Take session-wise tests or a full 100-question mock on Work, Energy & Power for NEET — with 90-second per-question timer, answer review, subtopic breakdown, and detailed solutions.
4 sectional sessions (25 Qs each) + 1 Full Mock (100 Qs). NEET 4/−1 marking. 90 sec per question timer.
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1. Work done by a force is zero when the angle between force and displacement is:
Explanation: W = Fs cosθ. When θ = 90° , cos90° = 0, so W = 0.
2. SI unit of work is:
Explanation: Work = Force × Displacement. SI unit is Joule (J) = N·m.
3. A force F = 5î + 3ĵ N displaces a body by d = 2î + 4ĵ m. Work done is:
Explanation: W = F·d = (5)(2) + (3)(4) = 10 + 12 = 22 J .
4. Work done by gravity on a body moving horizontally is:
Explanation: Gravity acts vertically; horizontal displacement is perpendicular. W = 0.
5. A 10 kg block is dragged 5 m by 20 N force at 60° to horizontal. Work done is:
Explanation: W = Fd cosθ = 20 × 5 × cos60° = 100 × 0.5 = 50 J .
6. Variable force F = 3x² N acts along x-axis. Work done from x = 1 m to x = 3 m:
Explanation: W = ∫₁³ 3x² dx = [x³]₁³ = 27 − 1 = 26 J .
7. Work done by friction on a block sliding on a rough surface is:
Explanation: Friction opposes motion (180° angle). W = Fd cos180° = negative .
8. Force F = (2x + 3) N acts along x-axis. Work done from x = 0 to x = 2 m:
Explanation: W = ∫₀²(2x+3)dx = [x²+3x]₀² = 4+6 = 10 J .
9. A coolie carrying load on head walking on level road does ______ work against gravity:
Explanation: Displacement is horizontal; gravity is vertical — perpendicular. Work = 0 .
10. Work done lifting 2 kg mass through 3 m at constant velocity (g = 10 m/s²):
Explanation: W = mgh = 2 × 10 × 3 = 60 J .
11. Kinetic energy of mass m moving with velocity v is:
Explanation: KE = ½mv² .
12. If speed of a body doubles, its kinetic energy becomes:
Explanation: KE ∝ v². If v → 2v, KE → 4 times .
13. Work-energy theorem states that net work done equals change in:
Explanation: W_net = ΔKE — the work-energy theorem .
14. A 2 kg body (v = 3 m/s) decelerates to rest over 3 m. Retarding force:
Explanation: −F×3 = −½×2×9 = −9. F = 3 N .
15. Two bodies (m and 4m) have equal KE. Their momenta ratio p₁:p₂:
Explanation: p = √(2mKE). p₁/p₂ = √(m/4m) = 1:2 .
16. Work done by gravity on ball dropped from height h:
Explanation: Force and displacement both downward. W = mgh .
17. Momentum increased by 20%. Percentage increase in KE:
Explanation: KE ∝ p². p→1.2p: KE→1.44KE. Increase = 44% .
18. Spring (k = 200 N/m) compressed by 0.1 m. Work done in compressing:
Explanation: W = ½kx² = ½ × 200 × 0.01 = 1 J .
19. Work done by normal force on a body on horizontal surface:
Explanation: Normal force is perpendicular to displacement. W = 0 .
20. Car (1000 kg) accelerates from 0 to 20 m/s. Work done by engine (frictionless):
Explanation: W = ΔKE = ½×1000×400 = 200,000 J .
21. KE of a body at rest is:
Explanation: At rest, v = 0. KE = ½mv² = 0 .
22. Force F = (3x² − 2x) N. Work from x = 0 to x = 2 m:
Explanation: W = ∫₀²(3x²−2x)dx = [x³−x²]₀² = 8−4 = 4 J .
23. Bullet mass doubles (same v). New KE:
Explanation: KE = ½mv². m→2m at same v: KE is doubled .
