§ 5 ✓NCERT Class 11 · Physics · Chapter 5

Chapter test: Work, Energy and Power

30 questions · Solve each question on paper first, then open "Show answer" to check your working.

QCC Notes · Class 11 Physics NCERT-aligned Updated 26 Sep 2026

Q1.
Find the kinetic energy of a 2 kg body moving at 18 km h⁻¹, in joules.
Show answer
25 J — 18 km/h = 5 m/s; K = ½ × 2 × 5² = 25 J
Q2.
A 25 N force acts at 60° to the displacement while its point of application moves 4 m. Find the work done, in joules.
Show answer
50 J — W = F s cos θ = 25 × 4 × 0.5 = 50 J
Q3.
A force starts at zero and grows linearly to 100 N over a displacement of 0.5 m. Find the work done, in joules.
Show answer
25 J — Linear rise → triangular F–x area: ½ × 0.5 × 100 = 25 J
Q4.
A spring of constant 200 N m⁻¹ is stretched by 15 cm from natural length. Find the energy stored, in joules.
Show answer
2.25 J — U = ½kx² = ½ × 200 × (0.15)² = 2.25 J
Q5.
A body falls freely from rest through 45 m. Find its speed on reaching the ground, in m s⁻¹.
Show answer
30 m s⁻¹ — mgh = ½mv² → v² = 2 × 10 × 45 = 900 → v = 30 (mass cancels)
Q6.
A vehicle is dragged at a steady 15 m s⁻¹ by a 400 N force along the motion. Find the power delivered, in watts.
Show answer
6000 W — P = F v = 400 × 15 = 6000 W = 6 kW
Q7.
A 2 kg body moving at 6 m s⁻¹ collides head-on with a 4 kg body at rest and they stick together. Find the kinetic energy lost, in joules.
Show answer
24 J — v = (2 × 6)/6 = 2 m/s; ΔK = ½ × 2 × 36 − ½ × 6 × 4 = 36 − 12 = 24 J
Q8.
Find the net work (in kJ) needed to accelerate an 800 kg car from rest to 15 m s⁻¹ on a level road.
Show answer
90 kJ — W_net = ΔK = ½ × 800 × 15² = 90,000 J — endpoints only
Q9.
A bead moves in a circular path. The work done by the radial (centripetal) force over any arc of the circle is: (a) positive (b) negative (c) zero (d) proportional to the arc length
Show answer
(c) — Radial force ⊥ tangential velocity at every instant → W = 0 over any arc
Q10.
A force acts on a body that slides 5 m along a line. The work done by the force is zero only if: (a) the force is small (b) the force is perpendicular to the displacement (c) the force is constant (d) the body is heavy
Show answer
(b) — θ = 90° kills the dot product; size and constancy of F are irrelevant
Q11.
Which of these forces is non-conservative? (a) gravitational force (b) ideal spring force (c) kinetic friction (d) electrostatic force
Show answer
(c) — Kinetic friction: work depends on path, no recoverable store
Q12.
The work done by a conservative force over any closed round trip is: (a) always positive (b) exactly zero (c) equal to the potential energy gained (d) path-dependent
Show answer
(b) — Closed-loop zero work is the defining property of conservatism
Q13.
In a head-on elastic collision between two equal masses, the second at rest, what happens afterwards? (a) both move together (b) the first stops and the second moves off with the first's velocity (c) both rebound backward (d) the first continues at half speed
Show answer
(b) — Equal-mass elastic = velocity exchange; the formulas collapse to it
Q14.
The kilowatt-hour is a unit of: (a) power (b) force (c) time (d) energy
Show answer
(d) — kWh = 1000 W × 3600 s = 3.6 × 10⁶ J — power × time = energy
Q15.
A ball is thrown straight up, ignoring air resistance. During the flight: (a) K + U decreases then increases (b) K + U stays constant (c) U stays constant (d) K stays constant
Show answer
(b) — Only gravity (conservative) works → K + U constant throughout
Q16.
A spring stores energy U when stretched by x from natural length. Stretched by 2x, it stores: (a) U (b) 2U (c) 4U (d) 8U
Show answer
(c) — U ∝ x²: doubling x quadruples the stored energy
Q17.
Assertion: In a perfectly inelastic collision, the loss of kinetic energy is the maximum consistent with momentum conservation. Reason: In a perfectly inelastic collision the bodies move with a common velocity after impact.
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(a) — Common velocity is exactly the condition that maximises the KE loss
Q18.
