The Work-Energy Theorem

The work-energy theorem states that the net work done on an object equals the change in its kinetic energy: W_net = ΔE_k = ½mv_f² − ½mv_i². In brief: positive net work speeds an object up, negative net work slows it down, and zero net work leaves its speed unchanged. The theorem follows from Newton’s second law, and it gives a way of finding speeds and distances without finding the acceleration or time.

Stopping distances of cars, the speed of a skier at the bottom of a hill and the penetration of a bullet into a block can be calculated by the work-energy theorem. This guide states the theorem and derives it, explains how to apply it to problems with several forces, compares it with the kinematic method in a table, works through examples and corrects common mistakes.

What does the theorem say?

The sum of the work done by all forces acting on a body (the net work, or the work done by the net force) equals the change in kinetic energy:

W_net = E_k,final − E_k,initial = ½mv² − ½mu²

It applies to forces that are constant or variable, and to any path. The net work is a scalar, so it is the sum of the signed works of the individual forces, or F_nets cos θ for a constant net force.

Where does it come from?

For a constant net force F along a line, Newton’s second law gives F = ma, and the kinematic equation v² = u² + 2as can be rearranged as as = (v² − u²) ÷ 2. Multiplying by m: Fs = ½mv² − ½mu². The left-hand side is the net work, and the right-hand side is the change in kinetic energy. For variable forces, the same result follows by adding up small steps (an integral).

How is it used?

1. Draw the object and list all forces acting on it.
2. For each force, calculate the work (Fs cos θ), with the sign.
3. Add them to find W_net.
4. Set W_net = ½mv² − ½mu² and solve for the unknown (speed, distance or force).

The method is especially useful when the force is not constant or the path is curved, since only the start and end speeds and the work matter.

Work-energy method and kinematic method compared

Worked example: braking and pushing

1. Car speeding up. A 1,000 kg car increases its speed from 10 m/s to 20 m/s. W_net = ½ × 1,000 × (20² − 10²) = 500 × 300 = 150,000 J = 150 kJ.
2. Braking distance. The same car moving at 20 m/s brakes with a constant force of 5,000 N until it stops. The work by the brake force is −5,000 × d. W_net = 0 − ½ × 1,000 × 20² = −200,000 J. So −5,000d = −200,000 and d = 40 m. At 40 m/s the kinetic energy is four times larger, and the distance is 160 m.
3. Pushing a block. A 3.0 kg block starts from rest on a smooth floor and is pushed by a constant horizontal force of 12 N through 5.0 m. W = 12 × 5.0 = 60 J = ½ × 3.0 × v², so v² = 40 and v = 6.3 m/s.
4. With friction. If the floor has a friction force of 4.0 N, the net force is 8.0 N, W_net = 8.0 × 5.0 = 40 J, and v = √(2 × 40 ÷ 3.0) = √26.7 = 5.2 m/s.

How was the theorem developed?

The relation between work and the change in vis viva was found by Leibniz and by Johann Bernoulli in the early 1700s. Coriolis formulated the modern form with the factor ½ in 1829, and it became a central principle of mechanics. In the 20th century it was extended to relativity, where the work changes the total energy including rest energy.

Which mistakes are common?

Mistake 1: Using the work of only one force

The theorem uses the net work of all forces, including friction and gravity.

Mistake 2: Forgetting the sign of work

Braking forces do negative work.

Mistake 3: Squaring the difference of speeds

Use v² − u², not (v − u)².

Mistake 4: Applying it as if it gave the time or direction

Energy is a scalar. The theorem gives the speed, but not the direction of the velocity.