NCERT Solutions for Class 11 Physics Chapter 5: Work, Energy and Power

Concept. Work done by a force F during a displacement d is W = F·d = Fd cosθ. The sign depends on the angle between the force and the displacement: positive if θ < 90° (force has a component along the displacement), negative if θ > 90° (force opposes the displacement), zero at 90°. A force that tends to speed the body up does positive work; one that slows it down does negative work.

1. (a) The bucket is lifted: the tension/pull acts upward and the displacement is upward, W = +Fd; this increases the PE (and KE) of the bucket.
2. (b) Gravity acts downward while the bucket moves upward: W = −mgh; the work done by the man is positive and the work done by gravity is negative (the net work is zero if the bucket is raised at constant speed).
3. (c) The kinetic friction on the sliding body acts up the incline, opposite to the displacement; W = −f s < 0 and it reduces the kinetic energy.
4. (d) Uniform velocity means zero net force: F_applied = f_friction. The applied force is along the displacement (positive work F s) and friction is opposite (negative work −f s); the sum is zero, consistent with ΔK = 0.
5. (e) Air resistance always opposes the velocity, so the work done is negative at all times and equal to the total energy lost; the oscillation dies down.

Check by energy: positive work increases the energy of the body (a, d: for the applied force), negative work decreases it (b, c, e).

Extra depth. The same force can do positive or negative work depending on the situation: friction usually does negative work on a moving body but does positive work on a box placed on an accelerating truck bed (static friction pulls it forward), and on a person walking (static friction on the foot is forward). Gravity does positive work when the body falls and negative work when it rises. For a force perpendicular to the displacement (centripetal force, the normal reaction on a horizontally moving body) the work is zero.

Conclusion: (a) positive; (b) negative; (c) negative; (d) positive; (e) negative.

Concept. First find the acceleration from the net force (applied force minus kinetic friction), then the displacement in 10 s from x = ½at² (starting from rest). The work done by each force is force × displacement × cosθ; the net work equals the change in kinetic energy (work–energy theorem). Take g = 9.8 m s⁻².

1. Mass m = 2 kg, F = 7 N, μ_k = 0.1, t = 10 s, u = 0.
2. Friction f_k = μ_k m g = 0.1 × 2 × 9.8 = 1.96 N (opposite to F).
3. Net force = 7 − 1.96 = 5.04 N ⇒ a = 5.04/2 = 2.52 m s⁻².
4. Displacement in 10 s: x = ½at² = ½ × 2.52 × 100 = 126 m. Final speed v = at = 25.2 m s⁻¹.
5. (a) W_applied = F x = 7 × 126 = 882 J.
6. (b) W_friction = −f_k x = −1.96 × 126 = −247 J (−246.96 J).
7. (c) W_net = F_net x = 5.04 × 126 = 635 J (= 882 − 247).
8. (d) ΔK = ½mv² − 0 = ½ × 2 × (25.2)² = 635 J.

Interpretation: (c) = (d): the work done by the net force equals the change in kinetic energy (the work–energy theorem). Of the 882 J supplied by the applied force, 247 J is dissipated as heat by friction and 635 J appears as kinetic energy of the body.

Extra depth. Energy bookkeeping: 882 J (input) = 635 J (KE gained) + 247 J (heat produced at the surfaces). The efficiency of the process in raising the kinetic energy is 635/882 = 72%. If the table were frictionless, a = 3.5 m s⁻², x = 175 m and the whole work W = 7 × 175 = 1225 J would appear as kinetic energy. The speed of 25.2 m s⁻¹ after 10 s is large for a real block on a table, but the problem is a numerical exercise on the work–energy theorem.

Conclusion: (a) 882 J; (b) −247 J; (c) 635 J; (d) 635 J: the net work equals the gain in kinetic energy.

Concept. The total energy of the particle is E = K + V(x) where K = ½mv² ≥ 0. So at any point the kinetic energy K = E − V(x) must be non-negative: the particle can be found only where V(x) ≤ E. The regions where V(x) > E are forbidden (classically) and the particle cannot enter them. The minimum possible total energy is the lowest value of V(x) in the accessible region, since K cannot be negative.

1. Draw the horizontal line E across the graph; the particle can exist only where the potential curve lies below this line.
2. At the points where V(x) = E the kinetic energy is zero: these are the turning points.
3. The minimum energy corresponds to K = 0 at the lowest point of V in the allowed region: E_min = V_min.

Remark on case (ii). The figure shows E lying below every shelf of the staircase potential, so for the energy marked no region is allowed. The particle can exist only if its energy is at least the potential of the lowest shelf (the left-hand step); the kinetic energy would then vanish there.

Extra depth. In classical mechanics the particle is strictly confined to the allowed regions. In quantum mechanics (beyond this course) there is a small probability to penetrate the forbidden regions (tunnelling), which explains alpha decay (case iv) and the operation of tunnel diodes. The potential-energy diagram is an excellent tool: slope of V = −force, and the particle is in stable equilibrium at the minima of V.

Conclusion: (i) forbidden x > a, E_min = 0; (ii) no allowed region for the energy shown; (iii) forbidden x < a and x > b, E_min = −V₁; (iv) forbidden a/2 < |x| < b/2, E_min = −V₁.

