# What is power?

Learn what power means and how we use it to describe the rate of energy transfer.

# What is power?

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.

# How do we measure power?

The standard unit used to measure power is the watt which has the symbol $\text{W}$. 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 $P$ represents power in watts, $\Delta E$ is the change in energy (number of joules) and $\Delta t$ is the time taken in seconds then:
$P = \frac{\Delta E}{\Delta t}$
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 $75~\text{kg}$ 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 $E_p = m\cdot g\cdot h$. So putting in the numbers we have:
$\frac{75~\mathrm{kg} \cdot ~ 9.807 ~\mathrm{m/s^2} \cdot 1~\mathrm{m}}{1 ~\mathrm{s}} = 735.5 ~\mathrm{W}$
There are, unfortunately, many different definitions of "horsepower". The value of $1\text{hp}=735.5 \text{ W}$ given above is called metric horsepower. There is another quantity called imperial horsepower that is defined such that $1\text{hp}=745.7 \text{ W}$.
People use the term "horsepower" loosely, so be careful. It isn't always clear which definition they are using.

# How do we measure varying power?

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 $P_\text{i}$, average power $P_\text{avg}$ and peak power $P_\text{pk}$. 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, $P = \Delta E / \Delta t$, then this is the measurement we get when $\Delta t$ 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 $\Delta t$ 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.
Figure 1 : Daily electricity usage of a typical house
Exercise 1 : Using figure 1, estimate the instantaneous power at 10 am, the average power for the entire twenty four hour time interval, and peak power.
Instantaneous power at 10am: $0.43~\text{kW}$
Average power : Each square is $0.1~\mathrm{kW \cdot hours}$. So the area is: .
Peak power : $0.94~\text{kW}$.
Exercise 2: One device in which there is a huge difference between peak and average power is known as an ultrashort pulse laser. These are used in physics research and can produce pulses of light which are extremely bright, but for extremely short periods of time. A typical device might produce a pulse of duration $100~\mathrm{fs}$ (note that $1 \text{ fs}=10^{-15}~\mathrm{s}$), with peak power of $350~\mathrm{kW}$ – that's about the average power demanded by 700 homes! If such a laser produces 1000 pulses per second, what is the average power output?
$(350\cdot 10^3 \mathrm{J/s}) \times (100 \times 10^{-15} \mathrm{s/pulse}) = 3.5 \cdot 10^{-8}\mathrm{J/pulse}$
$3.5 \cdot 10^{-8}~\mathrm{J} / \mathrm{pulse} \cdot 1000 ~\mathrm{pulse/s} = 0.000035~\mathrm{W} = 35~\mathrm{\mu W}$

# Can the concept of power help us describe how objects move?

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 $W=F\cdot\Delta x \text{ cos}\theta$ into the equation for power $P=\dfrac{W}{\Delta t}$ we find:
$P=\frac{F\cdot\Delta x \cdot \cos{\theta}}{\Delta t}$
If the force is along the direction of motion (as it is in many problems) then $\cos(\theta)=1$ and the equation can be re-written
$P=F\cdot v$
since a change in distance over time is a velocity. Or equivalently,
$P_i = m\cdot a\cdot v$
Note that in this equation we have made sure to specify that the power is the instantaneous power, $P_i$. 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.:
$P_\mathrm{avg} = m\cdot a \cdot \frac{1}{2}(v_\mathrm{final}+v_\mathrm{initial})$
This can be a particularly useful result. Suppose a car has a mass of $1000 \text{ kg}$ and has an advertised power output to the wheels of $75 \text{ kW}$ (around $100 \text{ hp}$). The advertiser claims that it has constant acceleration over the range of .
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.
$P_{avg} = m \cdot a \cdot \frac{1}{2} v_{final}$
Because acceleration is $\Delta v / \Delta t$:
\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}
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} \
Exercise 2: In the real world, we are unlikely to observe such a rapid acceleration. This is because work is also being done in the opposite direction (negative work) by the force of drag as the car pushes the air aside. Suppose we trust the manufacturer at their specification, but actually observe a time t=8 s. What fraction of the power of the engine is being used to overcome drag during the test?
If we put our newly measured time into our previous equation we can solve for the apparent power of the engine. This should be smaller because it is actually the power of the engine minus the average power supplied by the drag force.
\begin{aligned} \frac{mv_\mathrm{final}^2}{2t} &= P_\mathrm{engine} - P_\mathrm{drag} \\ \frac{(1000~\mathrm{kg})(25~\mathrm{m/s})^2}{(2)(8 \text{ s})}&= 75000~\mathrm{W} - P_\mathrm{drag} \\ 39~\mathrm{kW} &= 75~\mathrm{kW} - P_\mathrm{drag}\end{aligned}
This can be re-arranged to give the average power of the drag force.
$P_\mathrm{drag} = 36~\mathrm{kW}$
So relatively speaking, around half $(0.48)$ the power of the engine is being used in overcoming drag.