24. Block (4 kg) starts from rest on rough incline (h = 5 m, μ = 0.2, θ = 30°, g = 10 m/s²). Speed at bottom:
Explanation: L = h/sin30° = 10 m. W_net = mgh − μmg cos30°·L = 200 − 40√3. v ≈ √60 m/s .
25. Work done moving charge q through potential difference V:
Explanation: W = qV. Answer: qV .
26. Gravitational PE of mass m at height h is:
Explanation: PE = mgh .
27. 5 kg ball released from 20 m. Speed just before hitting ground (g = 10 m/s²):
Explanation: v = √(2gh) = √(400) = 20 m/s .
28. PE stored in spring constant k compressed by x:
Explanation: Spring PE = ½kx² .
29. Spring (k = 500 N/m) stretched 0.2 m. Elastic PE:
Explanation: PE = ½×500×0.04 = 10 J .
30. For a freely falling body, total mechanical energy:
Explanation: No non-conservative forces. Total ME = constant .
31. 0.5 kg ball thrown upward at 10 m/s. Max height (g = 10 m/s²):
Explanation: h = v²/2g = 100/20 = 5 m .
32. Pendulum (L = 1 m) released from 60°. Speed at lowest point (g = 10 m/s²):
Explanation: h = L(1−cos60°) = 0.5 m. v = √(2×10×0.5) = √10 m/s .
33. 10 kg mass falls 10 m. KE just before impact (g = 10 m/s²):
Explanation: KE = mgh = 10×10×10 = 1000 J .
34. Two springs (k₁=200, k₂=300 N/m) in series, total stretch 0.3 m. PE stored:
Explanation: k_eq = k₁k₂/(k₁+k₂) = 120 N/m. PE = ½×120×0.09 = 5.4 J .
35. In which situation is mechanical energy NOT conserved?
Explanation: Friction dissipates ME — not conserved on rough surface .
36. As object falls, its gravitational PE:
Explanation: PE = mgh; as h decreases, PE decreases .
37. Spring compressed by x has PE = E. Compressed by 2x, PE becomes:
Explanation: PE ∝ x². x→2x: PE → 4E .
38. 2 kg block slides 5 m on rough floor (μ = 0.3, g = 10 m/s²). Energy lost to friction:
Explanation: f = μmg = 6 N. E_lost = 6×5 = 30 J .
39. Work done by conservative force depends on:
Explanation: Conservative force work is path-independent — depends only on initial and final positions .
40. Which is a non-conservative force?
Explanation: Friction dissipates energy — it is non-conservative.
41. At midpoint of freely falling body from height H, its KE equals:
Explanation: At h = H/2: KE = Total − PE = mgH − mgH/2 = mgH/2 .
42. Total energy of a body executing SHM is:
Explanation: Total SHM energy = ½kA² = constant throughout motion.
43. Block on spring (k = 100 N/m) displaced 0.1 m. Restoring force:
Explanation: F = kx = 100 × 0.1 = 10 N .
44. Ball thrown at 30° with KE = 10 J. KE at maximum height:
Explanation: At max height, KE = KE₀cos²30° = 10×(3/4) = 7.5 J .
45. Work done by non-conservative force in a closed path is:
Explanation: Non-conservative forces dissipate energy; work in closed path is non-zero .
46. Potential energy is a property of:
Explanation: PE belongs to a system of interacting bodies .
47. 5 kg body slides frictionless incline (h = 4 m, g = 10 m/s²). Speed at bottom:
Explanation: v = √(2gh) = √80 m/s ≈ 8.94 m/s.
48. Spring (k=1000 N/m, x=0.1 m) fires 10 g ball. Ball's speed:
Explanation: ½kx² = ½mv². v² = kx²/m = 1000×0.01/0.01 = 1000. v = √1000 m/s ≈ 31.6 m/s.
49. PE at height h₁ = E₁ and at h₂ = E₂ (same mass). Ratio E₁/E₂:
Explanation: E = mgh. E₁/E₂ = h₁/h₂ .
