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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, $\mathrm{\Delta }E$ is the change in energy (number of joules) and $\mathrm{\Delta }t$ is the time taken in seconds then:
$P=\frac{\mathrm{\Delta }E}{\mathrm{\Delta }t}$
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 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:

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, $P=\mathrm{\Delta }E/\mathrm{\Delta }t$, then this is the measurement we get when $\mathrm{\Delta }t$ 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 $\mathrm{\Delta }t$ 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.
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.
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 (note that ), with peak power of – 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?

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 into the equation for power $P=\frac{W}{\mathrm{\Delta }t}$ we find:
$P=\frac{F\cdot \mathrm{\Delta }x\cdot \mathrm{cos}\theta }{\mathrm{\Delta }t}$
If the force is along the direction of motion (as it is in many problems) then $\mathrm{cos}\left(\theta \right)=1$ 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.:
${P}_{\mathrm{avg}}=m\cdot a\cdot \frac{1}{2}\left({v}_{\mathrm{final}}+{v}_{\mathrm{initial}}\right)$
This can be a particularly useful result. Suppose a car has a mass of and has an advertised power output to the wheels of (around ). The advertiser claims that it has constant acceleration over the range of $0–25\frac{\text{m}}{\text{s}}$.
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 $\mathrm{\Delta }v/\mathrm{\Delta }t$:
$\begin{array}{rl}{P}_{\mathrm{avg}}& =m\cdot \left({v}_{final}/t\right)\cdot \frac{1}{2}{v}_{\mathrm{final}}\\ & =\frac{m{v}_{\mathrm{final}}^{2}}{2t}\end{array}$
Which can be rearranged:
\
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?

Want to join the conversation?

• how is P=m.a.v instantaneous velocity?
• V= instantaneous velocity; Since acceleration and velocity are in the same equation, then the velocity is changing (accelerating) over time, so it would only make sense to pluck in the velocity of a specific time. However, Pavg= m . a . 1/2 (Vf+Vi) because you are using the AVERAGE velocity. So when using P = m . a . v , this is using the instantaneous velocity at a specific time and actually results in Pi (instantaneous power) not just P; you will need to use the average velocity in order to get the average power. In other words, to find Pavg you must specify that you are also using Vavg, and when using V (velocity at a specific time) you get Pi.
• How can we measure power
• If you know how much energy is being converted per unit time, that is all you need to calculate power. Ex. measure the current from the battery in a circuit and multiply by voltage to get power.
• In "Can the concept of power help us describe how objects move?", in the last few steps, why is Pavg = m a (1/2 vfinal). Isn't it 1/2 ( vfinal + vinitial) ? Thanks.
• Vinitial is 0, therefore it is 1/2Vfinal
• i cant understand the exercise 2(average power output) please someone explain me
• 1. power is all about converting whatever your work into the work with 1 second of window

2. in most cases, you do work for more than 1 sec. thus you have to do divide them by the time it take to do the work
e.g. work_of_pushing_a_box_right = 30J, time = 3s
power = work/time = 30J/3s = 10J/1s = 10W
meaning you do 10J amount of work per 1 sec on the box

3. but in exercise 2 (getting average power output of the laser device), the direction is opposite (converting less than 1 sec into 1 sec)
1) let's simplify the example
work_of_making_laser( pulse)s = 30J, time = 0.3s
power = work/time = 30/0.3 = 30 / 3/10 = 30 * 10/3 = 300/3 = 100W
or (more scientifically)
30/0.3 = 3*10^1 / 3*10^-1 = 3/3 * 10^1/10^-1 = 1 * 10^1*10^1 = 1 * 10^(1+1) = 1 * 10^(2) = 100W
2) the numbers in exercise 2 are simply having more 0's in them than this simple case
3) all you need to do is to plug different numbers (in scientific notation, preferably) into the equation of work/time

hope this to help clear some of your confusion
(1 vote)
• how does finding the area under, of a power vs time graph gives me an average power? shouldn't that be finding the slope?
• Area under a power vs. time graph is the equivalent of integrating power with respect to time, which gives work or change in energy. To find average power, you could divide the integral by the time interval.
• Power can be expressed as P = Work/Change in Time, instead of P = Change in Energy/Change in Time, because the change in energy is basically the work done. Is this correct?
• Hello Alex,

In general I would agree. In both cases you end up with joules / time. Watch out for context. It seems each field of study has it's own vocabulary.

Regards,

APD
• About the last exercise, the first thing I thought about was 4.17/8 = 0.52, therefore 48% of the time is being spent on something else (fighting air drag)
Then i applied the same ratios to the engine power
Is this a wrong way to think about it?

(if I sound somewhat weird, well,English is not my first language,sorry)
• This is actually a very good way to think about it. Its good to know equations sometimes, but you just solved a rather tricky problem in a very elegant way. Also, there were no English errors in your question in case you were wondering :)
• an engine of 4.9 kW power is used to pump water from a well 50 m deep.calculate the quantity of water in kilolitres which it can pump out in one hour.
• In exercise 1 where did they get the number 116.5 squares?
• sorry to say this
but you can simply count them with your hand (or eyes)

another way to do it roughly is cancelling out areas of uphill and downhill

if you hold the uphill area above ~0.45kW from hour 14 to 22.5, flip and split it into two, and then fit them into the downhill areas below ~0.45kW from hour 0 to 6 and 22.5 to 24, you may see a flat line along the 0.45W point from 0 to 24, and 0.45kW is not far from 0.485kW, the real solution

it's not mathematical
but faster than counting
and i do love estimating with eyes and mind-calculation, diving into actual calculation, and then comparing the two answers. and this case is one of the examples
(1 vote)
• If horsepower determines acceleration and not velocity, why do cars have a maximum speed? Why can't they accelerate to infinity? I know it's impossible, but why?