Work, energy and power for JEE Main

Newton's laws tell you the acceleration at one instant. Energy methods tell you how fast something is moving at the end of a journey without tracking every instant in between. That is why this chapter saves time: a block sliding down a curved, rough track is a nightmare with F = ma and a three-line problem with energy.

The chapter also runs through the rest of the paper. Vertical circles, collisions, SHM, gravitation and electrostatics all use the same ideas, so weak energy skills cost marks in several places.

Work by a constant force

For a constant force F and displacement s, W = F·s = Fs cos θ, where θ is the angle between them.

• θ < 90°: positive work (the force helps the motion).
• θ = 90°: zero work. The normal force on a block sliding on a floor, the tension in a string of a body in uniform circular motion and the magnetic force on a charge all do zero work.
• θ > 90°: negative work. Kinetic friction on a sliding block is the usual example.

Work depends on the frame. A person holding a bag while standing still does no work on it in the ground frame, however tired their arm gets.

Work by a variable force

When the force changes with position, add up small pieces: W = ∫F dx, which is the area under the F–x graph (areas below the axis count as negative). In vector form, W = ∫(Fₓ dx + F_y dy + F_z dz).

The work–energy theorem

Net work done by all forces = change in kinetic energy.

The derivation is two lines. Write a = dv/dt = v dv/dx, so F = mv dv/dx. Then ∫F dx = ∫mv dv = ½mv² − ½mu².

Because it uses the net work, the theorem holds for any force, constant or not, conservative or not. Include every force: gravity, normal, friction, tension, applied.

Conservative forces and potential energy

A force is conservative if the work it does between two points does not depend on the path. Equivalently, it does zero work around any closed path. Gravity, the spring force and the electrostatic force are conservative. Friction and air drag are not.

For a conservative force you can define potential energy U so that F = −dU/dx and W_conservative = −ΔU.

Two forms of energy conservation follow:

• Only conservative forces do work: KE + U stays constant.
• Non-conservative forces also act: W_nc = Δ(KE + U). Work done by friction is negative, so mechanical energy falls.

Reading a potential energy curve

Equilibrium points are where F = −dU/dx = 0, the flat points of the U–x graph.

• Stable: U is a minimum (d²U/dx² > 0). A small push produces a force back towards equilibrium.
• Unstable: U is a maximum (d²U/dx² < 0).
• Neutral: U is constant over a region.

A particle with total energy E can only be where U ≤ E. The points where U = E are turning points.

Springs

The spring force is F = −kx. The work you do to stretch a spring from x₁ to x₂ is ½k(x₂² − x₁²), not ½k(x₂ − x₁)². Stretching the second centimetre costs more than the first.

Power

Power is the rate of doing work: P = dW/dt = F·v. Average power is total work divided by total time.

Units: 1 W = 1 J/s; 1 horsepower ≈ 746 W; 1 kWh = 3.6 × 10⁶ J, which is energy, not power.

Standard problem types and traps

• Work by static friction is not always zero. A box on the floor of an accelerating truck is pushed forward by static friction, which does positive work on the box in the ground frame.
• KE and momentum are linked by KE = p²/2m. If KE rises by 44%, p rises by 20%, because √1.44 = 1.2.
• Chains and ropes: find the change in height of the centre of mass of the part that moves.
• Work done by gravity depends only on the vertical drop, whatever the path.

Worked problems

Problem 1 (numerical answer): variable force. A force F = (3x² − 2x) N acts along the x-axis. Find the work done as a particle moves from x = 1 m to x = 3 m.

W = ∫(3x² − 2x) dx from 1 to 3 = [x³ − x²] from 1 to 3 = (27 − 9) − (1 − 1) = 18 J.

Problem 2: rough incline. A 2 kg block is released from rest at the top of a 5 m long incline at 30°. The coefficient of kinetic friction is 1/(2√3). Find its speed at the bottom (g = 10 m/s²).

Work by gravity = mgL sin 30° = 2 × 10 × 5 × 0.5 = 50 J.
Friction force = μmg cos 30° = (1/(2√3)) × 20 × (√3/2) = 5 N, so work by friction = −5 × 5 = −25 J.
The normal force does no work. Net work = 25 J = ½ × 2 × v², so v = 5 m/s.

Problem 3: dropped onto a spring. A 1 kg block is dropped from rest 0.6 m above the top of a vertical spring (k = 400 N/m). Find the maximum compression (g = 10 m/s²).

At maximum compression x the block is momentarily at rest, having fallen (0.6 + x):
mg(0.6 + x) = ½kx², so 10(0.6 + x) = 200x², which gives 100x² − 5x − 3 = 0.
x = [5 + √(25 + 1200)] / 200 = (5 + 35)/200 = 0.2 m.

Problem 4: power on a slope. A 1,200 kg car climbs a road with sin θ = 1/20 at a steady 15 m/s against a resistance of 600 N (g = 10 m/s²). Find the engine's power.

Steady speed, so engine force = mg sin θ + resistance = 12,000/20 + 600 = 1,200 N.
P = Fv = 1,200 × 15 = 18 kW.

Practice set

1. U(x) = x² − 4x (SI units). Find the equilibrium position and say whether it is stable.
2. A spring has k = 200 N/m. Find the work needed to stretch it from 0 to 0.1 m, and then from 0.1 m to 0.2 m.
3. A body's kinetic energy increases by 44%. By what percentage does its momentum increase?
4. A force F = (2î + 3ĵ) N moves a particle through d = (3î − 2ĵ + k̂) m. Find the work done.
5. A 2 kg ball is thrown up at 20 m/s and reaches a maximum height of 18 m (g = 10 m/s²). How much work does air resistance do?
6. A pump lifts 600 kg of water through 20 m in one minute (g = 10 m/s²). Find the output power. What input power is needed if it is 80% efficient?
7. U(x) = 3x² − 2x³. Find the equilibrium points and classify them.
8. A uniform chain of mass M and length L lies on a smooth table with one-third of its length hanging over the edge. How much work is needed to pull it fully onto the table?

Answers

1. dU/dx = 2x − 4 = 0 gives x = 2 m. d²U/dx² = 2 > 0, so it is stable.
2. ½ × 200 × 0.01 = 1 J; then ½ × 200 × (0.04 − 0.01) = 3 J.
3. p ∝ √KE, so p becomes √1.44 = 1.2 times: a 20% increase.
4. W = F·d = 6 − 6 + 0 = 0 J. The force is perpendicular to the displacement.
5. Initial KE = ½ × 2 × 400 = 400 J; PE at the top = 2 × 10 × 18 = 360 J. Work by air = 360 − 400 = −40 J.
6. mgh/t = 600 × 10 × 20 / 60 = 2 kW output; input = 2/0.8 = 2.5 kW.
7. dU/dx = 6x − 6x² = 0 gives x = 0 and x = 1. d²U/dx² = 6 − 12x: at x = 0 it is +6 (stable); at x = 1 it is −6 (unstable).
8. The hanging part has mass M/3 and its centre of mass is L/6 below the table. W = (M/3) × g × (L/6) = MgL/18.

What to do next

• Derive the work–energy theorem and the spring work formula once, without notes.
• Solve 25 mixed problems, choosing energy or force methods before you start writing.
• Sketch three U–x curves and mark the stable, unstable and neutral points.
• Revise the vertical circle in circular motion, then move on to momentum and collisions, where energy is only sometimes conserved.