Why this chapter is a shortcut, not a topic
Consider a block sliding down a curved, frictionless track and being asked for its speed at the bottom. Using Newton's laws you would need the shape of the track, the normal force at every point, and an integration along a path you do not know. Using energy conservation it is one line: mgh = ½mv², so v = √(2gh). The shape is irrelevant.
That is the whole value of the chapter. Energy methods discard the intermediate motion and connect two states directly. The cost is that they cannot tell you anything about time or about the forces involved along the way - which is precisely why the exam mixes both kinds of question, to see whether you know which tool the question is asking for.
How JEE actually asks Work, Energy and Power
NTA does not publish chapter-wise weightage, so treat any figure circulating online as a coaching estimate derived from past papers, and confirm the syllabus and pattern for your session in the official NTA information bulletin.
The observable pattern is that this chapter rarely appears alone. Questions typically combine it with Laws of Motion (a block on a rough incline), with springs (a mass launched by a compressed spring), or with collisions (a bullet embedding in a block). Collision questions in particular are a reliable presence, and they are where the conservation rules get tested most sharply.
Key concepts, compressed
- ✓Work is the dot product F·d, so only the component of force along the displacement counts. A force perpendicular to motion does no work at all.
- ✓The work-energy theorem says the net work by all forces equals the change in kinetic energy. It is always true, with or without friction.
- ✓A conservative force is one whose work depends only on the endpoints, not the path - gravity and the spring force. Potential energy is defined only for these.
- ✓Mechanical energy, kinetic plus potential, is conserved only when non-conservative forces do no work.
- ✓Power is the rate of doing work. Average power is total work over total time; instantaneous power is the dot product F·v.
- ✓In all collisions momentum is conserved. Elastic collisions also conserve kinetic energy; inelastic ones do not, and perfectly inelastic ones lose the maximum possible while still conserving momentum.
Formulas you need before attempting the questions
| Quantity | Formula | Note |
|---|---|---|
| Work by a constant force | W = Fd cosθ | θ between force and displacement |
| Work by a variable force | W = ∫F dx | area under a force-displacement graph |
| Kinetic energy | KE = ½mv² = p²/2m | the second form is useful with momentum |
| Work-energy theorem | W(net) = ΔKE | always valid |
| Gravitational PE | U = mgh | near the Earth's surface |
| Spring PE | U = ½kx² | x measured from the natural length |
| Work to stretch a spring | W = ½k(x₂² − x₁²) | not ½k(x₂ − x₁)² |
| Average power | P = W/t | |
| Instantaneous power | P = F·v | |
| Coefficient of restitution | e = (relative speed after)/(relative speed before) | 1 elastic, 0 perfectly inelastic |
| Rebound height | e = √(h₂/h₁) | ball dropped and bouncing |
The five mistakes that cost the most marks
- ✓Applying conservation of mechanical energy when friction is present. Energy is still conserved overall, but mechanical energy is not - the friction work has to appear explicitly.
- ✓Assuming kinetic energy is conserved in every collision. It is conserved only in elastic collisions. Momentum, by contrast, is conserved in all of them.
- ✓Computing the work to stretch a spring from x₁ to x₂ as ½k(x₂ − x₁)². The correct expression is ½k(x₂² − x₁²), because the spring force depends on the absolute extension.
- ✓Getting the sign of friction work wrong. Friction opposes relative motion, so the work it does on the moving body is negative and removes kinetic energy.
- ✓Confusing average and instantaneous power. A question giving a total mass lifted over a total time wants W/t; one giving a force and a current speed wants F·v.
Practice set 1: work
1. A force of 10 N acts at 60° to the displacement of a body, which moves 5 m. How much work does it do?
25 J. W = Fd cosθ = 10 × 5 × cos60° = 10 × 5 × 0.5 = 25 J. Only the component of force along the displacement contributes; the perpendicular component does no work. Substituting sin60° instead of cos60° is the standard error and gives about 43 J, which will be one of the options.