Assertion: A man carrying a suitcase on his head walks along a level road; he does no work on the suitcase. Reason: The force he applies to the suitcase is vertical while its displacement is horizontal.
Show answer
(a) — Vertical force, horizontal displacement: θ = 90° → W = 0; R is the mechanism
Q19.
Statement I: The work-energy theorem remains valid even when the forces acting are non-conservative. Statement II: The theorem equates the net work of all forces to the change in kinetic energy, regardless of the forces' nature.
Show answer
(a) — W_net = ΔK holds for every force type; II is the statement of that generality
Q20.
Statement I: In every collision between two isolated bodies, total linear momentum is conserved. Statement II: In every collision between two isolated bodies, total kinetic energy is conserved.
Show answer
(c) — Momentum: always (isolated). KE: only elastic — II is false
B1.
[Pattern mirror: JEE Main F–x graph style] A block (2 kg, initially at rest, frictionless floor) is pushed by a force that stays 12 N for 2 m and then falls linearly to zero over the next 2 m. Its speed after 4 m is: (a) 4 m/s (b) 5 m/s (c) 6 m/s (d) 8 m/s
Show answer
(c) — Area = 12 × 2 + ½ × 12 × 2 = 36 J; ΔK = 36 → v = √(2 × 36/2) = 6 m/s
B2.
[Pattern mirror: NEET bullet-block spring style] A 10 g bullet moving at 200 m/s embeds itself in a 990 g block resting against a spring of constant 400 N m⁻¹. The maximum compression of the spring is: (a) 5 cm (b) 10 cm (c) 20 cm (d) 2.5 cm
Show answer
(b) — Momentum: 0.01 × 200 = 1 × v → v = 2 m/s; ½ × 1 × 4 = ½ × 400 x² → x = 0.1 m
B3.
[Pattern mirror: JEE Main elastic-collision style] A 1 kg body at 4 m s⁻¹ collides head-on and elastically with a 2 kg body at rest. The speed of the 2 kg body afterwards is about: (a) 2.7 m/s (b) 1.3 m/s (c) 4.0 m/s (d) 5.3 m/s
Show answer
(a) — v₂′ = 2m₁u₁/(m₁+m₂) = 8/3 ≈ 2.7 m/s (and v₁′ = −4/3 m/s)
B4.
[Pattern mirror: NEET braking style] A 1000 kg car moving at 20 m s⁻¹ is brought to rest by a net braking force of 4000 N. The braking distance is: (a) 25 m (b) 50 m (c) 80 m (d) 100 m
Show answer
(b) — ΔK = f s: ½ × 1000 × 400 = 4000 s → s = 200,000/4000 = 50 m
B5.
[Pattern mirror: JEE Main engine-power style] An engine develops 50 kW. Against a total resistive force of 2500 N, its maximum steady speed on a level road is: (a) 10 m/s (b) 15 m/s (c) 20 m/s (d) 25 m/s
Show answer
(c) — v_max = P/f = 50,000/2500 = 20 m/s = 72 km/h
B6.
[Pattern mirror: NEET spring-launch style] A spring of constant 500 N m⁻¹, compressed 20 cm, launches a 0.2 kg block along a frictionless floor. The launch speed is: (a) 5 m/s (b) 8 m/s (c) 10 m/s (d) 20 m/s
Show answer
(c) — U = ½ × 500 × 0.04 = 10 J = ½ × 0.2 × v² → v = 10 m/s
B7.
[Pattern mirror: JEE Main restitution style] A 1 kg body moving at 12 m s⁻¹ catches up with a 2 kg body moving at 4 m s⁻¹ in the same direction; the coefficient of restitution is 0.5. The velocity of the 1 kg body after impact is: (a) 4 m/s (b) 6 m/s (c) 8 m/s (d) 2 m/s
Show answer
(a) — p: v₁ + 2v₂ = 20; separation − approach: v₂ − v₁ = 0.5 × 8 = 4 → v₁ = 4 m/s
B8.
[Pattern mirror: NEET energy-chain style] A bead slides on a frictionless wire and must complete a vertical circular loop of radius R. The minimum release height above the loop's bottom is: (a) 2R (b) 5R/2 (c) 3R (d) 4R
Show answer
(b) — Loop top needs v² = gR (N = 0); mg(h − 2R) = ½mgR → h = 5R/2
B9.
[Pattern mirror: JEE Main electricity-bill style] A 2 kW heater runs 5 hours daily for 30 days. At ₹6 per kWh, the energy cost for the month is: (a) ₹180 (b) ₹900 (c) ₹1800 (d) ₹3600
Show answer
(c) — Energy = 2 × 5 × 30 = 300 kWh; cost = 300 × 6 = ₹1800
B10.
[Pattern mirror: NEET glancing-collision style] A moving billiard ball strikes an identical stationary ball with a glancing (off-centre) elastic blow. The angle between the two balls' velocities just after impact is: (a) 30° (b) 90° (c) 180° (d) dependent on the impact parameter
Show answer
(b) — Equal-mass elastic + one at rest: p⃗ and K conservation force perpendicular emergence (90°)

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