Original practice sets for Work, Power and Energy are being prepared. For now, use the chapter notes for concept mastery and revision.
Start with short chapter-wise drills or take the full chapter bank in one run. These questions are original and aligned to JEE-style concept checks and numerical thinking.
1. Work done by a force F over displacement s at angle θ between them is:
Explanation: W = F·s = Fs cosθ. Work is the dot product of force and displacement vectors.
2. A force acts perpendicular to the displacement. Work done is:
Explanation: W = Fs cos90° = 0. A centripetal force, for example, does zero work on a body in circular motion.
3. When is work done by a force negative?
Explanation: W = Fs cosθ. When 90°
4. The kinetic energy of a body of mass m moving at speed v is:
Explanation: KE = ½mv². It is always positive (scalar quantity).
5. The work-energy theorem states:
Explanation: The net work done on a body equals its change in kinetic energy: W_net = KE_f − KE_i.
6. The gravitational potential energy of a mass m at height h (above reference) is:
Explanation: PE = mgh, taking the reference level as h = 0.
7. The elastic potential energy stored in a spring compressed by x from its natural length is:
Explanation: Elastic PE = ½kx², where k is the spring constant.
8. Power is defined as:
Explanation: Power P = W/t (average) or P = dW/dt (instantaneous) = F·v.
9. A conservative force is one for which:
Explanation: For a conservative force, the work done in moving between two points is path-independent. Gravity and spring force are conservative; friction is not.
10. Work done by a variable force F(x) as x goes from x₁ to x₂ is given by:
Explanation: W = ∫F·dx for a varying force. For a constant force, this reduces to F×Δx.
11. In the absence of non-conservative forces, the total mechanical energy (KE + PE):
Explanation: Conservation of mechanical energy: ΔKE + ΔPE = 0, so KE + PE = constant (when only conservative forces do work).
12. A car engine exerts force F = 2000 N and moves at v = 20 m/s. Instantaneous power is:
Explanation: P = F × v = 2000 × 20 = 40000 W = 40 kW.
13. A block slides 4 m on a rough surface (μₖ = 0.3, m = 5 kg, g = 10 m/s²). Work done by friction is:
Explanation: Friction force = μₖmg = 0.3×5×10 = 15 N. W_friction = −15×4 = −60 J (negative since friction opposes motion).
14. The SI unit of work and energy is:
Explanation: 1 Joule = 1 Newton × 1 metre = 1 kg·m²/s².
15. If momentum of a body is doubled, its kinetic energy becomes:
Explanation: KE = p²/(2m). If p → 2p, KE → (2p)²/(2m) = 4p²/(2m) = 4×KE.
16. A spring (k = 300 N/m) is stretched 0.2 m from its natural length. Work done in stretching it is:
Explanation: W = ½kx² = ½×300×0.04 = 6 J.
17. A 2 kg book is lifted from the floor (0 m) to a shelf 1.5 m high. Change in gravitational PE is: (g = 10 m/s²)
Explanation: ΔPE = mgh = 2×10×1.5 = 30 J.
18. A force 10 N makes 60° with displacement of 5 m. Work done is:
Explanation: W = Fs cosθ = 10×5×cos60° = 50×0.5 = 25 J.
19. One horsepower (HP) equals approximately:
Explanation: 1 HP = 746 W (approximately 0.746 kW). This is the mechanical horsepower used in engineering.
20. The work done by friction on a body moving along a closed path (returning to the starting point) is:
Explanation: Friction always opposes motion, so work done by friction is always negative regardless of path direction. Over a closed path, total work by friction
21. A ball is thrown with KE = 40 J at 45°. At the highest point, its KE is:
Explanation: At the highest point, vertical velocity = 0. Horizontal component: v_x = v cos45° = v/√2. KE = ½m(v/√2)² = ½mv²/2 = 40/2 = 20 J.
22. A 3 kg block slides from rest down a smooth incline of height 5 m. Speed at the bottom is: (g = 10 m/s²)
Explanation: Using energy conservation: ½mv² = mgh → v = √(2gh) = √(2×10×5) = √100 = 10 m/s.
23. A 1000 kg car accelerates from 0 to 20 m/s in 10 s on a frictionless road. Average power of the engine is:
Explanation: W = ΔKE = ½×1000×400 = 200000 J. P_avg = W/t = 200000/10 = 20000 W = 20 kW.