2. A satellite moves in a circular orbit. How much work does gravity do on it over one revolution?
Zero. The gravitational force is always directed toward the centre while the velocity is tangential, so the two are perpendicular and cosθ = 0 at every instant. This is also why the satellite's speed does not change. The same reasoning applies to the tension in a whirled string and to the magnetic force on a moving charge.
3. A 2 kg body falls freely through 5 m. How much work does gravity do on it?
100 J. W = mgh = 2 × 10 × 5 = 100 J, and it is positive because the force and displacement are both downward. By the work-energy theorem this equals the gain in kinetic energy, so the body arrives at √(2 × 100/2) = 10 m/s - consistent with v = √(2gh).
4. The same 2 kg body is lifted 5 m at constant speed. What work is done by the applied force, and by gravity?
The applied force does +100 J; gravity does −100 J. At constant speed the acceleration is zero, so the applied force equals mg = 20 N and does 20 × 5 = 100 J. Gravity acts opposite to the displacement, so its work is negative. The two sum to zero net work, which matches the zero change in kinetic energy.
5. A variable force F = 3x newtons acts on a body as it moves from x = 0 to x = 2 m. Find the work done.
6 J. The force is not constant, so W = ∫F dx = ∫3x dx from 0 to 2 = [1.5x²] from 0 to 2 = 1.5 × 4 = 6 J. Using W = Fd with the final force of 6 N would give 12 J - exactly double, because the force grows linearly and the average value is 3 N, not 6 N.
6. A spring of force constant 200 N/m is stretched by 0.1 m from its natural length. How much work is done?
1 J. W = ½kx² = ½ × 200 × (0.1)² = ½ × 200 × 0.01 = 1 J. This expression is the integral of the spring force kx over the extension, which is why the answer is not simply force times distance - the force is zero at the start and 20 N at the end.
7. The same spring is now stretched further, from 0.1 m to 0.2 m. How much additional work is required?
3 J. W = ½k(x₂² − x₁²) = ½ × 200 × (0.04 − 0.01) = 100 × 0.03 = 3 J. Note that it takes three times as much work for the second 0.1 m as for the first, because the spring resists more as it extends. Computing ½k(x₂ − x₁)² = ½ × 200 × 0.01 = 1 J is the trap, and it wrongly suggests the two stages cost the same.
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Practice set 2: the work-energy theorem
8. A 2 kg body speeds up from 5 m/s to 10 m/s. What is the net work done on it?
75 J. By the work-energy theorem, W(net) = ΔKE = ½m(v² − u²) = ½ × 2 × (100 − 25) = 75 J. Do not compute ½m(v − u)² = ½ × 2 × 25 = 25 J - kinetic energy depends on the square of the speed, and differences of squares are not squares of differences.
9. A 4 kg body has kinetic energy 200 J. What is its speed?
10 m/s. From KE = ½mv², v = √(2 × KE/m) = √(2 × 200/4) = √100 = 10 m/s. Worth internalising the scaling: doubling the speed quadruples the kinetic energy, which is the basis of the next question.
10. If the momentum of a body is doubled, what happens to its kinetic energy?
It becomes four times larger. Using KE = p²/2m, kinetic energy is proportional to the square of momentum at constant mass, so doubling p multiplies KE by four. The p²/2m form is worth memorising precisely for questions that link momentum and energy, where ½mv² forces an extra step.
11. A 1 kg block sliding at 10 m/s on a rough horizontal surface with μ = 0.2 comes to rest. How far does it travel?
25 m. Friction is the only horizontal force, and its work equals the change in kinetic energy: −μmgd = −½mv², so d = v²/(2μg) = 100/(2 × 0.2 × 10) = 100/4 = 25 m. The mass cancels, so a 10 kg block would stop in the same distance. Doing this by forces and kinematics needs two steps; the work-energy route needs one.
12. A 10 g bullet travelling at 200 m/s embeds itself 0.1 m into a fixed block. What average resistive force does the block exert?
2000 N. The bullet's kinetic energy is ½ × 0.01 × 200² = ½ × 0.01 × 40000 = 200 J, and all of it is removed over 0.1 m. So F × 0.1 = 200, giving F = 2000 N. Note that the mass must be converted to kilograms first - leaving it as 10 grams is the most frequent error here.
