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Work, Energy and Power
Physics · Class 11NCERT pp. 71–862 sets × 25 questions
Key facts and formulas for Work, Energy and Power, three worked sample questions, and two full practice sets. Every question has a hint, a reason for every option and a step-by-step solution, with a calculator or animation where one helps. English and Hinglish.
Formula sheet for this chapter
| Idea | Rule or fact |
|---|
| Work by constant force | W = F d cos θ; positive for θ < 90°, zero at 90°, negative beyond 90° |
| Variable force | W = ∫F dx, the area under the force-position graph |
| Work-energy theorem | Net work = change in kinetic energy = ½ m (v² − u²) |
| Kinetic energy and momentum | K = ½ m v² = p² / 2m; p = √(2mK) |
| Potential energy | Gravity: U = m g h; spring: U = ½ k x²; F = −dU/dx |
| Conservative force | Zero work over a closed path; work independent of path (gravity, spring); friction is not |
| Vertical circle | Least speed at top v = √(gL); at bottom √(5gL); at horizontal √(3gL) |
| Power | P = W / t = F v; 1 kWh = 3.6 × 10⁶ J |
| Collisions | Momentum always conserved; elastic: KE conserved, e = 1; sticking: e = 0 |
| Elastic, target at rest | v₁ = (m₁ − m₂)u/(m₁ + m₂); v₂ = 2m₁u/(m₁ + m₂); equal masses exchange velocities |
Three sample questions
Q1. A force of 20 N acts at 60 degrees to a displacement of 5 m. The work done is
- 50 J
- 100 J
- 86.6 J
- 10 J
Hint and solution
Hint: Use work as force times displacement times the cosine of the angle between them. What is cos 60 degrees?
- W = F d cos θ — Work is the force component along the displacement times the displacement.
- cos 60° = 0.5 — The angle between force and displacement is 60 degrees, so the cosine is one half.
- W = 20 × 5 × 0.5 = 50 J — Multiply 20 by 5 to get 100, then by 0.5.
- Answer: 50 J — So the work done is 50 joule. Option A.
Answer: A. W = F d cos θ = 20 x 5 x 0.5 = 50 J.
Q2. The vectors 3i + 2j and λi - 6j are perpendicular. The value of λ is
- -4
- 9
- 4
- 2
Hint and solution
Hint: Perpendicular vectors have a dot product equal to zero. Multiply the i parts and the j parts and add.
- A·B = 0 for perpendicular vectors — The angle is 90 degrees and cos 90° is zero, so the dot product vanishes.
- A·B = 3 × λ + 2 × (-6) = 3λ - 12 — Multiply the i components together and the j components together, then add.
- 3λ - 12 = 0 → λ = 12 / 3 — Move 12 to the other side and divide by 3.
- λ = 4 — So λ equals 4. Option C.
Answer: C. A·B = 3λ - 12 = 0, so λ = 4.
Q3. Consider:
I. Work done by friction on a sliding body is negative.
II. Work done by the centripetal force in uniform circular motion is zero.
III. Gravity does positive work on a rising ball.
Which are correct?
- II and III only
- I and II only
- I and III only
- I, II and III
Hint and solution
Hint: Check the angle between each force and the displacement. Opposite gives negative, perpendicular gives zero.
- Work sign depends on angle: 0° +, 90° zero, 180° − — Work is positive when the force has a component along the motion, zero when perpendicular, negative when opposite.
- I: friction opposes sliding → negative (true) — Friction points opposite to the motion, so its work is negative.
- II: centripetal ⟂ velocity → zero (true) — The centripetal force always points to the centre, at 90 degrees to the displacement, so its work is zero.
- III: rising ball, gravity down → negative (false) — Gravity acts down while the ball moves up, so gravity's work is negative. Only I and II are correct. Option B.
Answer: B. Friction opposes motion (I true). Centripetal force is perpendicular to the displacement (II true). On a rising ball gravity points down and the ball moves up, so its work is negative (III false).
How to use this chapter
Read the NCERT pages first, then attempt Set 1 without the hints. Note every wrong answer in a mistake list, read the reason for each option, and retry Set 2 two days later. A topic is solid when you can say why each wrong option is wrong.