Power (Physics): Definition, Formula, Units, How To Find (W/ Examples)

A bodybuilder and a fifth grader could both carry all the books off a shelf up a flight of stairs, but they aren't likely to finish the task in the same amount of time. The bodybuilder will probably be faster because she has a higher ​power rating​ than the fifth grader.

Similarly, a race car with a high ​horsepower​ will be able to travel farther much faster than, well, a horse.

A quick note on horsepower: The term is meant to compare the output of a steam engine to that of a horse, as in a 700 horsepower engine could do about 700 times the work of a single horse. This dates back to when steam engines were new and one of the most prominent inventors working to improve their efficiencies, James Watt, coined the term as a way to convince the average person of their worth.

There are two ways to calculate power, depending on what information is available. Additionally, there are two units of power that are equally valid .

1\. Power in terms of work and time:

\(P=\frac{W}{t}\)

Where work ​W​ is measured in Newton-meters (Nm), and time ​t​ is measured in seconds (s).

2\. Power in terms of force and velocity:

\(P=Fv\)

Where force ​F​ is in Newtons (N), and velocity ​v​ is in meters/second (m/s).

These equations are not randomly equivalent. The second equation can be derived from the first one like so:

Note that ​work​ is the same as ​force times displacement:​

\(W=Fd\)

Substitute this in to the first power equation:

\(P=\frac{W}{t}=\frac{Fd}{t}=F\frac{d}{t}\)

Then, because displacement in any unit of time ​is velocity​ (v = d/t), rewrite the terms at the end as ​v​ to get the second power equation.

The SI unit of Power ​p​ is usually presented as ​Watts (W)​, named for the same James Watt who designed engines and compared them to horses. Light bulbs and other household appliances usually include this unit on their tags.

Looking at the second formula for power leads to another unit, however. Force times velocity gives a measurement in the units of Newton-meters per second (Nm/s). Then, because the unit of energy the Joule is also defined as one Newton-meter (Nm), the first part of that can be rewritten as a Joule instead, resulting in the second SI unit of power: ​Joules per second (J/s).​

Considering the definition of power and the two ways to find it yields multiple ways to ​increase something's power​: increase its strength (use more ​force​) or get the same work done faster (decrease ​t​ or increase ​v​). A powerful car is strong ​and​ fast, and a weak one is neither. The ​more easily and quickly work can be done​, the ​more powerful​ the entity doing the work.

This also implies that a very strong machine, say a highly muscular bodybuilder, could still ​lack power​. A person that can lift a very heavy load, but only very slowly, is less powerful than someone who can lift it fast.

Similarly, a very fast machine or person that doesn't get much done, someone rapidly flailing in place but getting nowhere, is not actually powerful.

1\. Usain Bolt generated about 25 W of power in his record-setting 100 m sprint, which took 9.58 seconds. How much work did he do?

Because ​P​ and ​t​ are given, and ​W​ is unknown, use the first equation:

\(P=\frac{W}{t}\implies 25=\frac{W}{9.58}\implies W=239.5\text{ Nm}\)

2\. With what average force was he pushing against the ground as he ran?

Since ​work​ in Nm is already known, as is the ​displacement​ in meters, dividing by the length of the race will give the ​force​ (put another way, ​work​ is the same as ​force times displacement:​ W = F × d):

\(\frac{239.5}{100}=2.395\text{ N}\)

3\. How much power does a 48 kg person who takes 6 seconds to race up a 3 meter set of stairs generate?

In this problem, the displacement and time are given, which quickly allows a velocity calculation:

\(v=\frac{d}{t}=\frac{3}{6}=0.5\text{ m/s}\)

The second power equation has velocity, but it also includes force. A person running up a flight of stairs is working to counter the force of gravity. So, the force in this case can be found using their mass and acceleration due to gravity, which on Earth is always equal to 9.8 m/s².

\(F_{grav}=mg=48\times 9.8=470.4\text{ N}\)

Now force and velocity fit into the second formula for power:

\(katex=Fv=470.4\times 0.5 = 235.2\text{ J/s}\)

Note the decision to leave the units here as J/s rather than Watts is arbitrary. An equally acceptable answer is 235.2 W.

4\. One horsepower in SI units is about 746 Watts, which is based on the load a fit horse could have carried for one minute. How much work did the example horse do in this time?

The only step before plugging the values for power and time into the first equation is to make sure time is in the proper SI units of seconds by rewriting one minute as 60 seconds. Then:

\(P=\frac{W}{t}\implies 746=\frac{W}{60}\implies W=44,670\text{ Nm}\)

Many electric utilities charge customers a fee based on their ​kilowatt-hours​ of usage. To understand the meaning of this common unit of electrical power, start by breaking down the units.

The prefix ​kilo​ means 1,000, so a ​kilowatt (kW)​ is equal to 1,000 Watts. Thus, a ​kilowatt-hour (kWh)​ is the amount of kilowatts used in one hour of time.

To count kilowatt-hours, multiply the number of kilowatts times the hours used. Thus, if someone uses a 100 Watt light bulb for 10 hours, they will have used a total of 1,000 Watt-hours, or 1 kWh of electricity.

1\. An electric utility charges $0.12 per kilowatt-hour. A very powerful 3,000 W vacuum is used for 30 minutes. How much does this amount of energy cost the homeowners?

3,000 W = 3 kW

30 minutes = 0.5 hours

\(3\text{ kW}\times 0.5\text{ h}= 1.5\text{ kWh}\text{ and }1.5\text{ kWh}\times 0.12\text{ dollars/kWh} = \$0.18\)

2\. The same utility credits a household with $10 for every 4 kWh of electricity it returns to the grid. The sun provides about 1,000 W of power per square meter. If a two-square-meter solar cell on a house collects energy for 8 hours, how much money does it generate?

Given the information in the problem, the solar cell must be able to collect 2,000 W from the Sun, or 2 kW. In 8 hours, that is 16 kWh.

\(\frac{\$10}{4\text{ kWh}}\times 16\text{ kWh}=\$40\)