24. A spring (k = 200 N/m) is compressed 0.1 m and releases a 0.5 kg block from rest on a frictionless surface. Speed of block when spring returns to natural length:
Explanation: ½kx² = ½mv² → v = x√(k/m) = 0.1×√(200/0.5) = 0.1×20 = 2 m/s.
25. If force F = −dU/dx (in 1D), then for F = −kx (spring-like), the PE is:
Explanation: F = −dU/dx = −kx → dU = kx dx → U = ½kx² + C. Taking U=0 at x=0: U = ½kx².
26. A 2 kg block starts at rest and slides 5 m down a 30° rough incline (μₖ = 0.2). Speed at bottom: (g = 10, sin30° = 0.5, cos30° = 0.866)
Explanation: W_net = W_gravity + W_friction = mg sinθ × L − μₖmg cosθ × L = 2×10×0.5×5 − 0.2×2×10×0.866×5 = 50 − 17.32 = 32.68 J. v = √(2×32.68/2) = √32.68 ≈ 5.72 m/s. (checking: mgL sinθ = 50, friction work = 0.2×2×10×0.866×5 = 17.32, net = 32.68, v = √(32.68) ≈ 5.72)
27. In a perfectly elastic collision, which quantity/quantities are conserved?
Explanation: Elastic collision: no deformation or heat generation. Both momentum and kinetic energy are conserved.
28. A crane lifts a 500 kg load at 2 m/s at constant speed. Power of the crane motor is: (g = 10)
Explanation: P = Fv = mgv = 500×10×2 = 10000 W = 10 kW.
29. At a point where the PE curve has zero slope (dU/dx = 0), the force on the particle is:
Explanation: F = −dU/dx. Where the slope of the PE curve is zero, the force is zero — this corresponds to a point of equilibrium.
30. A block (mass m) compresses a spring (constant k) by d and is released on a smooth surface. It reaches height h on an incline. Which equation gives h?
Explanation: Spring PE = ½kd² is fully converted to gravitational PE at height h: ½kd² = mgh.
31. Work done by the normal force on a block sliding along a horizontal surface is:
Explanation: Normal force is perpendicular to the horizontal surface, hence perpendicular to displacement → W = Fs cos90° = 0.
32. A 4 kg object accelerates from 2 m/s to 5 m/s. Work done by net force is:
Explanation: W = ΔKE = ½×4×(25−4) = ½×4×21 = 42 J.
33. A car (m = 1000 kg) moving at 30 m/s is braked to rest. Braking force = 15000 N. Stopping distance is:
Explanation: W = ΔKE: −15000×d = 0 − ½×1000×900 = −450000. d = 450000/15000 = 30 m.
34. A particle moves from A to B in a gravitational field by two different paths (path 1: straight; path 2: curved). The work done by gravity:
Explanation: Gravity is a conservative force. Work done by a conservative force depends only on initial and final positions, not the path taken.
35. A person of mass 60 kg climbs stairs (height 3 m) in 6 s. Power output is: (g = 10)
Explanation: P = mgh/t = 60×10×3/6 = 1800/6 = 300 W.
36. The minimum kinetic energy needed to project a body of mass m from Earth's surface to infinity (escape) is:
Explanation: Escape velocity v_e = √(2gR). KE = ½mv_e² = ½m×2gR = mgR.
37. A body accelerates from rest with F = 3t N (where t is time in seconds). Mass = 2 kg. Power at t = 4 s is:
Explanation: a = F/m = 3t/2. v = ∫a dt = 3t²/4. At t = 4: F = 12 N, v = 12 m/s. P = Fv = 12×12 = 144 W. (Alternatively: v = ∫₀⁴(3t/2)dt = [3t²/4]₀⁴ = 12 m/s. P = 12×12 = 144 W.)
38. Two equal masses (each m) collide head-on. One moves at v, the other at rest. After perfectly inelastic collision, energy lost is:
Explanation: After collision: v_f = mv/(2m) = v/2. KE_before = ½mv². KE_after = ½(2m)(v/2)² = ¼mv². Energy lost = ½mv² − ¼mv² = ¼mv².
39. A spring is first compressed by x then by 2x. Ratio of work done (second to first compression) is:
Explanation: W₁ = ½kx², W₂ = ½k(2x)² = 2kx². W₂/W₁ = 2kx²/(½kx²) = 4. Ratio = 4:1.
40. Which of these is true for a body moving at constant velocity on a rough horizontal surface?
Explanation: At constant velocity, net work = 0 = W_applied + W_friction. So W_applied = −W_friction.
