1. Chapter at a Glance
Work, energy and power give a second way — often much shorter than Newton's laws — to solve mechanics problems. Instead of tracking forces moment by moment, we compare energies at the start and end. The chapter covers work by constant and variable forces, the work–energy theorem, conservative forces and potential energy, conservation of mechanical energy, power, and collisions. Expect one or two questions in JEE Main.
- Work done by a variable force (area under F–x graph or integration).
- Work–energy theorem with friction; speed at a point of a track.
- Elastic and inelastic collisions, coefficient of restitution.
- Motion in a vertical circle; potential energy curves and equilibrium.
2. Work
- Constant force: W = F·s = Fs cosθ. Work is positive when force helps the motion, negative when it opposes it (friction), and zero when the force is perpendicular (centripetal force, normal force on a level surface).
- Variable force: W = ∫F dx = area under the F–x graph.
- Spring (Hooke's law F = −kx): work done in stretching from x₁ to x₂ is ½k(x₂² − x₁²).
- Unit: joule (J); 1 kWh = 3.6 × 10⁶ J; 1 eV = 1.6 × 10⁻¹⁹ J.
3. Kinetic Energy and the Work–Energy Theorem
The theorem holds for all forces — conservative, non-conservative, constant or variable. It is often the fastest way to find a speed when the path is complicated.
4. Conservative Forces and Potential Energy
- A force is conservative if the work it does depends only on the start and end points (zero around any closed path): gravity, spring force, electrostatic force. Friction and air drag are non-conservative.
- For a conservative force, F = −dU/dx. Gravitational PE near Earth: U = mgh; spring PE: U = ½kx².
- Conservation of mechanical energy: if only conservative forces do work, K + U stays constant. With friction: K₁ + U₁ = K₂ + U₂ + (work done against friction).
5. Power
Unit: watt (W); 1 horsepower = 746 W. A vehicle moving at constant speed against a resistive force R needs power P = Rv.
6. Collisions and the Vertical Circle
In every collision, momentum is conserved. Kinetic energy is conserved only in elastic collisions. The coefficient of restitution e = (speed of separation)/(speed of approach): e = 1 (elastic), 0 < e < 1 (partially inelastic), e = 0 (perfectly inelastic, bodies stick).
| 1D elastic collision (m₂ at rest) | Result |
|---|---|
| Final velocity of m₁ | v₁ = (m₁ − m₂)u₁/(m₁ + m₂) |
| Final velocity of m₂ | v₂ = 2m₁u₁/(m₁ + m₂) |
| Equal masses | velocities are exchanged |
| Perfectly inelastic | v = (m₁u₁ + m₂u₂)/(m₁ + m₂) |
A ball dropped from height h with restitution e rebounds to e²h. Vertical circle (bob on a string, radius r): minimum speed at the lowest point √(5gr), at the top √(gr); the tension at the bottom exceeds that at the top by 6mg.
7. Formula Master Table
| Quantity | Formula |
|---|---|
| Work | W = Fs cosθ; W = ∫F dx |
| Kinetic energy | K = ½mv² = p²/2m |
| Work–energy theorem | Wnet = ΔK |
| Spring PE | U = ½kx² |
| Force from PE | F = −dU/dx |
| Power | P = F·v |
| Restitution | e = (v₂ − v₁)/(u₁ − u₂) |
| Height after bounce | h′ = e²h |
| Vertical circle | vbottom,min = √(5gr); Tbottom − Ttop = 6mg |
8. Mind Map
- Constant force
- Variable force (area)
- Spring work
- Work–energy theorem
- Conservative forces
- U curves
- P = W/t
- P = F·v
- Elastic, inelastic
- Restitution
- Vertical circle
9. Tips and Tricks
- If a question asks for a speed and not a time, try energy first.
- In a perfectly inelastic collision with m₂ at rest, the fraction of KE lost is m₂/(m₁ + m₂).
- If momentum p is fixed, K ∝ 1/m; if K is fixed, p ∝ √m.
- A light body hitting a very heavy body elastically bounces back with nearly the same speed.