13. A body is dropped from 20 m. Using energy methods, find its speed on reaching the ground.
20 m/s. All the potential energy converts to kinetic: mgh = ½mv², so v = √(2gh) = √(2 × 10 × 20) = √400 = 20 m/s. The mass cancels, which is the energy-method statement of the fact that all bodies fall alike.
14. For the same body, what is its speed at half that height, 10 m above the ground?
About 14.1 m/s. It has fallen 10 m, so v = √(2 × 10 × 10) = √200 ≈ 14.14 m/s. Note that this is not half the final speed - speed goes as the square root of distance fallen, so at half the height the body has 1/√2 of its final speed, about 71% of it. Questions exploit the assumption that it would be 10 m/s.
Practice set 3: conservation of energy
15. A pendulum bob is released from rest at a height 0.45 m above its lowest point. What is its speed at the bottom?
3 m/s. Mechanical energy is conserved because the string tension is always perpendicular to the motion and does no work: mgh = ½mv², so v = √(2 × 10 × 0.45) = √9 = 3 m/s. The length of the string and the mass of the bob are irrelevant - only the vertical drop matters.
16. A block slides from rest down a frictionless incline through a vertical drop of 5 m. What is its speed at the bottom?
10 m/s. v = √(2gh) = √(2 × 10 × 5) = √100 = 10 m/s. The angle of the incline does not appear, and neither does the mass. This is the clearest demonstration of what energy methods buy you: the answer is the same for a steep ramp, a gentle ramp, or a curved track, provided there is no friction.
17. A spring of constant 100 N/m is compressed by 0.2 m and used to launch a 0.5 kg block horizontally. What is the block's speed as it leaves the spring?
About 2.83 m/s. The stored energy is ½kx² = ½ × 100 × 0.04 = 2 J, all of which becomes kinetic energy: 2 = ½ × 0.5 × v², so v² = 8 and v = 2√2 ≈ 2.83 m/s. On a horizontal surface there is no change in gravitational potential energy, which is what makes the transfer complete.
18. For a body performing vertical circular motion on the inside of a track, how do the minimum speeds at the top and the bottom compare?
The minimum speed at the top is √(gr) and at the bottom it is √(5gr), so the bottom speed is √5 times larger. At the top, gravity alone supplies the centripetal force when the track force is zero, giving v² = gr. Applying energy conservation between top and bottom over a height of 2r then gives v(bottom)² = gr + 4gr = 5gr. This combination of circular motion and energy conservation is a standard exam construction.
19. Under what condition is mechanical energy conserved?
When no non-conservative force does work on the system. Gravity and spring forces are conservative, so a system acted on only by these conserves mechanical energy. Friction, air resistance and applied pushes are non-conservative, and their presence means you must use the work-energy theorem with those contributions written in explicitly.
20. A ball is thrown vertically upward at 20 m/s. Using energy methods, how high does it rise?
20 m. All the kinetic energy converts to potential energy: ½mv² = mgh, so h = v²/2g = 400/20 = 20 m. Again the mass cancels. This is the same result kinematics gives through v² = u² + 2as, which is unsurprising - that equation is the work-energy theorem in disguise.
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Practice set 4: power
21. A force of 500 N drives a vehicle at a constant 4 m/s. What power is being delivered?
2000 W, or 2 kW. P = Fv = 500 × 4 = 2000 W. This is instantaneous power, and because the speed is constant it is also the average. At constant velocity the driving force equals the total resistive force, so this is the power spent overcoming resistance rather than accelerating.
22. A pump raises 100 kg of water through 10 m in 20 s. What is its output power?
500 W. The work done is mgh = 100 × 10 × 10 = 10000 J, and P = W/t = 10000/20 = 500 W. This is average power, which is what a total-work-over-total-time question always asks for. Note this is the useful output; a real pump's input power would be higher because of losses.
23. A 1000 W motor lifts a 50 kg load. What is the maximum steady speed it can sustain?
2 m/s. At maximum steady speed the load moves at constant velocity, so the lifting force equals mg = 500 N. Then P = Fv gives v = P/F = 1000/500 = 2 m/s. The phrase 'maximum steady speed' is the signal that acceleration is zero and the force is therefore just the weight.