41. A ball starts from rest at height 20 m. What is the minimum height of the top of a loop (radius 5 m) to complete the loop without losing contact? (g = 10)
Explanation: At the top of the loop: minimum speed v_top = √(gR). Energy: mgh = mg(2R) + ½mv_top² = mg(2R) + ½mgR = mg(5R/2). h = 5R/2 = 12.5 m.
42. A 2 kg block moving at 4 m/s hits a spring (k = 200 N/m) attached to a wall. Maximum compression of the spring is:
Explanation: At maximum compression, all KE converts to spring PE: ½mv² = ½kx². x = v√(m/k) = 4×√(2/200) = 4×0.1 = 0.4 m.
43. A 5 kg object is lifted 3 m, then moved 4 m horizontally (frictionless). Total work done against gravity is: (g = 10)
Explanation: Work against gravity = mgh = 5×10×3 = 150 J. Horizontal movement does no work against gravity. Total = 150 J.
44. A force F = 2x N acts on a particle (x in metres). Work done from x = 0 to x = 4 m is:
Explanation: W = ∫₀⁴ 2x dx = [x²]₀⁴ = 16 J.
45. A pendulum (length L) is released from horizontal position. Speed at the lowest point is:
Explanation: Height drop = L (from horizontal to lowest). Energy conservation: ½mv² = mgL → v = √(2gL).
46. A car engine delivers constant power P on a level road. As speed increases, acceleration:
Explanation: P = Fv. At constant P, F = P/v decreases as v increases. Since F = ma, acceleration a = P/(mv) also decreases with speed.
47. A 1 kg block on a smooth surface is attached to a spring (k = 100 N/m). Initially compressed by 0.3 m and released. Maximum KE of the block is:
Explanation: Maximum KE = initial spring PE = ½kx² = ½×100×0.09 = 4.5 J (at the natural length of the spring).
48. Two forces act on a 3 kg block on a smooth surface: F₁ = 10 N (forward), F₂ = 4 N (backward). Over 5 m displacement, net work done is:
Explanation: Net force = 10 − 4 = 6 N. W_net = 6×5 = 30 J. (Or W₁ + W₂ = 50−20 = 30 J.)
49. The escape speed from the Moon's surface (radius Rm = 1.74×10⁶ m, g_moon = 1.63 m/s²) is approximately:
Explanation: v_escape = √(2g_moon×R_m) = √(2×1.63×1.74×10⁶) = √(5.67×10⁶) ≈ 2381 m/s ≈ 2.38 km/s.
50. A ball dropped from height H bounces back to height h. The coefficient of restitution is:
Explanation: Speed before impact: v₁ = √(2gH). Speed after: v₂ = √(2gh). e = v₂/v₁ = √(2gh)/√(2gH) = √(h/H).
51. A car of mass 1000 kg moves up a slope (sin θ = 0.02) at constant speed 72 km/h. Friction coefficient = 0.01. Engine power is: (g = 10)
Explanation: v = 20 m/s. Resistance = mg sinθ + μmg cosθ ≈ mg(0.02 + 0.01) = 1000×10×0.03 = 300 N. P = Fv = 300×20 = 6000 W. (Note: cosθ ≈ 1 for small θ.)
52. A 3 kg clay ball moving at 10 m/s sticks to a 7 kg stationary clay block. Energy lost in the collision is:
Explanation: v_f = 3×10/10 = 3 m/s. KE_before = ½×3×100 = 150 J. KE_after = ½×10×9 = 45 J. Loss = 150−45 = 105 J.
53. A particle has PE U(x) = 4x² − 2x. The equilibrium position and force at x = 1 m are:
Explanation: F = −dU/dx = −(8x − 2). At equilibrium, F = 0, so 8x = 2 and x = 0.25 m. At x = 1 m, F = −(8 − 2) = −6 N.
54. A block pushed up a rough incline (θ=30°, μₖ=0.3, d=5m, m=2kg). Work done by friction is: (g=10, sin30°=0.5, cos30°=0.866)
Explanation: Normal force = mg cosθ = 2×10×0.866 = 17.32 N. Friction = μₖN = 0.3×17.32 = 5.196 N. W_friction = −5.196×5 = −25.98 J (opposing upward motion).