Concept. A particle ‘turns back’ where its velocity (and hence its kinetic energy) becomes zero. Since E = K + V, K = E − V(x) = 0 at the points where V(x) = E. Beyond these turning points V(x) > E would make K negative, which is impossible: the particle cannot go further.

1. Total energy E = 1 J (constant, since the spring force is conservative); V(x) = kx²/2 with k = 0.5 N m⁻¹, i.e. V(x) = 0.25 x² (J, with x in m).
2. Kinetic energy K = E − V(x) = 1 − 0.25x².
3. At a turning point K = 0 ⇒ 0.25 x² = 1 ⇒ x² = 4 ⇒ x = ± 2 m.
4. For |x| > 2 m: V = 0.25x² > 1 J = E, so K = E − V < 0, which is impossible. The particle therefore cannot go beyond x = ±2 m: it must stop momentarily and turn back there, since the restoring force F = −kx pulls it back towards the mean position.

Amplitude. The turning points x = ±2 m are the extreme positions of the motion, i.e. the amplitude is A = √(2E/k) = √(2 × 1/0.5) = 2 m. The maximum speed at x = 0 is v_max = √(2E/m) (the mass of the particle is not given).

Extra depth. This is the energy method for SHM: E = ½kA², so A = √(2E/k). Doubling the energy increases the amplitude by √2, not 2. The graph (Fig. 5.8) shows how kinetic and potential energies are complementary parabolas that exchange energy during the oscillation, their sum remaining E. The same argument gives the turning points of any one-dimensional motion in a potential V(x): solve V(x) = E.

Conclusion: At x = ±2 m the potential energy equals the total energy (1 J), so K = 0 and the particle must turn back; it cannot go beyond since K would be negative.

Concept. This exercise tests four central ideas: conservation of energy (a), the conservative nature of gravity (b), the relation between kinetic and potential energy in an orbit (c) and the definition of work as F·d (d). Answer each with a short explanation and the relevant formula.

(a) Burning of a rocket casing

The heat is obtained at the expense of the rocket. The friction (drag) between the air and the casing does negative work on the rocket, decreasing its kinetic energy; this lost mechanical energy of the rocket appears as heat in the casing, which burns up. The atmosphere does not supply the energy — its molecules gain energy from the rocket, not vice versa.

(b) Zero work by gravity over a complete orbit

The gravitational force is a conservative force: the work done by it depends only on the initial and final positions, not on the path, and in a closed path the initial and final positions coincide, so the net work is zero. For the elliptical orbit the force is not perpendicular to the velocity, so the work in individual parts of the orbit is not zero (positive when the comet approaches the Sun and speeds up; negative when it recedes and slows down), but the total over a complete revolution is zero: the gain and loss cancel.

(c) Satellite spirals inward and speeds up

For a satellite in a circular orbit of radius r: K = +GMm/2r, V = −GMm/r, so the total energy E = K + V = −GMm/2r = −K. When air drag decreases E (it becomes more negative), r decreases. Since K = −E, the kinetic energy and hence the speed increase as the total energy decreases: the potential energy decreases by twice as much as the total energy, the rest appears as increased kinetic energy. The drag does negative work, but gravity does more positive work as the satellite descends.

(d) Which man does more work?

1. Case (i): the man exerts an upward force mg = 15 × 9.8 = 147 N on the mass to support it, but the displacement (2 m) is horizontal, perpendicular to the force: W = F d cos90° = 0.
2. Case (ii): the rope is pulled horizontally with a force equal to the tension T = mg = 147 N (the hanging mass is raised at constant speed; the man’s force on the rope is along its displacement): W = T d = 147 × 2 = 294 J (positive work on the rope; it is used up in raising the hanging mass by 2 m: mgh = 15 × 9.8 × 2 = 294 J).
3. The work done is therefore greater in case (ii).

Extra depth. Statement (c) is a famous paradox: friction decreases the satellite’s energy, yet its speed rises. The resolution is that the drag is a small tangential force, while gravity (conservative) is the main force: the loss of total energy is converted by gravity into kinetic energy as r decreases (K = −E). In (d) the physical effort or muscular fatigue felt by the man in (i) is not ‘work’ in the physics sense: holding a weight uses chemical energy in muscles to maintain tension but no work is done on the load if there is no displacement along the force.

Conclusion: (a) at the expense of the rocket; (b) gravity is conservative; (c) E = −K so speed grows as E decreases; (d) work is greater in case (ii).

Concept. Each part tests a basic principle: (a) the relation W_conservative = −ΔV; (b) the nature of friction; (c) Newton’s second law for a system of particles (internal forces cancel in pairs); (d) conservation laws in collisions.

1. (a) Check with gravity: a ball falling from height h: gravity does +mgh; PE decreases from mgh to 0.
2. (b) Work done against friction = −(work by friction) > 0: ‘lost’ KE becomes thermal energy of the surfaces; for a block sliding on a table: ΔK = −f s.
3. (c) ΣF_i = F_ext + ΣF_internal = F_ext because ΣF_internal = 0; hence dP/dt = F_ext.
4. (d) Inelastic: P_i = P_f; E_total conserved; K_f < K_i.

Underlined choices: (a) decreases; (b) kinetic; (c) external force; (d) total linear momentum and total energy.

