Work, energy and power for NEET

Energy methods are the shortcut of mechanics. A question that needs three steps with Newton's laws often needs one line with the work–energy theorem, because energy is a scalar and you do not have to track directions. Learning when to switch from forces to energy is most of what this chapter teaches.

NEET usually asks it in a few recognisable forms: work from a graph or a variable force, speed after a fall or a slide, power of a pump or vehicle, the vertical circle, and collisions. Each has one idea you must be sure of.

Work

For a constant force, W = F·s = Fs cos θ, where θ is the angle between force and displacement. Work is a scalar, measured in joules.

• θ < 90°: positive work (gravity on a falling ball).
• θ > 90°: negative work (friction on a sliding block).
• θ = 90°: zero work. The centripetal force in uniform circular motion and the normal force on a block sliding along a level floor do no work.

For a variable force, W = ∫F dx, which is the area under the force–displacement graph. For a spring (F = −kx), the work done by an external agent to stretch it from 0 to x is ½kx². The spring itself does −½kx².

The work–energy theorem

The net work done by all forces on a body equals its change in kinetic energy: W_net = K_f − K_i.

A sketch of why: for a constant force, v² − u² = 2as. Multiply by m/2 to get ½mv² − ½mu² = mas = Fs. The result holds for variable forces too.

"All forces" includes friction, gravity, tension and any applied force. That is what makes the theorem powerful: you do not need the acceleration.

Kinetic energy and momentum are linked by K = p²/2m. For equal momenta, the lighter body has more kinetic energy. For equal kinetic energies, the heavier body has more momentum.

Conservative forces and potential energy

A force is conservative if the work it does depends only on the start and end points, not on the path, so the work round any closed loop is zero. Gravity and the spring force are conservative. Friction and air resistance are not.

Only conservative forces have a potential energy, related by F = −dU/dx. At equilibrium, dU/dx = 0. A minimum of U is stable equilibrium; a maximum is unstable.

Conservation of mechanical energy: if only conservative forces do work, K + U stays constant. When friction acts, the loss in mechanical energy equals the work done against friction.

Power

Power is the rate of doing work: P = dW/dt. For a force acting on a moving body, P = F·v. The SI unit is the watt; 1 horsepower = 746 W.

A vehicle moving at constant speed against a resistive force F needs power Fv from the engine. A pump lifting mass m through height h in time t needs at least mgh/t.

Motion in a vertical circle

A small body of mass m whirled on a string of length r in a vertical circle has different speeds at the top and bottom, because gravity does work as it moves.

• At the top, tension and weight both point down: T_top + mg = mv_top²/r. The string stays taut only if v_top ≥ √(gr).
• Energy conservation between top and bottom (height 2r): v_bottom² = v_top² + 4gr. So the minimum speed at the bottom is √(5gr).
• At the bottom: T_bottom − mg = mv_bottom²/r. With the minimum speeds, T_top = 0 and T_bottom = 6mg. In general, T_bottom − T_top = 6mg.

Collisions

Momentum is conserved in every collision if no external force acts during it. Kinetic energy is conserved only in an elastic collision.

Elastic collision in one dimension, with m₂ at rest:

• v₁ = (m₁ − m₂)u₁ / (m₁ + m₂)
• v₂ = 2m₁u₁ / (m₁ + m₂)

Special cases worth knowing: equal masses exchange velocities; a light ball hitting a very heavy one bounces back with almost the same speed; a heavy ball hitting a light one hardly slows and the light one moves off at nearly 2u₁.

Perfectly inelastic: common velocity v = (m₁u₁ + m₂u₂) / (m₁ + m₂).

In two dimensions, momentum is conserved separately along x and y. If a body hits an identical body at rest in an elastic glancing collision, the two move off at 90° to each other.

Worked numericals

Take g = 10 m s⁻².

Example 1: stopping distance with friction

A 2 kg block moving at 10 m s⁻¹ slides on a floor with μ_k = 0.5. How far does it go?

