Much like *energy*, the word *power* is something we hear a lot. In everyday life it has a wide range of meanings. In physics however, it has a very specific meaning. It is a measure of the rate at which work is done (or similarly, at which energy is transferred).

The ability to accurately measure power was one of the key abilities which allowed early engineers to develop the steam engines which drove the industrial revolution. It continues to be essential for understanding how to best make use of the energy resources which drive the modern world.

The standard unit used to measure power is the watt which has the symbol $\text{W}$Wstart text, W, end text. The unit is named after the Scottish inventor and industrialist James Watt. You have probably come across the watt often in everyday life. The power output of electrical equipment such as light bulbs or stereos is typically advertised in watts.

By definition, one watt is equal to one joule of work done per second. So if $P$PP represents power in watts, $\Delta E$ΔEdelta, E is the change in energy (number of joules) and $\Delta t$Δtdelta, t is the time taken in seconds then:

$P = \frac{\Delta E}{\Delta t}$P=ΔtΔEP, equals, start fraction, delta, E, divided by, delta, t, end fraction

There is also another unit of power which is still widely used: the horsepower. This is usually given the symbol hp and has its origins in the 17th century where it referred to the power of a typical horse when being used to turn a capstan. Since then, a *metric horsepower* has been defined as the power required to lift a $75~\text{kg}$75kg75, space, start text, k, g, end text mass through a distance of 1 meter in 1 second. So how much power is this in watts?

Well, we know that when being lifted against gravity, a mass acquires gravitational potential energy $E_p = m\cdot g\cdot h$Ep=m⋅g⋅hE, start subscript, p, end subscript, equals, m, dot, g, dot, h. So putting in the numbers we have:

$\frac{75~\mathrm{kg} \cdot ~ 9.807 ~\mathrm{m/s^2} \cdot 1~\mathrm{m}}{1 ~\mathrm{s}} = 735.5 ~\mathrm{W}$1s75kg⋅9.807m/s2⋅1m=735.5Wstart fraction, 75, space, k, g, dot, space, 9, point, 807, space, m, slash, s, squared, dot, 1, space, m, divided by, 1, space, s, end fraction, equals, 735, point, 5, space, W

[Wait. I thought one horsepower was equal to 746 W?]

In many situations where energy resources are being used, the rate of usage varies over time. The typical usage of electricity in a house (see Figure 1) is one such example. We see minimal usage during the day, followed by peaks when meals are prepared and an extended period of higher usage for evening lighting and heating.

There are at least three ways in which power is expressed which are relevant here: *Instantaneous power* $P_\text{i}$PiP, start subscript, start text, i, end text, end subscript, *average power* $P_\text{avg}$PavgP, start subscript, start text, a, v, g, end text, end subscript and *peak power* $P_\text{pk}$PpkP, start subscript, start text, p, k, end text, end subscript. It is important for the electricity company to keep track of all of these. In fact, different energy resources are often brought to bear in addressing each of them.

Instantaneous power is the power measured at a given instant in time. If we consider the equation for power, $P = \Delta E / \Delta t$P=ΔE/ΔtP, equals, delta, E, slash, delta, t, then this is the measurement we get when $\Delta t$Δtdelta, t is extremely small. If you are lucky enough to have a plot of power vs time, the instantaneous power is simply the value you would read from the plot at any given time.

Average power is the power measured over a long period, i.e., when $\Delta t$Δtdelta, t in the equation for power is very large. One way to calculate this is to find the area under the power vs time curve (which gives the total work done) and divide by the total time. This is usually best done with calculus, but it is often possible to estimate it reasonably accurately just using geometry.

Peak power is the maximum value the instantaneous power can have in a particular system over a long period. Car engines and stereo systems are example of systems which have the ability to deliver a peak power which is much higher than their rated average power. However, it is usually only possible to maintain this power for a short time if damage is to be avoided. Nevertheless, in these applications a high peak power might be more important to the driving or listening experience than a high average power.