Extra depth. In (d) ‘total energy’ includes the heat and other forms into which kinetic energy is transformed, which is why energy is conserved even though the kinetic energy is not; in (b) ‘always’ refers to the mechanical energy of the sliding body (kinetic), since friction is a non-conservative dissipative force and the sliding body does not store energy in the form of potential energy. Part (c) is the foundation for the centre-of-mass motion in Chapter 6: the centre of mass moves as though all mass were concentrated there and acted on by the external force alone.

Conclusion: (a) decreases; (b) kinetic; (c) external force; (d) total linear momentum and total energy.

Concept. Conservation laws apply to a system as a whole, under stated conditions: momentum is conserved if there is no external force; the total energy is conserved for an isolated system; the work done around a closed loop vanishes only for conservative forces; and in an inelastic collision some kinetic energy is lost.

1. (a) Example: a moving ball hitting an equal stationary ball: after the collision the first stops (its momentum and KE change to 0) and the second moves with the original velocity; the system totals are conserved, individual values are not.
2. (b) Example: when you push a box, your muscles (external agent) do work on it and its energy increases; the energy of the box+floor is not constant unless you include the agent.
3. (c) Example: friction around a closed loop on a rough table: the work is −f × (path length) ≠ 0.
4. (d) K_f < K_i; the difference is dissipated.

Summary: (a) False, (b) False, (c) False, (d) True.

Extra depth. In (d) the exception is the so-called super-elastic collision, where internal energy (chemical or nuclear) is released and K_f > K_i (e.g. an exploding shell, nuclear reactions); in the NCERT context of a normal inelastic collision the answer is ‘true’. Energy conservation (the first law of thermodynamics) remains valid in all of them: energy is never created or destroyed, only transformed. The distinction between a system with and without external forces is the most important idea of the exercise.

Conclusion: (a) False; (b) False; (c) False; (d) True.

Concept. During the very short time of contact the balls are deformed; part of the kinetic energy is temporarily stored as elastic potential energy of deformation. Momentum, on the other hand, is a conserved quantity at every instant as long as no external force acts (the internal forces of the two balls on each other are equal and opposite).

1. Elastic collision timeline: before contact K₀; during contact K decreases while V (deformation) increases; at maximum compression K is smallest; as the balls separate V returns to 0 and K returns to K₀.
2. The total momentum (vector sum) never changes in any time interval: before, during and after.
3. In an inelastic collision the deformation is partly permanent (heat), so even when the balls separate, K_f < K₀.

Answer in one line each: (a) no; (b) yes; (c) no and yes; (d) elastic.

Extra depth. The statement in (d) is the general criterion: a collision is elastic when the interaction is described by a conservative force (a potential energy function of position). Contact forces between billiard balls are nearly conservative, so billiard collisions are nearly elastic; collisions between clay balls (non-conservative internal forces: plastic deformation) are inelastic. Atomic collisions (gas molecules) can be exactly elastic. The same argument shows why scattering of charged particles by Coulomb forces (a conservative interaction) is elastic.

Conclusion: (a) No; (b) Yes; (c) No and Yes; (d) elastic.

Concept. Instantaneous power delivered by a force is P = F v. For a body of mass m moving with constant acceleration a from rest: the force F = ma is constant and the velocity v = at grows linearly with time. So the power P = F v = ma × at = ma² t is proportional to t.

1. Initial velocity u = 0; constant acceleration a ⇒ v = u + at = at.
2. Force F = ma (constant, by the second law).
3. Power P = F v = (ma)(at) = m a² t ∝ t¹.
4. Hence the correct alternative is (ii) t.

Cross-check with energy: K = ½mv² = ½m a² t²; P = dK/dt = m a² t ✓. The average power over time t is P_av = K/t = ½ m a² t, half of the instantaneous power at the end: P_inst = 2P_av.

Extra depth. Contrast with Ex 5.10 where the power is constant: then F = P/v decreases as the speed increases and the acceleration is not constant (v ∝ t^1/2). In a car with a constant-torque engine, the power grows linearly with speed until the engine’s maximum power is reached. The result P ∝ t for constant a is the reason why vehicles need more power as the speed rises during acceleration.

More to know — average and instantaneous power. For motion from rest with constant acceleration, the work done up to time t is W = ½mv² = ½ m a² t², so the average power is P_av = W/t = ½ m a² t, while the instantaneous power at that time is P = m a² t = 2P_av. A 1000 kg car accelerating uniformly at 2 m s⁻² needs P = 1000 × 4 × t W: 40 kW after 10 s (v = 20 m s⁻¹), and 80 kW after 20 s (v = 40 m s⁻¹). This explains why a car has limited acceleration at high speed: the engine’s maximum power is reached.

Conclusion: The power is proportional to t: alternative (ii).

Concept. Constant power means that the rate of doing work is constant: P = dK/dt = constant. Since K = ½mv², we first find the speed as a function of time and then integrate it to get the displacement.

1. Power P is constant: the work done in time t is W = Pt. By the work–energy theorem (starting from rest) ½mv² = Pt.
2. Therefore v = √(2Pt/m) ∝ t^1/2.
3. Displacement x = ∫v dt = √(2P/m) ∫t^1/2 dt = √(2P/m) (2/3) t^3/2, so x ∝ t^3/2.
4. The correct alternative is (iii) t^3/2.

