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Work, Power and Energy — IISER Physics MCQs with Solutions
Free IISER 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. 0° ≤ θ < 90°
C. θ = 0° only
D. 90° < θ ≤ 180° ✓ Correct
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. 50 J ✓ Correct
B. 86.6 J
C. 100 J
D. 25 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 Variable Force · easy · theory
For a force that varies with position, the work done between x₁ and x₂ equals:
A. F × (x₂ − x₁) using any F
B. The slope of the F–x graph
C. The area under the F–x graph between x₁ and x₂ ✓ Correct
D. Always zero
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. 6 J
B. 18 J
C. 12 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.
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. Depends on the speed of the particle
C. Is always positive
D. Is always zero
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. Depends on the loop's size
B. Zero ✓ Correct
C. +20 J
D. −20 J
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 = p/2m
B. KE = p²/2m ✓ Correct
C. KE = p²m/2
D. KE = 2m/p²
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. 20 J
B. 200 J
C. 100 J ✓ Correct
D. 50 J
Solution: KE = ½mv² = ½×2×100 = 100 J.
Q9 — Potential Energy · easy · theory
Potential energy can be defined only for:
A. Conservative forces ✓ Correct
B. Contact forces only
C. All forces including friction
D. Constant forces only
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. 150 J
C. 30 J
D. 600 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. Any force acts on it
C. The system is at rest
D. Friction 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. 90 m/s
B. 45 m/s
C. 21 m/s
D. 30 m/s ✓ Correct
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. At the extreme positions
C. The KE is constant
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. 2 m
B. 1 m
C. 0.25 m
D. 0.5 m ✓ Correct
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. The work done by gravity alone
B. Its total mechanical energy
C. The change in its kinetic energy ✓ Correct
D. The change in its potential energy
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. 50 J
B. 25 J
C. 100 J
D. 75 J ✓ Correct
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. Point of inflection
B. Constant value everywhere
C. Maximum
D. Minimum ✓ Correct
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. Neutral
B. Unstable
C. Stable ✓ Correct
D. Not an equilibrium
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 = 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.
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. 30 W
B. 300 W ✓ Correct
C. 3000 W
D. 1000 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. −100 J
C. 250 J
D. 150 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 · theory
A man holds a 20 kg load stationary on his shoulder for 5 minutes and gets tired. The work he does on the load, and the reason, are:
A. Zero — no displacement of the load; his fatigue is internal (muscular) energy expenditure ✓ Correct
B. Negative, since gravity acts down
C. ⁓10⁴ J, from the muscular effort
D. Positive, proportional to the holding time
Solution: Mechanical work needs displacement: W = Fs cosθ = 0 for s = 0. Physiological energy is consumed in repeated muscle-fibre contractions, but no work is done ON the load.
Q23 — Work by Constant Force · hard · numerical
Two forces F⃗₁ = (4î + ĵ) N and F⃗₂ = (−2î + 3ĵ) N act simultaneously on a particle displaced by d⃗ = (2î + 2ĵ) m. The total work done on the particle is:
A. 12 J ✓ Correct
B. 16 J
C. 8 J
D. 20 J
Solution: F⃗_net = (2î + 4ĵ); W = 2×2 + 4×2 = 12 J. (Equivalently W₁ + W₂ = 10 + 2 = 12 J — work adds like scalars.)
Q24 — 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. 50 J
B. 40 J ✓ Correct
C. 35 J
D. 30 J
Solution: Triangle: ½×2×10 = 10 J; rectangle: 10×3 = 30 J. Total = 40 J (area under the curve).
Q25 — Work by Variable Force · hard · theory
On an F–x graph, part of the curve lies below the x-axis. Over that section the force:
A. Cannot exist physically
B. Does negative work (opposes the displacement) ✓ Correct
C. Does no work
D. Does positive work
Solution: Area below the axis is negative area: F and dx have opposite signs there, so the force removes kinetic energy.
Q26 — Work by Variable Force · hard · numerical
A force F = F₀(1 − x/L) acts from x = 0 until it vanishes at x = L. The total work done by the force is:
A. F₀L/2 ✓ Correct
B. Zero
C. F₀L/3
D. F₀L
Solution: W = ∫₀ᴸ F₀(1 − x/L)dx = F₀(L − L/2) = F₀L/2 — the area of the triangle of height F₀ and base L.
Q27 — Conservative Force · hard · numerical
For U(x) = x³ − 3x² (J), the force on the particle at x = 1 m is:
A. +3 N ✓ Correct
B. −3 N
C. +9 N
D. Zero
Solution: F = −dU/dx = −(3x² − 6x) = 6x − 3x². At x = 1: 6 − 3 = +3 N.
Q28 — Conservative Force · hard · theory
Friction is NOT a conservative force because:
A. The work it does around a closed path is non-zero (always negative) ✓ Correct
B. It is too weak
C. It cannot do positive work in any frame
D. It acts only on rough surfaces
Solution: Round any closed loop friction continually opposes motion, so ΣW < 0 ≠ 0 — path matters, hence non-conservative. (In some frames friction CAN do positive work, so that option is wrong.)
Q29 — Conservative Force · hard · numerical
The potential energy of a particle in a plane is U(x, y) = 3x² + 2y² (J). The magnitude of the force on it at the point (1, 1) m is:
A. 5 N
B. √13 N
C. √52 ≈ 7.2 N ✓ Correct
D. 10 N
Solution: F⃗ = −(∂U/∂x î + ∂U/∂y ĵ) = −(6x î + 4y ĵ) = −(6î + 4ĵ) at (1,1); |F| = √(36+16) = √52 ≈ 7.2 N.
Q30 — Kinetic Energy · hard · numerical
The kinetic energy of a body is increased by 300%. The percentage increase in its momentum is:
A. 300%
B. 100% ✓ Correct
C. 200%
D. 73%
Solution: New KE = 4×old ⇒ p ∝ √KE doubles ⇒ +100%. Trap: 300% increase means ×4, not ×3.