55. Two blocks (2 kg and 3 kg) are connected by a spring (k = 100 N/m) on a frictionless surface. One is compressed towards the other by 0.2 m and released. Speed of the 2 kg block when spring regains natural length:
Explanation: PE = ½×100×0.04 = 2 J. Momentum: 2v₁ = 3v₂. Energy: ½×2×v₁² + ½×3×v₂² = 2. Solving: v₁² = 6/5, v₁ = √(6/5) ≈ 1.095 m/s.
56. Two equal masses (m) move toward each other at equal speeds v. The total KE in the centre of mass frame is:
Explanation: In the CM frame, both masses move at speed v (inward). KE = 2×½mv² = mv².
57. A skydiver (mass 70 kg) falls at terminal velocity 60 m/s. Air resistance equals weight. Power dissipated by air resistance is: (g = 10)
Explanation: At terminal velocity, F_air = mg = 700 N. P = F×v = 700×60 = 42000 W = 42 kW.
58. A spring (k = 50 N/m, natural length = 2 m) is attached to a wall. A block is pushed from 3 m to 1 m from the wall (compressing by 1 m). Work done by spring force on block:
Explanation: Spring does positive work when released from compression. W = ½k(x₁²−x₂²) from initial compression x₁ = 0 to x₂ = 1m: W_spring = ½×50×(0−1)² = −25 J (work done ON block pushing it away = +25 J in direction of motion... Spring force does positive work when compression decreases: 25 J).
59. A particle moves in a horizontal circle at constant speed. The work done by the centripetal force is:
Explanation: Centripetal force is always perpendicular to velocity (displacement direction). Work = F·ds = Fv cos90° = 0.
60. A block (mass m) slides off a smooth wedge (height H, angle θ, on frictionless floor). The block's speed when it leaves the wedge at the bottom is:
Explanation: On a frictionless wedge on a frictionless floor, the system's total energy is conserved. The block's PE = mgh is converted to its KE. Hence v = √(2gH) regardless of wedge angle.
61. A motor pumps water (density 1000 kg/m³) at 2 m³/min to a height of 5 m. Power needed (ignoring pipe friction): (g = 10)
Explanation: Mass flow rate = 1000 × 2/60 = 33.3 kg/s. Power = mgh/t = 33.3 × 10 × 5 = 1667 W.
62. A particle at the top of a PE energy hill (local maximum in U-x curve) is in:
Explanation: At a PE maximum: F = −dU/dx = 0 (equilibrium). A small displacement gives a force pushing the particle further away → unstable equilibrium.
63. Mass m₁ = m moving at v₀ collides elastically with m₂ = 3m at rest. Speed of m after collision:
Explanation: For elastic collision: v₁' = (m₁−m₂)v₀/(m₁+m₂) = (m−3m)v₀/(4m) = −v₀/2. Negative = reverses direction.
64. A block (mass m) is dropped from rest onto a vertical spring (k). The spring compresses maximally by d. Using energy conservation:
Explanation: Block falls height h above spring, then spring compresses by d. Total PE decrease = mg(h+d). PE stored in spring = ½kd². Setting equal: ½kd² = mg(h+d).
65. A resistor dissipates 100 W at voltage V. If voltage is doubled, power dissipated is:
Explanation: P = V²/R. If V → 2V: P → 4V²/R = 4×100 = 400 W.
66. A ball is projected horizontally with KE = E₀. At the lowest point of its trajectory (height H below launch): KE equals:
Explanation: By energy conservation: KE_final = KE_initial + ΔPE = E₀ + mgH.
67. A block-spring system oscillates. At equilibrium position, the PE is minimum and KE is maximum. If amplitude is A, the maximum KE is:
Explanation: At equilibrium, all energy is kinetic: KE_max = Total energy = ½kA².
68. A block slides along a closed path (loop) on a rough surface. Work done by friction is:
Explanation: Friction always opposes motion, so friction always does negative work. Over a closed path, total work by friction
69. A planet moves in an elliptical orbit around the Sun. As it moves from aphelion (farthest) to perihelion (nearest), the work done by gravity is:
Explanation: Moving from farther to nearer the Sun → force (gravity toward Sun) has a component along displacement → positive work by gravity → KE increases.
70. A machine does 500 J of useful work with 2000 J of input energy. Efficiency is:
Explanation: Efficiency = useful output/input × 100 = 500/2000 × 100 = 25%.
71. In a nuclear reaction, 0.1 g of mass is converted to energy. Energy released is: (c = 3×10⁸ m/s)
Explanation: E = mc² = 0.1×10⁻³ × (3×10⁸)² = 10⁻⁴ × 9×10¹⁶ = 9×10¹² J.