Check by the formula: x = (2/3)√(2P/m) t^3/2. For P = 1000 W, m = 100 kg, t = 10 s: v = √(2 × 1000 × 10/100) = 14.1 m s⁻¹ and x = (2/3)(4.47)(31.6) = 94 m.

Extra depth. In this case the acceleration is not constant (a ∝ t^-1/2), so the kinematic equations of Chapter 2 cannot be used; the work–energy theorem gives the answer in one step. For constant force (Ex 5.9) the speed is ∝ t, the displacement ∝ t² and the power ∝ t; for constant power the speed is ∝ t^1/2 and the displacement ∝ t^3/2. A car moving with its engine at constant power accelerates strongly at low speeds and gently at high speeds.

More to know — constant-power motion of a vehicle. For a vehicle of power P moving against negligible resistance, F = P/v: at low speed the force (and acceleration) is large, at high speed small. In practice resistive forces (air drag ∝ v²) set a maximum speed v_max where the driving force P/v equals the drag force: for a car with P = 50 kW and drag coefficient such that the force at 30 m s⁻¹ is 1000 N the balance occurs at about 50 m s⁻¹ (P = Fv). The exercise ignores drag; with drag the displacement grows more slowly than t^3/2 at late times.

Conclusion: The displacement is proportional to t^3/2: alternative (iii).

Concept. Work done by a constant force is the dot product of the force and the displacement vector, W = F·d. The body is constrained to move along the z-axis, so the displacement is d = 4 k̂ m; only the z-component of the force does work, because the dot product of k̂ with î and ĵ is zero.

1. F = (−1)î + 2ĵ + 3k̂ N; displacement d = 0î + 0ĵ + 4k̂ m.
2. W = F·d = (−1)(0) + (2)(0) + (3)(4) = 12 J.
3. The x- and y-components of F (−1 N and 2 N) are perpendicular to the displacement and do no work; they are balanced by the constraint (the guide/rail that keeps the body on the z-axis).

Check: W = F cosθ d with F = √(1 + 4 + 9) = √14 = 3.74 N and cosθ = F_z/F = 3/3.74 = 0.80: W = 3.74 × 0.80 × 4 = 12 J ✓.

Extra depth. The constraint forces (normal reactions from the guide) are perpendicular to the allowed displacement and therefore do no work; they simply cancel the x and y components of F. The work is positive because the z-component of the force is along the displacement; if the body had moved 4 m along the negative z-axis the work would be −12 J.

More to know — work with constraints. When a body is constrained to a line, rail or surface, the constraint (normal) force is perpendicular to the allowed displacement and does no work. For F = (−1, 2, 3) N and a displacement d = (0, 0, 4) m, W = F·d = 12 J; if the body moved instead 4 m along the x-axis the work would be −4 J, and along y it would be +8 J. The work done by a constant force depends only on the displacement vector, not on the path: W = F·(r_f − r_i) for any route between the end points, because a constant force is conservative.

Conclusion: The work done is 12 J.

Concept. From K = ½mv² the speed is v = √(2K/m). The particle with the smaller mass has the larger speed for a given energy, and the electron is also ~1836 times lighter than the proton. Convert the kinetic energies to joules before using the formula (1 keV = 10³ eV = 1.60 × 10⁻¹⁶ J).

1. Electron: K_e = 10 keV = 10 × 10³ × 1.60 × 10⁻¹⁹ J = 1.60 × 10⁻¹⁵ J; m_e = 9.11 × 10⁻³¹ kg.
2. v_e = √(2K_e/m_e) = √(2 × 1.60 × 10⁻¹⁵/9.11 × 10⁻³¹) = √(3.51 × 10¹⁵) = 5.93 × 10⁷ m s⁻¹.
3. Proton: K_p = 100 keV = 1.60 × 10⁻¹⁴ J; m_p = 1.67 × 10⁻²⁷ kg.
4. v_p = √(2K_p/m_p) = √(2 × 1.60 × 10⁻¹⁴/1.67 × 10⁻²⁷) = √(1.916 × 10¹³) = 4.38 × 10⁶ m s⁻¹.
5. Ratio v_e/v_p = 5.93 × 10⁷/4.38 × 10⁶ = 13.5.

Result: the electron is faster, about 13.5 times as fast as the proton, even though its kinetic energy is ten times smaller. Check of the ratio: v_e/v_p = √[(K_e/K_p)(m_p/m_e)] = √[(1/10)(1.67 × 10⁻²⁷/9.11 × 10⁻³¹)] = √(0.1 × 1833) = √183.3 = 13.5 ✓.

Extra depth. The electron’s speed, 5.9 × 10⁷ m s⁻¹, is about 20% of the speed of light (3 × 10⁸ m s⁻¹): relativistic corrections are small but not negligible at 10 keV (the correction to the kinetic energy is a few percent), and the non-relativistic formula K = ½mv² is used here as an approximation as in the NCERT answer. For the proton the speed is only 1.5% of c. The result illustrates that at the same kinetic energy the lighter particle is faster by √(m_p/m_e) = 42.8.