24. A car accelerates from rest. Is its average power over the first 10 s equal to its instantaneous power at t = 10 s?
No, and generally the instantaneous value is larger. Instantaneous power is F·v, which grows as the car speeds up from zero, so the average over the interval is lower than the final value. Questions that supply a total distance and time want W/t; those that supply a force and a current speed want F·v.
25. How many joules are there in one kilowatt-hour?
3.6 × 10⁶ J. One kilowatt is 1000 W and one hour is 3600 s, so 1000 × 3600 = 3.6 × 10⁶ J. The kilowatt-hour is a unit of energy, not power, despite containing 'watt' in its name - which is exactly the confusion the question is testing.
Practice set 5: collisions
26. A 2 kg body moving at 6 m/s collides with a stationary 4 kg body and they move off together. What is their common velocity?
2 m/s. This is a perfectly inelastic collision, and momentum is conserved: 2 × 6 + 4 × 0 = (2 + 4)v, so 12 = 6v and v = 2 m/s. Momentum conservation applies here even though the bodies stick together and energy is lost - the two facts are independent.
27. In that collision, how much kinetic energy is lost?
24 J. The initial kinetic energy is ½ × 2 × 6² = 36 J and the final is ½ × 6 × 2² = 12 J, so 24 J is lost - two-thirds of the original. It is converted to heat, sound and permanent deformation. A perfectly inelastic collision loses the maximum kinetic energy possible while still conserving momentum, which is a useful sanity check.
28. Two bodies of equal mass undergo a perfectly elastic head-on collision, with one initially at rest. What happens?
They exchange velocities: the moving body stops and the stationary one moves off with the original speed. Applying conservation of both momentum and kinetic energy to equal masses forces this outcome, and it is the principle behind a Newton's cradle. This result holds only for equal masses in an elastic collision.
29. What are the values of the coefficient of restitution for perfectly elastic and perfectly inelastic collisions?
e = 1 for perfectly elastic and e = 0 for perfectly inelastic. The coefficient is the ratio of relative separation speed to relative approach speed. When e = 1 the bodies separate as fast as they approached and kinetic energy is conserved; when e = 0 they do not separate at all and move together. Real collisions fall between the two.
30. A ball dropped from 5 m rebounds to a height of 1.25 m. What is the coefficient of restitution?
0.5. For a ball bouncing off the ground, e = √(h₂/h₁) = √(1.25/5) = √0.25 = 0.5. That relation follows from the fact that both impact and rebound speeds go as √(2gh), so the speed ratio is the square root of the height ratio. Answering 0.25 comes from taking the height ratio directly and skipping the square root.
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How to study this chapter efficiently
- ✓Before starting any mechanics question, ask whether it mentions time or force. If it does not, and it asks about speed or position, try energy first - you will usually save two steps.
- ✓Write down explicitly whether friction is present. That single note decides whether you may use conservation of mechanical energy or must use the work-energy theorem with a friction term.
- ✓For collisions, write momentum conservation first, always. Only then ask whether kinetic energy is also conserved. Doing it in that order prevents the most common error in the chapter.
- ✓Practise spring questions with two different extensions, not just from the natural length. The ½k(x₂² − x₁²) form is where marks are lost.
- ✓Check your answers against limiting cases: does the energy lost go to zero when e = 1? Does the speed at the bottom of an incline stop depending on the angle when friction is removed?
- ✓Combine this chapter with Laws of Motion in your practice rather than studying it separately, because that is how it appears in the paper.
Turn this into active practice
The skill this chapter needs - choosing between forces and energy - only develops by facing mixed questions where the choice is not signposted. Working through a set on springs alone never trains it, because you already know what tool to use.
The JEE Work, Energy and Power quiz on QUFF generates fresh questions across the whole chapter each time, scores them instantly and explains each answer. Do a mixed timed set, and for every question you get wrong write down whether the error was choosing the wrong method or executing the right one badly. Those two failures need completely different fixes.