- Assuming kinetic energy is conserved in every collision.
- Forgetting that normal force does no work on a fixed surface but friction does.
- Using v = √(2gh) for rough surfaces, where part of the energy is lost to friction.
10. Important One-liners
- Work done by a centripetal force is always zero.
- Work done by a conservative force around a closed path is zero.
- If momentum doubles, kinetic energy becomes four times.
- Momentum is conserved in all collisions; kinetic energy only in elastic ones.
- In an elastic head-on collision of equal masses, velocities are exchanged.
- Stable equilibrium corresponds to minimum potential energy.
11. Solved Numericals: Average to Advanced
Five average-level and five advanced-level problems. Click to reveal each answer and its full working. Take g = 10 m s⁻².
Average level
A force F = (3î + 4ĵ) N moves a body through a displacement s = (2î + 3ĵ) m. Find the work done.
Show answer & solution
Concept: W = F·s = Fxsx + Fysy.
- W = 3 × 2 + 4 × 3.
- = 6 + 12 = 18 J.
Answer: 18 J
A pump lifts 500 kg of water to a height of 20 m in 10 s. Find its power output.
Show answer & solution
Concept: P = work/time = mgh/t.
- Work = 500 × 10 × 20 = 1.0 × 10⁵ J.
- P = 10⁵/10 = 10⁴ W = 10 kW.
Answer: 10 kW
A spring (k = 200 N m⁻¹) is compressed by 0.1 m and then launches a 0.1 kg ball on a smooth level surface. Find the ball's speed.
Show answer & solution
Concept: spring PE becomes kinetic energy.
- U = ½kx² = ½ × 200 × 0.01 = 1 J.
- ½ × 0.1 × v² = 1 ⇒ v² = 20.
- v ≈ 4.47 m s⁻¹.
Answer: about 4.47 m s⁻¹
The speed of a 2 kg body increases from 4 m s⁻¹ to 10 m s⁻¹. Find the net work done on it.
Show answer & solution
Concept: work–energy theorem W = ΔK.
- K₁ = ½ × 2 × 16 = 16 J; K₂ = ½ × 2 × 100 = 100 J.
- W = 100 − 16 = 84 J.
Answer: 84 J
A car moves at a steady 20 m s⁻¹ against a total resistance of 500 N. What power must its engine deliver?
Show answer & solution
Concept: at constant speed, engine force = resistance; P = Fv.
- F = 500 N.
- P = 500 × 20 = 10 000 W = 10 kW.
Answer: 10 kW
Advanced level
A bob on a 1 m string is to complete a vertical circle. Find the minimum speed at the lowest point and the difference between the string tensions at the bottom and the top (mass 0.2 kg).
Show answer & solution
Concept: at the top, the least speed makes T = 0, so mg = mvtop²/r; then use energy conservation.
- vtop² = gr = 10.
- Energy: vbottom² = vtop² + 2g(2r) = 10 + 40 = 50 ⇒ vbottom ≈ 7.07 m s⁻¹ (= √(5gr)).
- Bottom: Tb = mg + mvb²/r; top: Tt = mvt²/r − mg.
- Tb − Tt = 2mg + m(vb² − vt²)/r = 2mg + 4mg = 6mg = 6 × 0.2 × 10 = 12 N.
Answer: 7.07 m s⁻¹; 12 N
A 2 kg ball moving at 6 m s⁻¹ hits a stationary 4 kg ball head-on. Find the final velocities if the collision is elastic, and the kinetic energy lost if instead the balls stick together.
Show answer & solution
Concept: elastic formulas for m₂ at rest; momentum conservation for sticking.
- v₁ = (2 − 4)/(6) × 6 = −2 m s⁻¹ (bounces back).
- v₂ = 2 × 2/6 × 6 = 4 m s⁻¹. Check KE: 36 J before; 4 + 32 = 36 J after.
- Sticking: common velocity = (2 × 6)/6 = 2 m s⁻¹.
- KE after = ½ × 6 × 4 = 12 J, so loss = 36 − 12 = 24 J.
