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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 · theory
The work done by a force is negative when the angle θ between the force and the displacement satisfies:
A. θ = 90° only
B. 90° < θ ≤ 180° ✓ Correct
C. θ = 0° only
D. 0° ≤ θ < 90°
Solution: W = Fs cosθ; cosθ < 0 for obtuse angles, so work is negative when the force has a component opposite to the displacement (e.g. friction).
Q2 — Work by Constant Force · easy · numerical
A force of 20 N acts on a block at 60° to the horizontal while it moves 5 m horizontally. The work done by the force is:
A. 100 J
B. 86.6 J
C. 25 J
D. 50 J ✓ Correct
Solution: W = Fs cosθ = 20 × 5 × cos60° = 100 × 0.5 = 50 J. Trap: use the angle with the displacement, not the vertical.
Q3 — Work by Variable Force · easy · theory
For a force that varies with position, the work done between x₁ and x₂ equals:
A. The slope of the F–x graph
B. Always zero
C. F × (x₂ − x₁) using any F
D. The area under the F–x graph between x₁ and x₂ ✓ Correct
Solution: W = ∫F dx = area under the force–displacement curve (areas below the axis count negative).
Q4 — Work by Variable Force · easy · numerical
A force F = 2x (N) acts on a particle. The work done in moving it from x = 0 to x = 3 m is:
A. 18 J
B. 6 J
C. 9 J ✓ Correct
D. 12 J
Solution: W = ∫₀³ 2x dx = [x²]₀³ = 9 J. Trap: don't use F(3)×3 = 18 J — the force varies.
Q5 — Conservative Force · easy · theory
A force is conservative if the work it does:
A. Is independent of the path (zero over any closed loop) ✓ Correct
B. Is always zero
C. Depends on the speed of the particle
D. Is always positive
Solution: Path-independence (equivalently, zero net work around any closed path) defines a conservative force — gravity, spring and electrostatic forces qualify; friction does not.
Q6 — Conservative Force · easy · numerical
A 2 kg particle is carried around a full closed loop in a uniform gravitational field. The work done by gravity over the loop is:
A. −20 J
B. +20 J
C. Zero ✓ Correct
D. Depends on the loop's size
Solution: Gravity is conservative: over any closed path the net work is exactly zero.
Q7 — Kinetic Energy · easy · theory
Kinetic energy in terms of the momentum p and mass m is:
A. KE = 2m/p²
B. KE = p²/2m ✓ Correct
C. KE = p²m/2
D. KE = p/2m
Solution: KE = ½mv² and p = mv give KE = p²/2m. KE is a scalar and can never be negative.
Q8 — Kinetic Energy · easy · numerical
A 2 kg body moves at 10 m/s. Its kinetic energy is:
A. 50 J
B. 200 J
C. 20 J
D. 100 J ✓ Correct
Solution: KE = ½mv² = ½×2×100 = 100 J.
Q9 — Potential Energy · easy · theory
Potential energy can be defined only for:
A. Contact forces only
B. Constant forces only
C. Conservative forces ✓ Correct
D. All forces including friction
Solution: U is defined through ΔU = −W_cons, which requires path-independent (conservative) forces. Its zero level is an arbitrary choice.
Q10 — Potential Energy · easy · numerical
A 2 kg block is raised through a height of 15 m. The gain in its potential energy is:
A. 300 J ✓ Correct
B. 30 J
C. 600 J
D. 150 J
Solution: ΔU = mgh = 2×10×15 = 300 J.
Q11 — Law of Conservation of Mechanical Energy · easy · theory
The total mechanical energy (KE + PE) of a system is conserved when:
A. Only conservative forces do work on it ✓ Correct
B. Friction acts on it
C. The system is at rest
D. Any force acts on it
Solution: If only conservative forces (gravity, spring…) do work, KE + PE = constant. Friction/air resistance drain mechanical energy into heat.
Q12 — Law of Conservation of Mechanical Energy · easy · numerical
A ball is dropped from a height of 45 m. Its speed on reaching the ground is:
A. 30 m/s ✓ Correct
B. 90 m/s
C. 21 m/s
D. 45 m/s
Solution: mgh = ½mv² ⇒ v = √(2gh) = √(2×10×45) = √900 = 30 m/s.
Q13 — Conservation of Mechanical Energy · easy · theory
A mass oscillating on a spring (no friction) has its maximum kinetic energy:
A. At the mean (equilibrium) position ✓ Correct
B. The KE is constant
C. At the extreme positions
D. Halfway to the extreme
Solution: KE + ½kx² = constant; KE is largest where the spring PE is least — at x = 0 (the mean position).
Q14 — Conservation of Mechanical Energy · easy · numerical
A 2 kg block moving at 10 m/s on a smooth floor hits a spring of k = 800 N/m. The maximum compression of the spring is:
A. 0.25 m
B. 1 m
C. 0.5 m ✓ Correct
D. 2 m
Solution: ½mv² = ½kx² ⇒ x = v√(m/k) = 10√(2/800) = 10×0.05 = 0.5 m.
Q15 — Work Energy Theorem · easy · theory
The work–energy theorem states that the NET work done on a particle equals:
A. Its total mechanical energy
B. The change in its potential energy
C. The change in its kinetic energy ✓ Correct
D. The work done by gravity alone
Solution: W_net = ΔKE = ½mv² − ½mu². It holds for ALL forces — conservative or not.
Q16 — Work Energy Theorem · easy · numerical
A 2 kg body accelerates from 5 m/s to 10 m/s. The net work done on it is:
A. 100 J
B. 50 J
C. 75 J ✓ Correct
D. 25 J
Solution: W = ½m(v² − u²) = 1×(100 − 25) = 75 J.
