Work, Energy and Power

Push a box across the floor and you have done work on it. Lift it onto a shelf and you have stored energy in it. Do either one quickly and you have used a lot of power. These three ideas are really one story about how energy moves from place to place, and once you see that energy is never lost, only swapped from one form to another, almost every problem becomes a tidy bit of bookkeeping.

Work is force times distance

In everyday speech “work” means effort. In physics it means something exact: you do work on an object only when a force moves it through a distance. Hold a heavy bag still and you feel tired, but you have done zero work on the bag, because it did not move.

In symbols, work is the force times the displacement in the direction of the force:

$W = Fs = \Delta E$

Here $F$ is the force and $s$ is the displacement along the line of the force. Doing work transfers energy, $\Delta E$, measured in joules. If the force acts along the motion the work is positive and energy goes in; if it acts against the motion, like friction, the work is negative and energy is taken away. A force at right angles to the motion does no work at all.

Two kinds of energy to keep track of

Energy is the capacity to do work, and for motion problems there are two flavours you will meet again and again.

• Kinetic energy is the energy of moving, $KE = \tfrac{1}{2}mv^{2}$. Because the speed is squared, doubling the speed gives four times the kinetic energy.
• Potential energy is stored energy waiting to be released. Lift something up and you store gravitational potential energy, $GPE = mgh$. Stretch or squash a spring and you store strain (elastic) potential energy, $\tfrac{1}{2}k(\Delta x)^{2}$, where the spring obeys Hooke’s law $F = -k\Delta x$ (the restoring force pulls it back toward its natural length).

The key move in any problem is watching energy change from one of these forms into another.

Energy trades off as things move

Watch a ball drop and you see the swap happen in real time. At the top it is high and slow, so it is all gravitational potential energy and almost no kinetic energy. As it falls it speeds up: the potential energy shrinks and the kinetic energy grows by exactly the same amount. At the bottom it is fast and low, so it is all kinetic energy. The total never changes.

The left bars are the ball at the top: tall blue potential energy and a sliver of red kinetic energy. The right bars are the same ball at the bottom: the blue has shrunk to almost nothing and the red has grown to match. Add the two bars in each pair and you get the same total every time.

The work energy theorem

There is a clean shortcut that links work straight to motion. The work energy theorem says that the net work done on an object equals its change in kinetic energy:

$W_{\text{net}} = \Delta KE = \tfrac{1}{2}mv^{2} - \tfrac{1}{2}mu^{2}$

So if you know the net force and the distance, you can find the change in speed without ever touching the time. Push a trolley and the work you do shows up directly as extra kinetic energy. If friction does negative work, it removes kinetic energy and the trolley slows down.

See it for yourself

Build a track, let the skater roll, and watch the energy bars rise and fall. Notice how the kinetic and potential bars trade off as the skater drops and climbs, while the total stays fixed. Then switch on friction and watch some of that energy leak away as heat, so the skater no longer climbs back to the same height.

Power is how fast you do work

Two cranes can lift the same beam to the same height and do exactly the same work. The faster one has more power. Power is the rate of doing work, measured in watts, where one watt is one joule per second:

$P = \frac{W}{t} = \frac{\Delta E}{\Delta t}$

That is why a car needs more engine power to hold a high speed: the faster it goes, the more energy it must pour out every second against air resistance.

How to actually solve one

Most work and energy questions fall into a short recipe.

1. Decide which energies are present at the start and at the end: kinetic, gravitational, elastic.
2. If there is no friction, set the total energy at the start equal to the total at the end and solve.
3. To find a change in speed from a force over a distance, use the work energy theorem $W_{\text{net}} = \Delta KE$.
4. If a question asks how fast energy is transferred, you want power: $P = \dfrac{W}{t}$.

Watch your masses and your squares. The half and the squared speed in $\tfrac{1}{2}mv^{2}$ are the two most common slip ups in examiner reports.

See the recipe in action in the Worked Examples tab, then test yourself in Try It.

Lock it in with active recall

Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.