The work-energy theorem

What is the work-energy theorem?

In plain words: add up the work done on a body by every force acting on it. That total is exactly how much its kinetic energy goes up (or, if the total is negative, goes down). The form used here assumes constant mass, nonrelativistic motion and an inertial frame. For a particle or a rigid body in pure translation, kinetic energy is ½mv^². A rotating rigid body also has rotational kinetic energy: total work must be calculated at the actual points where the forces act, not simply from the net force times centre-of-mass displacement. For a system of several bodies, the work done by the internal forces between them must be added too: two blocks pushed apart by a spring gain kinetic energy from forces inside the system.

Here K_i and K_f are the initial and final kinetic energies, and W_net is the work done by the net force, which is the same as the sum of the works done by all the individual forces. The kinetic energy of a body of mass m moving at speed v is $K = \tfrac{1}{2}mv^{2}$.

The theorem connects forces with speeds without needing the time taken, which makes many problems short.

Where the theorem comes from

Start with a body of mass m moving in a straight line under a constant net force F, so its acceleration a is constant. From the equations of motion,

where u and v are the initial and final speeds and s is the displacement. Multiply both sides by m/2:

By Newton’s second law of motion, ma = F, so the right side is Fs:

The left side is the change in the quantity ‘half the mass times the square of the speed’: this is the kinetic energy. The right side, force times the displacement along it, is the work done by the net force.

In three dimensions the same steps use vectors. The equation of motion becomes

and multiplying by m/2, with $\vec F = m\vec a$, gives

so only the component of the force along the displacement does work. The theorem also holds for a force that varies; that proof uses calculus and comes in a later section of this chapter.

Using the theorem

• Add the works of all the forces. Gravity, friction, a push or a pull each do their own work, with its own sign. The net work is their sum, and it equals the change in kinetic energy.
• Positive net work speeds a body up; negative net work slows it down. Kinetic friction on a body sliding over a fixed surface does negative work.
• Zero net work over an interval gives equal initial and final speeds. The speed may change in between, as a ball rises and returns to its launch height. Speed stays constant throughout only if net power is zero at every instant; uniform circular motion is an example, although its velocity direction changes.
• No time is needed. Questions that ask “how far” or “how fast”, and not “how long”, are often one line with the theorem.
• Stopping distance grows with the square of the speed. With a constant net retarding force of magnitude F along a straight stopping path, the work −Fd must remove the kinetic energy $K = \tfrac{1}{2}mv^{2}$, so the stopping distance is K/F, that is mv²/(2F). Doubling the speed makes it four times as long, and two bodies with the same kinetic energy stop in the same distance under the same force.

The theorem in daily life

• Braking distance and speed limits. With the same braking force, a car at 60 km/h needs four times the braking distance it needs at 30 km/h (not counting the distance covered during the driver’s reaction time). This is one reason speed limits near schools are low.
• Bullets and armour. A bullet stopped by a fixed block loses all its kinetic energy to the work done by the resistive force over the distance it penetrates.
• A raindrop’s fall. Gravity does work on a falling drop, but air resistance does negative work, so the drop lands far slower than it would in a vacuum.
• Crumple zones and airbags. They make the distance over which a passenger is stopped longer, so a smaller force is enough to remove the same kinetic energy.

Solved examples

Common mistakes to avoid

Quick revision summary

• The work done by the net force on a body equals the change in its kinetic energy.
• It follows from the equation of motion for constant acceleration multiplied by m/2, and holds for varying forces too.
• The net work is the sum of the works done by all the forces, each with its sign.
• Positive net work increases kinetic energy; negative net work decreases it.
• With a constant braking force, the stopping distance is proportional to the square of the speed.

Questions students ask

What is the difference between the work-energy theorem and conservation of energy?

The work-energy theorem says that the net work done by all the forces on a body equals the change in its kinetic energy, W_net = K_f − K_i, whatever the forces are. Conservation of mechanical energy, K + V = constant, follows from it when the forces doing work are conservative, by writing each one’s work as a fall in potential energy, −ΔV. The wider law of conservation of energy says that the total energy of an isolated system, including heat, never changes, so kinetic energy lost to friction reappears as heat.

Does the work-energy theorem apply when friction acts?

Yes, the work-energy theorem applies when friction acts: friction is one of the forces on the body, so its work, usually negative, is part of the net work, W_gravity + W_friction + … = K_f − K_i. Unlike conservation of mechanical energy, the theorem does not need the forces to be conservative.

Can internal forces change the kinetic energy of a system?

Yes, internal forces can change the kinetic energy of a system, so the work-energy theorem for a system must count their work as well as that of the external forces: W_ext + W_int = K_f − K_i. Internal forces cancel in pairs, but their works need not, because the parts can move through different distances. So a compressed spring pushing two blocks apart gives them kinetic energy with no external work at all, and a person jumping up gains kinetic energy though the floor’s push does no work.

How much work is needed to increase a body’s speed from u to v?

The net work needed equals the gain in kinetic energy: W = ½m(v^² − u^²). For example, taking a 5 kg body from 3 m s^⁻¹ to 7 m s^⁻¹ needs ½ × 5 × (49 − 9) = 100 J of net work. Doubling a body’s speed needs extra work equal to three times its original kinetic energy.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. The speed of a 3 kg body increases from 2 m s^⁻¹ to 4 m s^⁻¹. How much net work is done on it?

   Show answerHide answer

   $$\begin{aligned} W &= \tfrac{1}{2}(3)(4^{2} - 2^{2}) \\ &= 1.5 \times 12 = 18\ \text{J} \end{aligned}$$

2. Q2. A ball of mass 0.5 kg is thrown straight up at 10 m s^⁻¹. Ignoring air resistance, find the work done by gravity from the throw to the highest point, and the height reached. Take g = 9.8 m s⁻².

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   At the top the ball is momentarily at rest, so gravity does work equal to the change in kinetic energy:

   $$\begin{aligned} W_g &= 0 - \tfrac{1}{2}(0.5)(10)^{2} \\ &= -25\ \text{J} \end{aligned}$$

   Since W_g = −mgh:

   $$h = \frac{25}{0.5 \times 9.8} \approx 5.10\ \text{m}$$

3. Q3. A bullet of mass 20 g moving at 300 m s^⁻¹ is stopped by a fixed wooden block after penetrating 5 cm. Find the distance-averaged magnitude of the resistive force; neglect other work during penetration.

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   The resistive force removes all the kinetic energy over 0.05 m:

   $$\begin{aligned} F &= \frac{\tfrac{1}{2}mv^{2}}{d} \\ &= \frac{0.5 \times 0.02 \times 300^{2}}{0.05} \\ &= 1.8 \times 10^{4}\ \text{N} \end{aligned}$$