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Work, Power and Energy — JEE Main Physics MCQs with Solutions
Free JEE Main Physics Work, Power and Energy MCQs with step-by-step solutions covering Work by Constant Force, Work by Variable Force, Conservative Force, Kinetic Energy, Potential Energy, Law of Conservation of Mechanical Energy. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Work by Constant Force · easy · numerical
A porter drags a 20 kg trunk 4 m across a platform with a 50 N force at 37° above the horizontal (cos37° = 0.8). The work done by the porter is:
A. 640 J
B. 200 J
C. 120 J
D. 160 J ✓ Correct
Solution: W = Fs cosθ = 50 × 4 × 0.8 = 160 J.
Q2 — Work by Variable Force · easy · numerical
The work required to stretch a spring of k = 400 N/m by 5 cm from its natural length is:
A. 0.5 J ✓ Correct
B. 0.05 J
C. 1.0 J
D. 10 J
Solution: W = ½kx² = ½ × 400 × (0.05)² = 200 × 0.0025 = 0.5 J.
Q3 — Conservative Force · easy · numerical
A 2 kg stone falls from a height of 10 m. The work done by the gravitational force on it is:
A. −200 J
B. +200 J ✓ Correct
C. Zero
D. +100 J
Solution: W_gravity = +mgh = 2×10×10 = 200 J (force and displacement are both downward). ΔU = −200 J.
Q4 — Kinetic Energy · easy · numerical
A 0.5 kg ball is dropped from a height of 20 m. Its kinetic energy just before hitting the ground is:
A. 50 J
B. 100 J ✓ Correct
C. 10 J
D. 200 J
Solution: KE = mgh = 0.5×10×20 = 100 J (all PE converts on a free fall).
Q5 — Potential Energy · easy · numerical
A 50 kg person climbs 40 stairs, each 0.25 m high. The gain in potential energy is:
A. 5000 J ✓ Correct
B. 2500 J
C. 500 J
D. 10000 J
Solution: h = 40×0.25 = 10 m; ΔU = 50×10×10 = 5000 J.
Q6 — Law of Conservation of Mechanical Energy · easy · numerical
A ball is thrown vertically downward at 10 m/s from a height of 15 m. Its speed on hitting the ground is:
A. 15 m/s
B. 25 m/s
C. 20 m/s ✓ Correct
D. 17.3 m/s
Solution: v² = u² + 2gh = 100 + 300 = 400 ⇒ v = 20 m/s (energy method: ½mv² = ½mu² + mgh).
Q7 — Conservation of Mechanical Energy · easy · numerical
A child starts from rest at the top of a smooth slide of height 3.2 m. Her speed at the bottom is:
A. 16 m/s
B. 8 m/s ✓ Correct
C. 4 m/s
D. 5.7 m/s
Solution: v = √(2gh) = √(2×10×3.2) = √64 = 8 m/s.
Q8 — Work Energy Theorem · easy · numerical
Starting from rest, a net work of 64 J is done on a 2 kg body. Its final speed is:
A. 32 m/s
B. 8 m/s ✓ Correct
C. 4 m/s
D. 16 m/s
Solution: ½mv² = 64 ⇒ v = √(64) = 8 m/s.
Q9 — Types of Equilibrium · easy · numerical
For U(x) = 5 + (x − 1)² (J), the stable equilibrium is at:
A. x = 1 m ✓ Correct
B. x = −1 m
C. x = 5 m
D. x = 0
Solution: U is a parabola with minimum at x = 1 (U = 5 J there); F = −2(x−1) is restoring about it.
Q10 — Power · easy · numerical
A 60 kg person climbs stairs of total height 3 m in 6 s. The average power developed is:
A. 30 W
B. 300 W ✓ Correct
C. 600 W
D. 180 W
Solution: P = mgh/t = 60×10×3/6 = 300 W.
Q11 — Work by Constant Force · medium · numerical
A 10 kg suitcase is lifted from the floor onto a shelf 1.5 m high, then carried 10 m horizontally at the same height. The total work done against gravity is:
A. 150 J ✓ Correct
B. 1000 J
C. 1150 J
D. Zero
Solution: Against gravity: W = mgh = 10×10×1.5 = 150 J. The horizontal carry does no work against gravity (force ⊥ displacement) — the classic coolie trap.
Q12 — Work by Variable Force · medium · numerical
A spring stores 4 J when stretched by 2 cm. The energy stored when it is stretched by 4 cm is:
A. 4 J
B. 32 J
C. 8 J
D. 16 J ✓ Correct
Solution: U = ½kx² ∝ x². Doubling x quadruples the energy: 4 × 4 = 16 J.
Q13 — Conservative Force · medium · numerical
A ball is thrown up to a height of 10 m and returns to the throw point. The total work done by gravity over the entire trip is:
A. +200 J
B. −200 J
C. Zero ✓ Correct
D. −100 J
Solution: Up: −mgh; down: +mgh. Net = 0 — a closed path for a conservative force.
Q14 — Kinetic Energy · medium · numerical
A 10 g bullet travels at 400 m/s. Its kinetic energy is:
A. 400 J
B. 1600 J
C. 800 J ✓ Correct
D. 80 J
Solution: KE = ½×0.01×400² = ½×0.01×160000 = 800 J.
Q15 — Potential Energy · medium · numerical
A pump raises 100 kg of water to an overhead tank at a height of 10 m. The potential energy given to the water is:
A. 1 kJ
B. 100 kJ
C. 10 J
D. 10 kJ ✓ Correct
Solution: ΔU = mgh = 100×10×10 = 10⁴ J = 10 kJ.
Q16 — Law of Conservation of Mechanical Energy · medium · numerical
A ball thrown vertically up at 20 m/s has a speed of 10 m/s at height h. Then h equals:
A. 10 m
B. 15 m ✓ Correct
C. 5 m
D. 20 m
Solution: mgh = ½m(u² − v²) ⇒ h = (400 − 100)/(2×10) = 15 m.
Q17 — Conservation of Mechanical Energy · medium · numerical
A 0.1 kg ball dropped from 10 m rebounds to 5 m. The energy lost in the bounce is:
A. 0.5 J
B. 5 J ✓ Correct
C. 10 J
D. 2.5 J
Solution: Loss = mg(h₁ − h₂) = 0.1×10×5 = 5 J (50% of the initial energy).
Q18 — Work Energy Theorem · medium · numerical
A 20 g bullet moving at 500 m/s penetrates 10 cm into a wooden block before stopping. The average resistive force is:
A. 25000 N ✓ Correct
B. 5000 N
C. 2500 N
D. 50000 N
Solution: KE = ½×0.02×250000 = 2500 J; F = KE/d = 2500/0.1 = 25000 N.
Q19 — Types of Equilibrium · medium · theory
A cylinder lying on its curved side on a horizontal table is in ______ equilibrium:
A. Unstable
B. Neutral ✓ Correct
C. Stable
D. No equilibrium
Solution: Rolling it sideways neither raises nor lowers its centre of gravity ⇒ U constant ⇒ neutral. (On its flat end: stable; balanced on an edge: unstable.)
Q20 — Power · medium · numerical
A 2 kW motor pump lifts water to a height of 20 m. The mass of water raised per second is:
A. 2 kg
B. 100 kg
C. 20 kg
D. 10 kg ✓ Correct
Solution: P = (dm/dt)gh ⇒ dm/dt = 2000/(10×20) = 10 kg/s.