50. Ball slides down rough incline (mgh = 50 J, friction loss = 20 J). KE at bottom:
Explanation: KE = 50 − 20 = 30 J .
51. Power is defined as:
Explanation: P = W/t = Work / Time .
52. SI unit of power is:
Explanation: Power SI unit is Watt (W) = J/s.
53. Motor lifts 1000 kg water to 20 m in 10 s (g = 10 m/s²). Power of motor:
Explanation: P = mgh/t = 1000×10×20/10 = 20,000 W = 20 kW .
54. Car (1000 kg) accelerates from 0 to 30 m/s in 10 s. Average power:
Explanation: W = ½×1000×900 = 450,000 J. P = 450,000/10 = 45,000 W .
55. 100 N force acts on body moving at 5 m/s. Instantaneous power:
Explanation: P = Fv = 100 × 5 = 500 W .
56. In a perfectly inelastic collision:
Explanation: Perfectly inelastic: bodies stick together . Momentum conserved, KE not.
57. Ball A (2 kg, 5 m/s) hits stationary ball B (2 kg). After perfectly inelastic collision, speed:
Explanation: 2×5 = 4×v. v = 2.5 m/s .
58. Elastic collision of equal masses (one at rest). After collision:
Explanation: Equal-mass elastic: first stops, second takes initial speed .
59. Coefficient of restitution e for perfectly elastic collision:
Explanation: For perfectly elastic collision, e = 1 .
60. Which quantity is ALWAYS conserved in any collision?
Explanation: Total momentum is always conserved in any collision.
61. Pump (efficiency 80%) raises 1000 L water/min to 10 m (g = 10 m/s²). Input power required:
Explanation: P_out = 1000×10×10/60 = 1667 W. P_in = 1667/0.8 = 2083 W .
62. 10 g bullet embeds in 1 kg block (at rest). Bullet speed 300 m/s. Block speed after:
Explanation: 0.01×300 = 1.01×v. v = 3/1.01 ≈ 2.97 m/s .
63. 1 horsepower equals approximately:
Explanation: 1 HP = 746 W .
64. Ball (m, velocity v) bounces off wall with same speed. Magnitude of impulse:
Explanation: Δp = m(−v) − m(v). |Impulse| = 2mv .
65. Power of machine doing 600 J work in 2 minutes:
Explanation: P = 600/120 = 5 W .
66. KE lost in perfectly inelastic collision (m₁=m₂=m, u₂=0):
Explanation: KE_i = ½mu₁². KE_f = ¼mu₁². ΔKE = ¼mu₁² .
67. Body (m) moves at constant velocity v against friction f. Power required:
Explanation: At constant v, applied force = f. P = fv = fv .
68. Ball (m) dropped from H, bounces with e = 0.5. Height of first bounce:
Explanation: v_after = e√(2gH). h = v²/2g = e²H = 0.25H = H/4 .
69. Neutron (m) elastically hits carbon nucleus (12m) at rest. Fractional KE loss:
Explanation: Fractional KE loss = 4Mm/(M+m)² = 4×12/169 = 48/169 .
70. Elevator lifts 800 kg load at 2 m/s (g = 10 m/s²). Power required:
Explanation: P = mgv = 800×10×2 = 16,000 W .
71. Which unit is equivalent to Watt?
Explanation: Watt = J/s = kg·m²/s³.
72. Bodies A (3 kg, 4 m/s) and B (2 kg, −2 m/s) collide and stick. Final velocity:
Explanation: p = 3×4 + 2×(−2) = 8 kg·m/s. v = 8/5 = 1.6 m/s .
73. Perfectly elastic collision at macroscopic scale is:
Explanation: At macroscopic scale, truly elastic collisions are impossible — there is always some energy loss .
74. 1 kWh equals:
Explanation: 1 kWh = 1000 W × 3600 s = 3.6×10⁶ J .
75. 60 W bulb runs for 10 hours. Energy consumed in kWh:
Explanation: E = 60W × 10h = 600 Wh = 0.6 kWh .