Q17 — Types of Equilibrium · easy · theory
In STABLE equilibrium, the potential energy of the body is at a:
A. Minimum ✓ Correct
B. Constant value everywhere
C. Point of inflection
D. Maximum
Solution: Stable: U minimum (displacement raises U, restoring force pulls back). Unstable: U maximum. Neutral: U constant.
Q18 — Types of Equilibrium · easy · numerical
For U(x) = x² (J), the equilibrium at x = 0 is:
A. Stable ✓ Correct
B. Unstable
C. Not an equilibrium
D. Neutral
Solution: dU/dx = 2x = 0 at x = 0; d²U/dx² = 2 > 0 ⇒ minimum ⇒ stable (this is the spring potential).
Q19 — Power · easy · theory
Instantaneous power delivered by a force F⃗ to a body moving with velocity v⃗ is:
A. P = F⃗ × v⃗
B. P = Fv always (any angle)
C. P = F⃗·v⃗ ✓ Correct
D. P = F/v
Solution: P = F⃗·v⃗ = Fv cosθ. It is a scalar; 1 W = 1 J/s and 1 hp ≈ 746 W.
Q20 — Power · easy · numerical
A motor lifts a 100 kg load through 3 m in 10 s at constant speed. The power delivered is:
A. 300 W ✓ Correct
B. 1000 W
C. 30 W
D. 3000 W
Solution: P = mgh/t = 100×10×3/10 = 300 W.
Q21 — Work by Constant Force · hard · numerical
A 2 kg block is dragged 10 m along a floor (μ = 0.5) by a horizontal force of 15 N. The NET work done on the block is:
A. 50 J ✓ Correct
B. 150 J
C. 250 J
D. −100 J
Solution: Friction = μmg = 0.5×2×10 = 10 N. W_net = (15 − 10) × 10 = 50 J. (W_applied = 150 J, W_friction = −100 J.) Trap: net work uses the net force.
Q22 — Work by Constant Force · hard · numerical
A 2 kg block is pushed 5 m up a 37° incline (μ = 0.25, sin37° = 0.6, cos37° = 0.8) at constant speed by a force along the incline. The work done by the applied force is:
A. 60 J
B. 100 J
C. 40 J
D. 80 J ✓ Correct
Solution: F = mg sinθ + μmg cosθ = 12 + 4 = 16 N. W = 16 × 5 = 80 J (60 J against gravity + 20 J against friction).
Q23 — Work by Variable Force · hard · numerical
An F–x graph rises linearly from 0 to 10 N over x = 0 → 2 m, then stays constant at 10 N up to x = 5 m. The total work done is:
A. 35 J
B. 30 J
C. 50 J
D. 40 J ✓ Correct
Solution: Triangle: ½×2×10 = 10 J; rectangle: 10×3 = 30 J. Total = 40 J (area under the curve).
Q24 — Work by Variable Force · hard · numerical
A force F = A/x² acts on a particle moving along x. The work done in moving from x = a to x = 2a is:
A. A/2a ✓ Correct
B. A ln 2
C. A/a
D. 3A/2a
Solution: W = ∫ₐ²ᵃ A x⁻² dx = A[−1/x]ₐ²ᵃ = A(1/a − 1/2a) = A/2a. Trap: the ln form belongs to F ∝ 1/x, not 1/x².
Q25 — Conservative Force · hard · numerical
For U(x) = x³ − 3x² (J), the force on the particle at x = 1 m is:
A. +9 N
B. −3 N
C. Zero
D. +3 N ✓ Correct
Solution: F = −dU/dx = −(3x² − 6x) = 6x − 3x². At x = 1: 6 − 3 = +3 N.
Q26 — Conservative Force · hard · numerical
A spring force F = −100x acts on a particle. The work done BY this conservative force as the particle moves from x = 0.2 m to x = 0 is:
A. +2 J ✓ Correct
B. Zero
C. −2 J
D. +4 J
Solution: W_cons = −ΔU = U(0.2) − U(0) = ½×100×0.04 − 0 = +2 J. Moving towards equilibrium, the spring force does positive work.
Q27 — Kinetic Energy · hard · numerical
The kinetic energy of a body is increased by 300%. The percentage increase in its momentum is:
A. 200%
B. 100% ✓ Correct
C. 73%
D. 300%
Solution: New KE = 4×old ⇒ p ∝ √KE doubles ⇒ +100%. Trap: 300% increase means ×4, not ×3.
Q28 — Kinetic Energy · hard · numerical
The kinetic energy of a particle decreases by 19%. The percentage decrease in its momentum is:
A. 9.5%
B. 19%
C. 10% ✓ Correct
D. 81%
Solution: KE becomes 0.81× ⇒ p ∝ √KE becomes 0.9× ⇒ 10% decrease.
Q29 — Potential Energy · hard · numerical
A uniform rod of mass 2 kg and length 1 m lies flat on the ground. The work needed to stand it vertically on one end is:
A. 5 J
B. 10 J ✓ Correct
C. 20 J
D. 40 J
Solution: Only the centre of mass matters: it rises by L/2 = 0.5 m. W = mg(L/2) = 2×10×0.5 = 10 J.
Q30 — Potential Energy · hard · numerical
A spring stores 25 J at stretch x. The ADDITIONAL work required to stretch it from x to 2x is:
A. 50 J
B. 75 J ✓ Correct
C. 100 J
D. 25 J
Solution: U(2x) = 4×25 = 100 J. Additional work = 100 − 25 = 75 J. Trap: not 25 J — the force grows with stretch.