Answer: −2 m s⁻¹ and 4 m s⁻¹; 24 J lost if they stick
A 1 kg body at rest is acted on by a force F = 3x² N along x. Find the work done from x = 0 to x = 2 m and the speed at x = 2 m.
Show answer & solution
Concept: W = ∫F dx, then W = ΔK.
- W = ∫₀² 3x² dx = [x³]₀² = 8 J.
- ½ × 1 × v² = 8 ⇒ v² = 16.
- v = 4 m s⁻¹.
Answer: 8 J; 4 m s⁻¹
A 2 kg block slides from rest down a 10 m long incline at 30° with μk = 0.2. Find the work done by friction and the speed at the bottom.
Show answer & solution
Concept: loss of PE = gain of KE + work against friction.
- Height dropped h = 10 sin30° = 5 m; PE lost = 2 × 10 × 5 = 100 J.
- Friction = μmg cos30° = 0.2 × 20 × 0.866 = 3.46 N; work against it = 3.46 × 10 ≈ 34.6 J.
- KE at bottom = 100 − 34.6 = 65.4 J.
- v = √(2 × 65.4/2) ≈ 8.08 m s⁻¹.
Answer: −34.6 J by friction; about 8.08 m s⁻¹
A 10 g bullet moving at 500 m s⁻¹ gets embedded in a 2 kg wooden block hanging from a long string. How high does the block rise, and what fraction of the bullet's kinetic energy is lost?
Show answer & solution
Concept: momentum is conserved during the impact (not energy); after that, mechanical energy is conserved as the block swings.
- Common velocity V = (0.01 × 500)/(2.01) ≈ 2.49 m s⁻¹.
- Rise h = V²/2g = 6.19/20 ≈ 0.31 m.
- Fraction lost = M/(m + M) = 2/2.01 ≈ 0.995.
- So about 99.5% of the bullet's KE becomes heat and deformation.
Answer: about 31 cm; about 99.5% lost
12. PYQ-pattern Questions with Solutions
These questions are original and modelled on concepts frequently tested in JEE Main.
If the momentum of a body is doubled, its kinetic energy becomes:
- Double
- Four times
- Half
- Unchanged
Show solution
K = p²/2m, so doubling p multiplies K by 4.
Answer: (B)
The work done by a conservative force along a closed path is:
- Always positive
- Always negative
- Zero
- Depends on the path
Show solution
For a conservative force, work depends only on end points; for a closed path they coincide, so W = 0.
Answer: (C)
The potential energy of a particle is U(x) = x² − 4x (SI units). The particle is in:
- Stable equilibrium at x = 2 m
- Unstable equilibrium at x = 2 m
- Stable equilibrium at x = 4 m
- No equilibrium
Show solution
dU/dx = 2x − 4 = 0 at x = 2; d²U/dx² = 2 > 0, so U is minimum: stable.
Answer: (A)
A constant force acts on a body that starts from rest. The power delivered by the force varies with time t as:
- P ∝ t⁰
- P ∝ t
- P ∝ t²
- P ∝ √t
Show solution
Constant force ⇒ constant acceleration ⇒ v = at. P = Fv = Fat ∝ t.
Answer: (B)
A ball dropped from 20 m hits the floor with coefficient of restitution 0.5. Find the height (in m) of its first rebound.
Show solution
h′ = e²h = 0.25 × 20 = 5 m.
Answer: 5
A 1 kg body has kinetic energy 50 J. Find the magnitude of its momentum in kg m s⁻¹.
Show solution
p = √(2mK) = √(2 × 1 × 50) = 10 kg m s⁻¹.
Answer: 10
13. Quick Revision Summary
- W = Fs cosθ; variable force: area under F–x graph.
- Wnet = ΔK for any force; K = p²/2m.
- F = −dU/dx; stable equilibrium at minimum U.
- P = F·v; at constant speed P = (resistance) × v.
- Momentum always conserved in collisions; e = separation/approach speed; h′ = e²h.
- Vertical circle: √(5gr) at bottom, √(gr) at top, tension difference 6mg.