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Work, Power and Energy — NEET Physics MCQs with Solutions
Free NEET 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. 90° < θ ≤ 180° ✓ Correct
B. θ = 0° only
C. 0° ≤ θ < 90°
D. θ = 90° only
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. 50 J ✓ Correct
C. 25 J
D. 86.6 J
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 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.
Q4 — Work by Variable Force · easy · theory
For a force that varies with position, the work done between x₁ and x₂ equals:
A. The area under the F–x graph between x₁ and x₂ ✓ Correct
B. Always zero
C. F × (x₂ − x₁) using any F
D. The slope of the F–x graph
Solution: W = ∫F dx = area under the force–displacement curve (areas below the axis count negative).
Q5 — 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. 12 J
B. 18 J
C. 6 J
D. 9 J ✓ Correct
Solution: W = ∫₀³ 2x dx = [x²]₀³ = 9 J. Trap: don't use F(3)×3 = 18 J — the force varies.
Q6 — 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.
Q7 — Conservative Force · easy · theory
A force is conservative if the work it does:
A. Is always positive
B. Is independent of the path (zero over any closed loop) ✓ Correct
C. Is always zero
D. Depends on the speed of the particle
Solution: Path-independence (equivalently, zero net work around any closed path) defines a conservative force — gravity, spring and electrostatic forces qualify; friction does not.
Q8 — 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.
Q9 — 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.
Q10 — Kinetic Energy · easy · theory
Kinetic energy in terms of the momentum p and mass m is:
A. KE = p²m/2
B. KE = p²/2m ✓ Correct
C. KE = p/2m
D. KE = 2m/p²
Solution: KE = ½mv² and p = mv give KE = p²/2m. KE is a scalar and can never be negative.
Q11 — Kinetic Energy · easy · numerical
A 2 kg body moves at 10 m/s. Its kinetic energy is:
A. 100 J ✓ Correct
B. 50 J
C. 20 J
D. 200 J
Solution: KE = ½mv² = ½×2×100 = 100 J.
Q12 — 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).
Q13 — Potential Energy · easy · theory
Potential energy can be defined only for:
A. Constant forces only
B. All forces including friction
C. Conservative forces ✓ Correct
D. Contact forces only
Solution: U is defined through ΔU = −W_cons, which requires path-independent (conservative) forces. Its zero level is an arbitrary choice.
Q14 — Potential Energy · easy · numerical
A 2 kg block is raised through a height of 15 m. The gain in its potential energy is:
A. 600 J
B. 30 J
C. 300 J ✓ Correct
D. 150 J
Solution: ΔU = mgh = 2×10×15 = 300 J.
Q15 — 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.
Q16 — Law of Conservation of Mechanical Energy · easy · theory
The total mechanical energy (KE + PE) of a system is conserved when:
A. Friction acts on it
B. Any force acts on it
C. Only conservative forces do work on it ✓ Correct
D. The system is at rest
Solution: If only conservative forces (gravity, spring…) do work, KE + PE = constant. Friction/air resistance drain mechanical energy into heat.
Q17 — 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. 45 m/s
B. 90 m/s
C. 21 m/s
D. 30 m/s ✓ Correct
Solution: mgh = ½mv² ⇒ v = √(2gh) = √(2×10×45) = √900 = 30 m/s.
Q18 — 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).
Q19 — 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).
Q20 — 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. 2 m
B. 0.25 m
C. 0.5 m ✓ Correct
D. 1 m
Solution: ½mv² = ½kx² ⇒ x = v√(m/k) = 10√(2/800) = 10×0.05 = 0.5 m.
Q21 — 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.
Q22 — Work Energy Theorem · easy · theory
The work–energy theorem states that the NET work done on a particle equals:
A. The change in its kinetic energy ✓ Correct
B. The change in its potential energy
C. Its total mechanical energy
D. The work done by gravity alone
Solution: W_net = ΔKE = ½mv² − ½mu². It holds for ALL forces — conservative or not.
Q23 — 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. 50 J
B. 75 J ✓ Correct
C. 25 J
D. 100 J
Solution: W = ½m(v² − u²) = 1×(100 − 25) = 75 J.
Q24 — 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.
Q25 — Types of Equilibrium · easy · theory
In STABLE equilibrium, the potential energy of the body is at a:
A. Constant value everywhere
B. Minimum ✓ Correct
C. Maximum
D. Point of inflection
Solution: Stable: U minimum (displacement raises U, restoring force pulls back). Unstable: U maximum. Neutral: U constant.
Q26 — Types of Equilibrium · easy · numerical
For U(x) = x² (J), the equilibrium at x = 0 is:
A. Unstable
B. Stable ✓ Correct
C. Neutral
D. Not an equilibrium
Solution: dU/dx = 2x = 0 at x = 0; d²U/dx² = 2 > 0 ⇒ minimum ⇒ stable (this is the spring potential).
Q27 — 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.
Q28 — Power · easy · theory
Instantaneous power delivered by a force F⃗ to a body moving with velocity v⃗ is:
A. P = F⃗ × v⃗
B. P = F⃗·v⃗ ✓ Correct
C. P = F/v
D. P = Fv always (any angle)
Solution: P = F⃗·v⃗ = Fv cosθ. It is a scalar; 1 W = 1 J/s and 1 hp ≈ 746 W.
Q29 — Power · easy · numerical
A motor lifts a 100 kg load through 3 m in 10 s at constant speed. The power delivered is:
A. 1000 W
B. 30 W
C. 3000 W
D. 300 W ✓ Correct
Solution: P = mgh/t = 100×10×3/10 = 300 W.
Q30 — 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.