Conclusion: The electron is faster: v_e = 5.9 × 10⁷ m s⁻¹, v_p = 4.4 × 10⁶ m s⁻¹, ratio v_e/v_p = 13.5.

Concept. Gravity is a constant force mg on the drop, so the work done by gravity in falling a height h is simply mgh, independent of the resistive force. The work done by the resistive force is found from the work–energy theorem: W_gravity + W_resistive = ΔK.

1. Radius r = 2 mm = 2 × 10⁻³ m; volume V = (4/3)πr³ = (4/3)(3.14)(2 × 10⁻³)³ = 3.35 × 10⁻⁸ m³.
2. Mass m = density × volume = 10³ kg m⁻³ × 3.35 × 10⁻⁸ m³ = 3.35 × 10⁻⁵ kg; weight mg = 3.35 × 10⁻⁵ × 9.8 = 3.28 × 10⁻⁴ N.
3. First half (250 m): W₁ = mgh = 3.35 × 10⁻⁵ × 9.8 × 250 = 0.082 J.
4. Second half (250 m): the displacement is the same and gravity is the same: W₂ = 0.082 J (the work by gravity is the same in the two halves even though the speed pattern differs).
5. Total work by gravity W_g = 0.164 J. Kinetic energy on reaching the ground K_f = ½mv² = ½ × 3.35 × 10⁻⁵ × 10² = 1.68 × 10⁻³ J; K_i = 0.
6. WE theorem: W_g + W_r = K_f − K_i ⇒ W_r = 1.68 × 10⁻³ − 0.164 = −0.162 J (negative: the resistive force opposes the motion).

Energy interpretation. Almost all (99%) of the 0.164 J of gravitational work is dissipated as heat in the air by the viscous drag; only 1.7 mJ remains as kinetic energy. In the second half of the journey the drop moves with terminal speed (K constant) so the work done by gravity (0.082 J) is exactly cancelled by the resistive force (−0.082 J).

Extra depth. If there were no air resistance, the drop would reach the ground with speed √(2gh) = 99 m s⁻¹, K = 0.164 J and could damage crops and roofs; it is the drag that reduces this speed to only 10 m s⁻¹ (terminal speed). Terminal speed occurs when the resistive force equals the weight (net force zero, zero acceleration). The assumption of the problem that the terminal speed is attained at half the height is used only for the statement that the drop moves uniformly afterwards; it does not affect the work done by gravity.

Conclusion: Work by gravity: 0.082 J in each half; work by the resistive force over the whole journey: −0.162 J.

Concept. In the collision the molecule exerts a force on the wall and the wall on the molecule (third law). The wall is attached to the container (and hence to the Earth), so its mass is effectively infinite. Momentum is conserved for the system molecule + wall (+ container); the kinetic energy is judged by the speed of the molecule before and after.

1. Let the wall be horizontal (the normal is vertical). Before the collision the velocity components of the molecule are: normal v cos30° (towards the wall) and tangential v sin30°. After the collision, the speed is the same (200 m s⁻¹) and the angle with the normal is 30°: the normal component reverses (−v cos30°) and the tangential component is unchanged.
2. Change in momentum of the molecule: Δp = 2mv cos30° (along the normal, away from the wall), so the momentum of the molecule alone is not conserved.
3. By the third law the wall receives an equal and opposite momentum 2mv cos30° (towards the wall). The total momentum of molecule + wall is conserved. Because the wall (with the container and the Earth) has a huge mass, its velocity change is negligible (momentum p/M ≈ 0), and its kinetic energy change is negligible.
4. Kinetic energy of the molecule before and after: ½mv² is the same (same speed), so there is no loss of kinetic energy: the collision is elastic.

Answer: Yes, the momentum of the system (molecule + wall) is conserved; the collision is elastic since the speed (kinetic energy) of the molecule is unchanged.

Extra depth. This simple picture is the basis of the kinetic theory of gases (Chapter 12): each elastic collision transfers momentum 2mv_x to the wall, and the sum over a huge number of molecules per second gives the pressure of the gas. The wall does not move because its mass is huge, and the tangential component of the velocity is unaltered because the wall is smooth (no tangential force); a rough wall would change it (a non-elastic effect).

Conclusion: The momentum of the molecule+wall system is conserved (the molecule alone changes momentum by 2mv cos30°); the collision is elastic.

Concept. The pump does work against gravity, raising the water through 40 m. The useful (output) power is the work done per second on the water; the electric (input) power consumed is larger because the efficiency is only 30%: efficiency = output power/input power.

1. Volume V = 30 m³ ⇒ mass of water m = 30 × 10³ kg (density 10³ kg m⁻³) = 3.0 × 10⁴ kg.
2. Work done to lift the water through h = 40 m: W = mgh = 3.0 × 10⁴ × 9.8 × 40 = 1.176 × 10⁷ J.
3. Time t = 15 min = 900 s: output (useful) power P_out = W/t = 1.176 × 10⁷/900 = 1.31 × 10⁴ W.
4. Efficiency = 30% = 0.30 = P_out/P_in ⇒ P_in = 1.31 × 10⁴/0.30 = 4.36 × 10⁴ W ≈ 43.6 kW.

Energy consumed: 43.6 kW × (15/60) h = 10.9 kWh per tank filling; at an electricity cost of ₹6 per kWh this would be about ₹65.