The equation for power connects work done and time. Since we know that work is done by forces, and forces can move objects, we might expect that knowing the power can allow us to learn something about the motion of a body over time.

If we substitute the work done by a force $W=F\cdot\Delta x \text{ cos}\theta$W=F⋅ΔxcosθW, equals, F, dot, delta, x, start text, space, c, o, s, end text, theta into the equation for power $P=\dfrac{W}{\Delta t}$P=ΔtWP, equals, start fraction, W, divided by, delta, t, end fraction we find:

$P=\frac{F\cdot\Delta x \cdot \cos{\theta}}{\Delta t}$P=ΔtF⋅Δx⋅cosθP, equals, start fraction, F, dot, delta, x, dot, cosine, theta, divided by, delta, t, end fraction

If the force is along the direction of motion (as it is in many problems) then $\cos(\theta)=1$cos(θ)=1cosine, left parenthesis, theta, right parenthesis, equals, 1 and the equation can be re-written

$P=F\cdot v$P=F⋅vP, equals, F, dot, v

since a change in distance over time is a velocity. Or equivalently,

$P_i = m\cdot a\cdot v$Pi=m⋅a⋅vP, start subscript, i, end subscript, equals, m, dot, a, dot, v

Note that in this equation we have made sure to specify that the power is the instantaneous power, $P_i$PiP, start subscript, i, end subscript. This is because we have both acceleration and velocity in the equation and therefore the velocity is changing over time. It only makes sense if we take the velocity at a given instant. Otherwise, we need to use the average velocity, i.e.:

$P_\mathrm{avg} = m\cdot a \cdot \frac{1}{2}(v_\mathrm{final}+v_\mathrm{initial})$Pavg=m⋅a⋅21(vfinal+vinitial)P, start subscript, a, v, g, end subscript, equals, m, dot, a, dot, start fraction, 1, divided by, 2, end fraction, left parenthesis, v, start subscript, f, i, n, a, l, end subscript, plus, v, start subscript, i, n, i, t, i, a, l, end subscript, right parenthesis

This can be a particularly useful result. Suppose a car has a mass of $1000 \text{ kg}$1000kg1000, start text, space, k, g, end text and has an advertised power output to the wheels of $75 \text{ kW}$75kW75, start text, space, k, W, end text (around $100 \text{ hp}$100hp100, start text, space, h, p, end text). The advertiser claims that it has constant acceleration over the range of $0–25 \dfrac{\text m}{\text s}$0–25sm0, –, 25, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction.

Using only this information we can find out the time the car should take under ideal conditions to accelerate from zero to a speed of 25 m/s.

$P_{avg} = m \cdot a \cdot \frac{1}{2} v_{final}$Pavg=m⋅a⋅21vfinalP, start subscript, a, v, g, end subscript, equals, m, dot, a, dot, start fraction, 1, divided by, 2, end fraction, v, start subscript, f, i, n, a, l, end subscript

Because acceleration is $\Delta v / \Delta t$Δv/Δtdelta, v, slash, delta, t:

$\begin{aligned}P_\mathrm{avg} &= m \cdot (v_{final} / t) \cdot \frac{1}{2} v_\mathrm{final} \\&= \frac{mv_\mathrm{final}^2}{2t}\end{aligned}$Pavg=m⋅(vfinal/t)⋅21vfinal=2tmvfinal2

Which can be rearranged:

$\begin{aligned} t &= \frac{v_\mathrm{final}^2 \cdot m}{2 \cdot P_\mathrm{avg}} \\ &= \frac{(25~\mathrm{m/s})^2 \cdot 1000~\mathrm{kg}}{2\cdot 75000~\mathrm{W}} \\ &= 4.17~\mathrm{s} \end{aligned}$t=2⋅Pavgvfinal2⋅m=2⋅75000W(25m/s)2⋅1000kg=4.17s\