Class XI · Chapter 05
Work, Energy and Power — Exam Practice
Practice questions in CBSE, NEET, and JEE Main style covering the work-energy theorem, power, and elastic collisions.
A note on these questions: these are original practice questions, written to match the exact style, format, and difficulty of CBSE Board, NEET, and JEE Main papers — not verbatim reproductions of specific past papers.
1CBSE Board Exam Style
Q1. Is the work done by centripetal force on a body moving in a circle zero? Justify your answer.
Answer
Yes. Centripetal force is always directed toward the centre, exactly perpendicular to the object’s velocity (which is tangential) at every instant. Since W = Fd cosθ and θ = 90° here, cosθ = 0, so the work done is always zero.
Yes. Centripetal force is always directed toward the centre, exactly perpendicular to the object’s velocity (which is tangential) at every instant. Since W = Fd cosθ and θ = 90° here, cosθ = 0, so the work done is always zero.
Derive the work-energy theorem starting from Newton’s second law.
Answer
F = ma = m(dv/dt). Work done: W = ∫F dx = ∫m(dv/dt)dx = ∫mv dv (using dx/dt = v).
F = ma = m(dv/dt). Work done: W = ∫F dx = ∫m(dv/dt)dx = ∫mv dv (using dx/dt = v).
Integrating from initial velocity u to final velocity v: W = m∫uv v dv = ½m(v² − u²) = KEf − KEi.
Q3. (a) Derive the expression for the power delivered by a constant force acting on an object moving with velocity v, and relate it to the work-energy theorem. (b) A pump lifts 200 kg of water per minute from a well 10 m deep. Find the power delivered by the pump. (g = 10 m/s²)
Answer
(a) Power = rate of doing work = dW/dt = d(F·x)/dt = F(dx/dt) = F·v, for constant force F. This directly connects to the work-energy theorem, since the work being done at this rate is exactly what changes the object’s kinetic energy over time.
(a) Power = rate of doing work = dW/dt = d(F·x)/dt = F(dx/dt) = F·v, for constant force F. This directly connects to the work-energy theorem, since the work being done at this rate is exactly what changes the object’s kinetic energy over time.
(b) P = mgh/t = (200 × 10 × 10) / 60 = 20000/60 ≈ 333.3 W.
2NEET Style (Single-Correct MCQ)
Q4. Kinetic energy can be expressed in terms of momentum p and mass m as:
Solution
KE = ½mv², and p = mv → v = p/m. Substituting: KE = ½m(p/m)² = p²/2m.
KE = ½mv², and p = mv → v = p/m. Substituting: KE = ½m(p/m)² = p²/2m.
Q5. Potential energy can be meaningfully defined only for:
Solution
Potential energy requires work done to be path-independent (depending only on start/end position) — true only for conservative forces like gravity and spring force, not for friction or drag.
Potential energy requires work done to be path-independent (depending only on start/end position) — true only for conservative forces like gravity and spring force, not for friction or drag.
Q6. One watt is equivalent to:
Solution
Power = work/time, so its SI unit, the watt, is exactly one joule per second.
Power = work/time, so its SI unit, the watt, is exactly one joule per second.
Q7. In a perfectly elastic collision, which quantities are conserved?
Solution
Elastic collisions are specifically defined by the conservation of both quantities — the defining feature that distinguishes them from inelastic collisions.
Elastic collisions are specifically defined by the conservation of both quantities — the defining feature that distinguishes them from inelastic collisions.
3JEE Main Style
Q8. A 2 kg block is dropped from a height of 5 m. Using energy conservation, find its velocity just before hitting the ground. (g = 10 m/s²)
Solution
Loss in PE = gain in KE: mgh = ½mv² → v = √(2gh) = √(2×10×5) = √100 = 10 m/s.
Loss in PE = gain in KE: mgh = ½mv² → v = √(2gh) = √(2×10×5) = √100 = 10 m/s.
Q9. The work done by gravity on an object moving horizontally at constant height is:
Solution
Gravity acts vertically; horizontal displacement is perpendicular to it, so W = Fd cos90° = 0.
Gravity acts vertically; horizontal displacement is perpendicular to it, so W = Fd cos90° = 0.
Q10. A spring of spring constant 100 N/m is compressed by 0.2 m. Find the elastic potential energy stored.
Solution
U = ½kx² = ½ × 100 × (0.2)² = ½ × 100 × 0.04 = 2 J.
U = ½kx² = ½ × 100 × (0.2)² = ½ × 100 × 0.04 = 2 J.
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