = 0.4 m.

Q3. Why is the work done by a centripetal force on a body in uniform circular motion zero?

ANSWERThe centripetal force always points toward the centre, perpendicular to the velocity (and hence the displacement). Since θ = 90° and cos 90° = 0, the work done is zero, which is why the speed stays constant.

Q4. Distinguish between conservative and non-conservative forces with one example each.

ANSWERA conservative force (e.g. gravity, spring force) does path-independent work that is zero over a closed loop, and a potential energy can be defined for it. A non-conservative force (e.g. friction) does path-dependent work that is non-zero over a closed loop and dissipates mechanical energy as heat.

Q5. A 60 W bulb operates for 5 hours. Express the energy used in kWh and in joules.

ANSWEREnergy = power × time = 60 W × 5 h = 300 Wh = 0.3 kWh. In joules: 0.3 × 3.6×10⁶ = 1.08×10⁶ J.

Long Answer Type Questions

Q1. Derive the work–energy theorem for a constant force acting on a body moving in a straight line.

ANSWERFor rectilinear motion under constant acceleration a, v² − u² = 2as. Multiply both sides by m/2: ½mv² − ½mu² = mas. By Newton’s second law F = ma, so the right side is Fs, which is the work W done by the net force. Therefore ½mv² − ½mu² = W, i.e. Kf − Ki = Wnet. This is the work–energy theorem: the change in kinetic energy of a body equals the net work done on it. The same result can be generalised to a variable force by integrating dK = F dx between the initial and final positions.

Q2. Two bodies of masses m₁ and m₂ undergo a one-dimensional elastic collision, m₂ initially at rest. Derive the final velocities and discuss the cases of equal masses and a very heavy target.

ANSWERConserving momentum: m₁v1i = m₁v1f + m₂v2f. Conserving KE: m₁v1i² = m₁v1f² + m₂v2f². Solving these gives v1f = [(m₁−m₂)/(m₁+m₂)]v1i and v2f = [2m₁/(m₁+m₂)]v1i. Equal masses (m₁ = m₂): v1f = 0 and v2f = v1i — the first body stops and the second moves off with the incoming speed. Very heavy target (m₂ >> m₁): v1f ≈ −v1i and v2f ≈ 0 — the light body bounces straight back while the heavy body stays nearly at rest.

Q3. Explain the conservation of mechanical energy for a body falling freely from a height H, finding the speed at the ground.

ANSWERFor a body dropped from rest at height H, only gravity (a conservative force) acts, so K + V is constant. At the top: K = 0, V = mgH, so E = mgH. At a height h: E = mgh + ½mvh², giving vh² = 2g(H − h). At the ground (h = 0): all energy is kinetic, ½mvf² = mgH, so vf = √(2gH). Thus the potential energy at the top is converted fully into kinetic energy at the ground, and the total mechanical energy stays constant throughout the fall.

MCQs & Assertion–Reason

1. The work done by a force is zero when the angle between the force and displacement is:

(a) 0°    (b) 45°    (c) 90°    (d) 180°

2. The SI unit of power, the watt, is equal to:

(a) 1 J    (b) 1 J s−¹    (c) 1 J s    (d) 1 N m−¹

3. The dimensions of work and energy are:

(a) [MLT−²]    (b) [ML²T−²]    (c) [ML²T−³]    (d) [MLT−¹]

4. In a completely inelastic one-dimensional collision, the colliding bodies:

(a) bounce back    (b) move together after collision    (c) exchange velocities    (d) stop instantly

5. The elastic potential energy stored in a spring of constant k stretched by x is:

(a) kx    (b) ½kx    (c) ½kx²    (d) kx²

6. 1 kWh of energy is equal to:

(a) 3.6×10⁶ J    (b) 3.6×10³ J    (c) 1000 J    (d) 746 J

7. A body moves with constant power from rest; its displacement is proportional to:

(a) t    (b) t1/2    (c) t3/2    (d) t²

8. The work done by friction on a body moving over a closed loop is:

(a) zero    (b) positive    (c) negative (non-zero)    (d) infinite

9. In an elastic collision of two equal masses, one at rest, after collision the moving mass:

(a) keeps moving with the same speed    (b) comes to rest    (c) reverses direction    (d) speeds up

10. The kinetic energy of a body of momentum p and mass m is:

(a) p²/m    (b) p²/2m    (c) 2p²/m    (d) pm/2

Answer key: 1-(c), 2-(b), 3-(b), 4-(b), 5-(c), 6-(a), 7-(c), 8-(c), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: When you push hard against a rigid wall, you do no work on the wall.

Reason: Work done is zero if there is no displacement, however large the force.

A-R 2. Assertion: The total linear momentum is conserved in every collision.

Reason: Kinetic energy is also always conserved in every collision.

A-R 3. Assertion: The work done by a conservative force over a closed path is zero.

Reason: A conservative force can be derived from a potential energy function.

A-R 4. Assertion: A satellite spiralling toward Earth due to air drag speeds up.

Reason: The decrease in its gravitational potential energy exceeds the energy lost to drag, so its kinetic energy increases.

A-R 5. Assertion: kWh is a unit of power.

Reason: kWh equals 3.6×10⁶ joules.

Answer key: 1-(A), 2-(C), 3-(A), 4-(A), 5-(D).

Common Mistakes to Avoid

Watch out for these

  • Forgetting the cosθ factor — work is Fd cosθ, not always Fd; a perpendicular force does no work.
  • Treating kinetic energy as a vector — KE, work and power are all scalars (though work and power can be negative).
  • Assuming kinetic energy is conserved in every collision — only momentum is always conserved; KE is conserved only in elastic collisions.
  • Confusing kWh (a unit of energy) with kW (a unit of power).
  • Mixing units — convert km/h to m/s (36 km/h = 10 m/s) and mm to m before substituting.
  • Adding pump/efficiency wrongly — input power = output power ÷ efficiency, not × efficiency.
  • Forgetting that gravitational work depends only on height, so it is the same in each half of a fall regardless of changing speed.

How to score full marks in this chapter

Always write the relevant formula first, then substitute with units, and box the final answer with its unit. For sign questions, state the angle between force and displacement. For collision problems, write the momentum-conservation equation, and add the KE equation only if the collision is elastic. Use the work–energy theorem (ΔK = Wnet) whenever a force is variable or unknown — it often turns a hard problem into a one-line calculation. Convert all data to SI units before calculating, and remember 1 hp = 746 W and 1 kWh = 3.6×10⁶ J.

Frequently Asked Questions

What is Class 11 Physics Chapter 5 Work, Energy and Power about?

Chapter 5 defines work (W = F·d), kinetic and potential energy, and power (P = F·v). Its central idea is the work–energy theorem — the net work on a body equals its change in kinetic energy — followed by conservation of mechanical energy for conservative forces and the physics of elastic and inelastic collisions.

How many exercise questions are there in Chapter 5 and are all solved here?

There are 23 NCERT Exercise questions (5.1 to 5.23). Every one is reproduced verbatim and solved on this page, with each numerical worked out step by step and cross-checked against the official NCERT answer key.

State the work–energy theorem.

The work–energy theorem states that the change in kinetic energy of a body equals the net work done on it: Kf − Ki = Wnet. It holds for both constant and variable forces and in all inertial frames.

Are these Class 11 Physics Chapter 5 solutions free?

Yes. All ClearStudy NCERT Solutions for Class 11 Physics are free and follow the official NCERT textbook for the 2026–27 session.

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