Extra depth. The efficiency of a machine is the ratio of useful output to input: η = P_out/P_in. Here 70% of the electric power is lost as heat in the motor and friction in pipes and bearings. The kinetic energy of the water leaving the pipe is neglected in this calculation (the water is delivered slowly into the tank). If the tank were at 80 m, the power would double; if the pump efficiency were 60% it would halve.

More to know — efficiency and power ratings. A pump is rated by the electrical power it draws (43.6 kW), but only the useful hydraulic power (13.1 kW) lifts the water. Efficiency η = useful output/input × 100%. Typical efficiencies: electric motor 70–95%, pump 50–80%, human muscle about 20%, petrol engine about 25–30%, power station about 35–40%. For the pump of the exercise, an efficiency of 70% would reduce the electrical power to 18.7 kW (about 7.5 kWh of energy for one filling of the tank).

Conclusion: The pump consumes about 43.6 kW (4.4 × 10⁴ W) of electric power.

Concept. An elastic collision must conserve both the total momentum and the total kinetic energy. Before the collision ball 1 (mass m) moves with speed V and balls 2 and 3 are at rest: p_i = mV and K_i = ½mV². Check each alternative of Fig. 5.14 against both conditions.

1. Momentum is conserved in all three alternatives, so it cannot discriminate.
2. Kinetic energy: only (ii) conserves it: K_f = ½mV² = K_i.
3. Hence only (ii) is possible for an elastic collision. Physically, the moving ball transfers all its momentum and energy to the first ball at rest, which passes it on through the chain to ball 3 (Newton’s cradle): the middle ball stays at rest.

Why not (i)? (i) would require 50% of the kinetic energy to disappear; (iii) would require 83% to disappear. Such collisions occur if the balls stick together (completely inelastic collisions), not for steel ball bearings.

Extra depth. The result is general: for equal masses in a head-on elastic collision the velocities are exchanged (v_1f = 0, v_2f = V). In a chain of identical balls the ‘exchange’ is passed along until the last ball moves off with V (Newton’s cradle). One equation (momentum) leaves many possible outcomes; the second equation (energy) selects the unique result: elastic ⇒ two equations ⇒ unique solution for the two unknown speeds.

Conclusion: Only (ii) is possible: balls 1 and 2 at rest and ball 3 moving with speed V.

Concept. Two steps: (1) the bob A swings down from rest at 30° and reaches the lowest point with speed v given by energy conservation; (2) at the lowest point it strikes bob B (same mass, at rest) in a head-on elastic collision, in which equal masses exchange velocities.

1. Speed of A at the bottom: ½mv² = mgL(1 − cos30°) ⇒ v = √[2gL(1 − cos30°)] = √(0.268 gL).
2. Elastic head-on collision of two equal masses, B initially at rest: v_Af = (m − m)/(m + m) v = 0 and v_Bf = 2m/(2m) v = v. A stops and B moves off with A’s speed v.
3. Since A is at rest at the lowest point after the collision, it has no kinetic energy to rise: A does not rise at all (height = 0); it stays at rest at the lowest point.
4. Bob B slides along the table with speed v = √(0.268 gL) (e.g. for L = 1 m, v = 1.62 m s⁻¹).

Check by conservation: momentum before = mv; after = m × 0 + m × v = mv ✓; kinetic energy before = ½mv², after = ½mv² ✓.

Extra depth. In this arrangement (the first bob at the lowest point of its arc, the second on the table at the same level) the collision happens at the bottom. For unequal masses A would rebound or continue forward with v_Af = (m_A − m_B)v/(m_A + m_B) and rise to a height h = v_Af²/2g. The perfect transfer of motion between equal masses is the principle behind Newton’s cradle and billiards (a straight-on shot stops the cue ball dead).

Conclusion: Bob A does not rise (height 0); it comes to rest at the lowest point and B moves off with A’s speed.

Concept. A bob released from a horizontal position falls through a height equal to the length L of the pendulum when it reaches the lowest point. Without friction all of its potential energy mgL would become kinetic energy; with air resistance 5% of the initial energy is dissipated, so only 95% is converted to kinetic energy.

1. Initial potential energy (taking the lowest point as zero): E_i = mgL, with L = 1.5 m, g = 9.8 m s⁻².
2. Energy dissipated = 5% of E_i; hence the kinetic energy at the bottom K = 0.95 mgL.
3. ½mv² = 0.95 mgL ⇒ v² = 1.9 gL = 1.9 × 9.8 × 1.5 = 27.93 m² s⁻².
4. v = √27.93 = 5.3 m s⁻¹.

Check: v_ideal = √(2 × 9.8 × 1.5) = √29.4 = 5.42 m s⁻¹; the actual speed is √0.95 = 97.5% of this value: 5.42 × 0.975 = 5.29 m s⁻¹ ✓.

Extra depth. The speed is independent of the mass of the bob (m cancels). The 5% loss of energy changes the speed by only 2.5% because v ∝ √E. The work done by air resistance is −0.05 mgL, so the work–energy theorem reads mgL − 0.05mgL = ½mv². The tension in the string does no work because it is always perpendicular to the displacement.

