Power: how fast work is done

Two people each carry an identical box up the same flight of stairs. They do exactly the same work, because they lift the same weight through the same height. But one runs up in 5 seconds and the other walks up in 20 seconds. The runner is more powerful: they do the work faster.

Power is the rate of doing work, meaning the work done per second:

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

The unit is the watt: $1\ \text{W} = 1\ \text{J s}^{-1}$, one joule of work every second. Engines are usually measured in kilowatts: $1\ \text{kW} = 1000\ \text{W}$. Always convert kW to W before calculating.

If the work isn't done at a steady rate, $\dfrac{W}{t}$ gives the average power over that time.

Suppose a force $F$ pushes an object along, in the direction it is moving, at speed $v$.

• In one second the object moves $v$ metres (that's what a speed of $v\ \text{m s}^{-1}$ means).
• Work done $=$ force $\times$ distance moved in the direction of the force, so in that one second the force does $F \times v$ joules of work.
• Work done per second is power. So:

$P = Fv$

This is the formula you will use most. It links the power of an engine to the force it produces and the speed it is going. Notice the trade-off: for a fixed power, the faster you go, the smaller the force the engine can produce, because $F = \dfrac{P}{v}$. That is why a car accelerates hard at low speed but struggles to gain speed near its top speed.

Most questions involve a car, lorry or cyclist. Two forces matter along the road: the engine's driving force $D$ pushing forwards, and the resistance $R$ holding it back.

At constant speed the forces balance. Constant speed means no acceleration, so by Newton's first law the forces along the road cancel. On level ground that simply means $D = R$.

On a hill, part of the weight also pulls the vehicle back down the slope. On a slope at angle $\theta$, that part (the component of the weight down the slope) is $mg\sin\theta$. Going up the hill at constant speed, the engine has to beat both: $D = R + mg\sin\theta$.

Then the engine's power is $P = Dv$.

Top speed. An engine has a maximum power. As speed increases, the driving force $D = \dfrac{P}{v}$ gets smaller, until it only just equals the resistance. Then the vehicle can't accelerate any more: that is its top speed. So at top speed, set $D$ equal to the resistance and solve $v = \dfrac{P}{D}$.