72. Two springs (k₁, k₂) are in parallel. Compressed by x. Total energy stored is:
Explanation: In parallel, both springs have the same compression x. Total PE = ½k₁x² + ½k₂x² = ½(k₁+k₂)x². Both expressions A and B are mathematically identical.
73. Mass m collides elastically with 2m at rest. Fraction of KE transferred to 2m is:
Explanation: v₂' = 2mv/(m+2m) = 2v/3. KE_transferred = ½(2m)(2v/3)² = ½×2m×4v²/9 = 4mv²/9. Fraction = (4mv²/9)/(½mv²) = 8/9.
74. In an oblique elastic collision of equal masses where the target is at rest, the angle between the velocities after collision is:
Explanation: For elastic collision of equal masses, one initially at rest: using conservation of momentum and energy, the two masses move at 90° to each other after the collision.
75. A particle in a 1D potential U(x) = x⁴ − 4x². The stable equilibrium positions are at:
Explanation: F = −dU/dx = −(4x³−8x) = 0 → x(4x²−8) = 0 → x=0 or x=±√2. U''(x) = 12x²−8. At x=±√2: U'' = 24−8 = 16 > 0 → stable. At x=0: U'' = −8
76. A mass m on a spring (k) on a surface inclined at θ (rough, μ). Energy dissipated per oscillation when amplitude reduces from A to A' is:
Explanation: Per half-cycle, friction force μmg cosθ acts over distance (A+A')/2... over a full cycle, displacement of the block is 2(A+A') approximately, and friction does work 2μmg cosθ×(A+A')/2×2... The energy dissipated per full cycle = 4μmg cosθ × A_avg. A more careful analysis gives the dissipation per cycle = 4μmg cosθ×A for amplitude A.
77. A rocket of initial mass M ejects gas at relative velocity u. When mass reduces to M/2, thrust force on rocket is equal to:
Explanation: Thrust = −u(dM/dt). Here dM/dt is the rate of mass ejection (negative since mass decreases). Thrust magnitude = u|dM/dt|, independent of current mass.
78. A bead slides on a frictionless wire from height h to height 0 (along any shape of wire). Speed at the bottom depends on:
Explanation: Only gravity (conservative) does work. By conservation of energy: v = √(2gh), independent of path shape or wire length.
79. For a conservative system, W_conservative = −ΔU. If a particle moves from A to B in a gravitational field and W_gravity = 30 J, the change in gravitational PE is:
Explanation: W_gravity = −ΔPE → ΔPE = −W_gravity = −30 J. PE decreases by 30 J (particle moved to lower height).
80. A ball (mass m) is thrown vertically with KE = 3mgR at Earth's surface (radius R). The maximum height above Earth's surface it reaches (using exact energy conservation with 1/r² gravity) is:
Explanation: KE = GMm/R − GMm/(R+h) → 3mgR = GMm[1/R − 1/(R+h)] = mgR[1 − R/(R+h)]. 3 = 1 − R/(R+h) → R/(R+h) = −2... This gives imaginary answer, suggesting enough KE to escape. The calculation for 3mgR versus escape KE (= mgR for surface approx.) indicates the exact calculation with h.
81. Force on a particle: F = (4 − 2x) N, where x is in metres. Work done from x = 0 to x = 3 m is:
Explanation: W = ∫₀³ (4−2x) dx = [4x − x²]₀³ = 12 − 9 = 3 J.
82. In a perfectly inelastic collision between m₁ = 2 kg (v = 6 m/s) and m₂ = 4 kg (v = 3 m/s in opposite direction), the final speed and energy lost are:
Explanation: Momentum: 2×6 − 4×3 = 12 − 12 = 0. Final velocity = 0. Energy lost = ½×2×36 + ½×4×9 = 36 + 18 = 54 J... Wait: KE_before = ½×2×36 + ½×4×9 = 36+18 = 54 J. KE_after = 0. Loss = 54 J. (Option C is correct.)
83. A motor rotates a wheel (I = 0.5 kg·m²) from 0 to 60 rad/s in 5 s uniformly. Average power of the motor is:
Explanation: Work done = ΔKE = ½×0.5×3600 = 900 J. Power = 900/5 = 180 W.
84. In a potential energy curve, a particle is said to be 'bound' when:
Explanation: A particle is bound if its total energy E is less than the potential energy at the boundaries — it cannot escape. If E > U at all points, it is unbound (can go to infinity).