76. Minimum speed at top of vertical circle (radius r) to maintain contact:
Explanation: At top: mg = mv²/r. v_min = √(rg) .
77. Minimum speed at bottom to complete full vertical loop (radius r):
Explanation: Using energy conservation + min speed at top: v_bottom_min = √(5rg) .
78. Stone (0.5 kg) in vertical circle (r=1 m), v_bottom = √(5g), g=10 m/s². Tension at bottom:
Explanation: T = mg + mv²/r = 0.5×10 + 0.5×50/1 = 5+25 = 30 N .
79. At top of vertical circle, centripetal force is provided by:
Explanation: At top, T and mg both point toward center. Centripetal = T + mg .
80. Work done by centripetal force on a body in circular motion:
Explanation: Centripetal force ⊥ velocity. W = Fd cos90° = 0 .
81. Bullet (m, v) fired into suspended block (M, rest). Height block rises:
Explanation: v' = mv/(M+m). h = v'²/2g = m²v²/2(M+m)²g .
82. Tension difference T_bottom − T_top in vertical circular motion:
Explanation: T_b − T_t = m(v²_b−v²_t)/r + 2mg = 4mg + 2mg = 6mg .
83. Pendulum (m, length L) given horizontal velocity v at bottom. Minimum v for full rotation:
Explanation: v_min = √(5gL) .
84. Motorcycle in vertical loop (r=5 m, g=10 m/s²). Minimum speed at top:
Explanation: v = √(gr) = √(10×5) = √50 m/s ≈ 7.07 m/s.
85. Person pushes wall for 1 hour; wall doesn't move. Work done on wall:
Explanation: Displacement = 0. W = F × 0 = 0 .
86. Relation between kinetic energy K and momentum p:
Explanation: K = ½mv² = p²/2m. K = p²/2m .
87. Spring (k=400 N/m) compressed 0.1 m releases 0.2 kg ball vertically. Height reached:
Explanation: ½kx² = mgh. ½×400×0.01 = 0.2×10×h. 2 = 2h. h = 1 m .
88. Machine: 800 J output for 1000 J input. Efficiency:
Explanation: η = (800/1000)×100 = 80% .
89. Body most likely to lose contact with vertical circular track at:
Explanation: At the top , minimum speed is needed; insufficient speed means contact lost.
90. Work done by net force on body moving along any path equals:
Explanation: Work-energy theorem: W_net = ΔKE .
91. Power developed by gravity on falling mass m with velocity v:
Explanation: P = F·v. Gravity and v both downward. P = mgv .
92. Area under force-displacement graph gives:
Explanation: W = ∫F dx = area under F-x graph = Work done .
93. Force F at angle θ above horizontal moves mass m on rough surface (μ) distance d. Work by friction:
Explanation: N = mg − Fsinθ. W_friction = −μ(mg−Fsinθ)d .
94. Potential energy depends on:
Explanation: PE depends on position in the force field .
95. 2 kg block compresses spring (k=2000 N/m) by 0.1 m. Max speed on frictionless floor:
Explanation: ½kx² = ½mv². v² = kx²/m = 2000×0.01/2 = 10. v = √10 m/s .
96. Stone (0.1 kg) in vertical circle (r=2 m), v=6 m/s at lowest point (g=10 m/s²). Tension at bottom:
Explanation: T = mg + mv²/r = 0.1×10 + 0.1×36/2 = 1 + 1.8 = 2.8 N .
97. KE of 2 kg body moving at 5 m/s:
Explanation: KE = ½×2×25 = 25 J .
98. In oblique elastic collision of equal masses (one at rest), after collision they move:
Explanation: Oblique elastic collision of equal masses (one at rest): they move at right angles to each other.
99. Constant force F gives body acceleration a (start from rest). Instantaneous power at time t:
Explanation: v = at = (F/m)t. P = Fv = F×(F/m)t = F²t/m .
100. Which unit is equivalent to Joule?
Explanation: 1 Joule = 1 N·m = 1 kg·m²/s².