More to know — speed at the lowest point with different losses. v = √[2gL(1 − f)] where f is the fraction of energy lost. With f = 0, 5%, 10%, 20%, the speeds are 5.42, 5.28, 5.14 and 4.85 m s⁻¹, so a 5% energy loss reduces the speed by 2.5% only. The tension at the lowest point is T = mg + mv²/L = mg(1 + 2(1 − f)) = 2.9mg for f = 0.05, which is why a rope must withstand almost three times the weight. The same energy method is used for roller coasters, swings and ski jumps.

Conclusion: The bob reaches the lowest point with a speed of about 5.3 m s⁻¹.

Concept. The track is frictionless, so no external horizontal force acts on the trolley or the sand. The sand that leaks out leaves the trolley with the same horizontal velocity as the trolley at that moment (it falls vertically relative to the trolley, with no horizontal push), and therefore carries away exactly its share of the horizontal momentum, with no change in the velocity of the remaining system.

1. Initial speed v = 27 km h⁻¹ = 27 × 5/18 = 7.5 m s⁻¹.
2. Consider the trolley and the sand in it at some instant. A small mass Δm of sand falls through the hole with horizontal velocity v (same as the trolley). Momentum conservation in the horizontal direction (no external horizontal force): (M + m)v = Δm v + (M + m − Δm) v′ ⇒ v′ = v.
3. The horizontal speed of the trolley therefore does not change during the leakage; no force acts on the trolley to alter its speed (the sand leaves without pushing it).
4. After all the sand has leaked out the speed of the trolley is the same as before: 7.5 m s⁻¹ = 27 km h⁻¹.

Time for the sand to run out: 25 kg ÷ 0.05 kg s⁻¹ = 500 s (about 8 min 20 s); the trolley covers 7.5 × 500 = 3750 m in that time. The momentum of the trolley decreases (mass falls) but the sand carries away the lost momentum.

Extra depth. Note the contrast with a rocket, where the exhaust leaves the vehicle with a backward velocity relative to it and so changes the velocity of the vehicle. Here the sand falls with zero velocity relative to the trolley (horizontally), so no thrust arises. A common wrong answer is to apply conservation of momentum to the trolley alone (300 × v′ = 325 × 7.5) which would give v′ = 8.1 m s⁻¹: this is wrong because the sand also carries momentum and remains in motion with the original speed.

Conclusion: The speed of the trolley remains 27 km/h (7.5 m s⁻¹) after all the sand has leaked out.

Concept. The velocity depends on position as v = a x^3/2. We do not need the force: by the work–energy theorem the work done by the net force equals the change in kinetic energy, W = K_f − K_i = ½m(v_f² − v_i²).

1. At x = 0: v_i = a(0)^3/2 = 0, so K_i = 0.
2. At x = 2 m: v_f = a x^3/2 = 5 × (2)^3/2 = 5 × 2.828 = 14.14 m s⁻¹; v_f² = a² x³ = 25 × 8 = 200 m² s⁻².
3. K_f = ½ m v_f² = ½ × 0.5 × 200 = 50 J.
4. W = K_f − K_i = 50 − 0 = 50 J.

Cross-check by integrating the force. From v² = a²x³: F = m v dv/dx = m (½ d(v²)/dx) = ½ m a² (3x²) = (3/2)(0.5)(25)x² = 18.75 x² N. W = ∫₀² 18.75x² dx = 18.75 × (8/3) = 50 J ✓.

Extra depth. This is a nice example of a position-dependent force (F ∝ x²): the work is the area under the F–x curve, the integral of 18.75x² from 0 to 2. The work–energy theorem avoids the integral and the need to know the force, which is the main advantage of energy methods. Dimensional check: [a] = m^-1/2 s⁻¹, [a x^3/2] = m s⁻¹ = velocity ✓.

More to know — force and work for a position-dependent velocity. From v = a x^3/2: a (the acceleration) = v dv/dx = a² x^3/2 × (3/2)x^1/2 = (3/2)a² x² = 37.5 x² m s⁻², so the force F = 0.5 × 37.5x² = 18.75x² N (zero at x = 0, 75 N at x = 2 m): the force grows rapidly with displacement. The work 50 J equals the area under F–x: ∫18.75x² dx = 6.25x³ = 50 J. At x = 2 m the body is moving at 14.1 m s⁻¹ and the power delivered is P = Fv = 75 × 14.1 = 1060 W.

Conclusion: The work done by the net force is 50 J.

Concept. The air that passes through the blade area A in time t is contained in a cylinder of base area A and length vt (the distance moved by the air in time t). Its mass is the density times that volume; its kinetic energy is ½mv² (all the air moves with speed v). The power is energy per unit time; the electrical power is the fraction (25%) converted.

1. (a) Mass: volume of air passing in time t is V = A v t; mass m = ρV = ρ A v t.
2. (b) Kinetic energy: K = ½ m v² = ½ (ρ A v t) v² = ½ ρ A v³ t.
3. Power available in the wind = K/t = ½ ρ A v³ (proportional to v³ and to the swept area).
4. (c) Numbers: v = 36 km h⁻¹ = 10 m s⁻¹; A = 30 m²; ρ = 1.2 kg m⁻³. Power in the wind P = ½ × 1.2 × 30 × (10)³ = 1.8 × 10⁴ W = 18 kW.
5. Electrical power produced = 25% of 18 kW = 0.25 × 1.8 × 10⁴ = 4.5 × 10³ W = 4.5 kW.