• Work done by friction = −μ_k mg d = −0.5 × 2 × 10 × d = −10d.
• Change in KE = 0 − ½ × 2 × 100 = −100 J.
• −10d = −100, so d = 10 m. (In general d = v²/2μg, independent of mass.)

Example 2: work by a variable force

A force F = 3x² N acts along x. Work done from x = 0 to x = 2 m?

• W = ∫₀² 3x² dx = [x³]₀² = 8 J.

Example 3: vertical circle

A 0.5 kg stone on a 1 m string is whirled in a vertical circle, just completing it.

• Minimum speed at the top = √(10 × 1) ≈ 3.16 m s⁻¹.
• Speed at the bottom = √(5 × 10 × 1) = √50 ≈ 7.07 m s⁻¹.
• Tension at the bottom = mg + mv²/r = 5 + 0.5 × 50 = 30 N, which is 6mg ✓.

Example 4: elastic versus sticky collision

A 2 kg ball at 6 m s⁻¹ hits a 4 kg ball at rest.

• Elastic: v₁ = (2 − 4) × 6 / 6 = −2 m s⁻¹ (bounces back). v₂ = 2 × 2 × 6 / 6 = 4 m s⁻¹.
• Check momentum: 2 × 6 = 12 and 2 × (−2) + 4 × 4 = 12 ✓. Check KE: 36 J before; 4 + 32 = 36 J after ✓.
• Perfectly inelastic: v = 12/6 = 2 m s⁻¹. KE after = ½ × 6 × 4 = 12 J, so 24 J (two-thirds) is lost.

Practice MCQs

1. A 1 kg body moving at 4 m s⁻¹ hits and sticks to an identical body at rest. The common velocity is: (a) 1 m s⁻¹ (b) 2 m s⁻¹ (c) 4 m s⁻¹ (d) 8 m s⁻¹
2. The work done by gravity on a 5 kg body falling 3 m is: (a) 15 J (b) 50 J (c) 150 J (d) 1,500 J
3. A body's momentum increases by 50%. Its kinetic energy increases by: (a) 50% (b) 100% (c) 150% (d) 125%
4. A car moves at a steady 20 m s⁻¹ against a total resistance of 500 N. The engine's power is: (a) 10 kW (b) 5 kW (c) 2.5 kW (d) 25 kW
5. A spring (k = 800 N m⁻¹) is stretched from 5 cm to 10 cm. The work done is: (a) 1 J (b) 2 J (c) 3 J (d) 4 J
6. The potential energy of a particle is U = 4x² − 8x (SI units). The particle is in equilibrium at: (a) x = 0 (b) x = 1 m (c) x = 2 m (d) x = 4 m
7. A pump lifts 100 kg of water through 10 m in 20 s. Its minimum power is: (a) 500 W (b) 200 W (c) 50 W (d) 1,000 W
8. A ball hits an identical ball at rest in an elastic, glancing collision. The angle between their paths afterwards is: (a) 0° (b) 45° (c) 90° (d) 180°

Answers

1. (b) 1 × 4 = 2 × v, so v = 2 m s⁻¹.
2. (c) W = mgh = 5 × 10 × 3 = 150 J.
3. (d) K ∝ p², and 1.5² = 2.25, an increase of 125%.
4. (a) P = Fv = 500 × 20 = 10,000 W.
5. (c) ½ × 800 × (0.10² − 0.05²) = 400 × 0.0075 = 3 J. Not ½k(0.05)², which gives 1 J.
6. (b) F = −dU/dx = −(8x − 8) = 0 at x = 1 m.
7. (a) mgh/t = 100 × 10 × 10 / 20 = 500 W.
8. (c) Equal masses, one at rest, elastic: 90°.

What to do next

• For every mechanics numerical this week, ask first: can energy do this in one line?
• Derive the √(5gr) result and the 6mg tension difference once from scratch.
• Memorise the two elastic-collision formulas and test the three special cases on them.
• Solve previous-year NEET questions on collisions and power, timed.

Next: rotational motion for NEET. For the forces behind these problems, see laws of motion for NEET.