Cube law. The power is proportional to v³: doubling the wind speed to 72 km h⁻¹ gives eight times the power (36 kW electrical); that is why wind farms are placed where the wind is strong and steady. (The Betz limit of 59% is the theoretical maximum fraction extractable; the 25% efficiency here is realistic.)

Extra depth. The method of ‘volume swept in time t’ is a general technique (flow of air, water, electric charge): mass flow rate = ρAv. The same calculation is used for hydroelectric power with a dam, where the energy is potential. In the windmill the wind slows down after passing the blades, so not all kinetic energy can be extracted; the 25% efficiency of the question accounts for this and for mechanical and electrical losses. 4.5 kW can supply a few households (average Indian house about 0.3–0.5 kW).

Conclusion: (a) ρAvt; (b) ½ρAv³t; (c) 4.5 kW of electrical power.

Concept. Lifting the mass raises its potential energy by mgh each time; this is the work she does against gravity. When she lowers the mass the potential energy is dissipated (not recovered), so every lift needs fresh energy. For the second part, only 20% of the chemical energy supplied by fat appears as mechanical work.

1. (a) Work in one lift = mgh = 10 × 9.8 × 0.5 = 49 J. For 1000 lifts: W = 1000 × 49 = 4.9 × 10⁴ J (49 kJ).
2. (b) The mechanical work 4.9 × 10⁴ J is 20% of the energy drawn from fat: energy from fat E = W/0.20 = 4.9 × 10⁴/0.20 = 2.45 × 10⁵ J.
3. Mass of fat used = E/(energy per kg) = 2.45 × 10⁵/3.8 × 10⁷ = 6.45 × 10⁻³ kg = 6.45 g (about 6.5 g).

Perspective. 6.45 g of fat is a very small amount: to lose 1 kg of fat the dieter would have to lift the mass about 155 times as many repetitions (155,000 lifts), which shows how little weight is lost by such exercise alone; the remaining 80% of the energy appears as heat in the body.

Extra depth. The result also illustrates the difference between the chemical energy consumed (245 kJ, about 59 kcal) and the useful work done (49 kJ): the 20% efficiency is typical of human muscles. Because the lowering of the mass dissipates the PE as heat inside the body and does not return energy, each repetition costs the same mgh. Dietary energy (1 kg of fat ≈ 9000 kcal = 3.8 × 10⁷ J) is huge compared with ordinary physical exertion, which is why exercise alone burns fat slowly.

More to know — food energy. The energy value of fat is about 3.8 × 10⁷ J per kg (9 kcal per g); carbohydrates and proteins give about 1.7 × 10⁷ J per kg (4 kcal per g). A person doing the lifting exercise (49 kJ of work) actually uses 245 kJ of chemical energy — about the energy in two slices of bread. The rest of the energy is released as heat, which is why exercise makes us warm and sweat. A daily food intake of about 2000–2500 kcal (8–10 MJ) is needed by an adult, far more than the work of this exercise.

Conclusion: (a) 4.9 × 10⁴ J; (b) about 6.45 × 10⁻³ kg (6.45 g) of fat.

Concept. The solar power that can be converted into useful electrical energy per unit area is the incident power per unit area times the efficiency. The area required is the power needed divided by the useful power per square metre.

1. Incident solar power per m² = 200 W m⁻²; efficiency 20% = 0.20.
2. Useful electrical power per m² = 0.20 × 200 = 40 W m⁻².
3. Required area A = P/(power per m²) = 8000 W / 40 W m⁻² = 200 m².
4. (b) A typical house roof is of area about 100 m² (for example, 10 m × 10 m). So the solar panel area needed (200 m²) is about twice the area of an average roof: a rooftop alone is not enough to supply 8 kW all the time (and sunshine is available only part of the day).

Practical considerations. Solar energy is available only in the daytime (about 5–6 peak-sun hours per day) and varies with weather, so batteries or a grid connection are needed; the average power of a family is usually lower than the peak 8 kW. Also, 200 W m⁻² is an average over day and night; at noon the incident power may be about 1000 W m⁻².

Extra depth. The comparison shows both the promise and the limitation of solar energy: the energy is free and clean but dilute, so large collecting areas are needed (1 m² produces only 40 W, i.e. less than a small bulb). Improvements in efficiency (e.g. 20% to 30%) reduce the area; rooftop systems in India typically provide 1–5 kW for houses, so an 8 kW demand needs the roof plus extra area or demand reduction.

More to know — solar constant. Above the atmosphere the solar power per square metre perpendicular to the rays is 1.36 kW m⁻² (the solar constant); at the ground, at noon on a clear day it is about 1 kW m⁻², and the average over a 24-hour day is about 200 W m⁻² (as used in the exercise). With 20% efficient photovoltaic panels each square metre supplies about 40 W on average, or about 1 kWh per day; a house using 8 kW continuously (192 kWh per day) would need about 200 m² of panels, as found. Concentrating collectors and higher efficiencies (30–40% for multi-junction cells) can reduce this area.

Conclusion: (a) 200 m²; (b) this is about twice the roof area of a typical house (about 100 m²).