<!-- @@author="Courtesy of Ryan Hagerty, U.S. Fish and Wildlife Service" -->

Typical Pressurized Water Reactors (PWR), a type of nuclear power reactor originally built in the 1970's, produce 1100 to 1500 megawatts, or about 1,500,000,000 Joules/second. By comparison, a windmill farm with hundreds of individual windmills produces about 5 megawatts (5,000,000 Joules/second).

### Power

**Power** is defined as the rate at which work is done, or the rate at which energy is transformed.

\begin{align*}Power=\frac{Work}{Time}\end{align*}

In SI units, power is measured in Joules per second, which is given a special name: the watt**, \begin{align*}W\end{align*}.**

1.00 watt = 1.00 J/s

Another unit for power that is fairly common is horsepower.

1.00 horsepower = 746 watts

**Example Problem: ** A 70.0 kg man runs up a long flight of stairs in 4.0 s. The vertical height of the stairs is 4.5 m. Calculate the power output of the man in watts and horsepower.** **

**Solution: **The force exerted must be equal to the weight of the man:\begin{align*}mg = (70.0 \ \text{kg})(9.80 \ \text{m/s}^2) = 686 \ \text{N}\end{align*}

\begin{align*}W = Fd = (686 \ \text{N})(4.5 \ \text{m}) = 3090 \ \text{N m} = 3090 \ \text{J}\end{align*}

\begin{align*}P=\frac{W}{t}=\frac{3090 \ \text{J}}{4.0 \ \text{s}}=770 \ \text{J/s}=770 \ \text{W}\end{align*}

\begin{align*}P = 770 \ \text{W} = 1.03 \ \text{hp}\end{align*}

Since \begin{align*}P = \frac{W}{t}\end{align*} and \begin{align*}W = Fd\end{align*}, we can use these formulas to derive a formula relating power to the speed of the object that is produced by the power.

\begin{align*}P=\frac{W}{t}=\frac{Fd}{t}=F \frac{d}{t}=Fv\end{align*}

The velocity in this formula is the average speed of the object during the time interval.

**Example Problem: **Calculate the power required of a 1400 kg car if the car climbs a 10° hill at a steady 80. km/h.

**Solution:** First convert 80. km/h to m/s: 22.2 m/s.

In 1.00 s, the car would travel 22.2 m on the road surface but the distance traveled upward would be (22.2 m)(sin 10°) = (22.2 m)(0.174) = 3.86 m. The force in the direction of the upward motion is the weight of the car: (1400 kg)(9.80 m/s^{2}) = 13720 N.

\begin{align*}W = Fd = (13720 \ \text{N})(3.86 \ \text{m}) = 53,000 \ \text{J}\end{align*}

Since this work was done in 1.00 second, the power would be 53,000 W.

This problem can be solved a different way; by calculating the upward component of the velocity of the car. The process would be similar, and start with finding the vertical component of the velocity vector: (22.2 m/s)(sin 10°) = (22.2 m/s)(0.174) = 3.86 m/s. Again, calculate the weight of the car: (1400 kg)(9.80 m/s^{2}) = 13720 N. Finally, we could use the formula relating power to average speed to calculate power.

\begin{align*}P = Fv = (13720 \ \text{N})(3.86 \ \text{m/s}) = 53,000 \ \text{W} \end{align*}

#### Summary

- Power is defined as the rate at which work is done or the rate at which energy is transformed.
- \begin{align*}\text{Power}=\frac{\text{Work}}{\text{Time}}\end{align*}
- \begin{align*}\text{Power} = \text{Force} \times \text{velocity}\end{align*}

#### Practice

Use the video below to answer the following questions about work and power?

https://www.youtube.com/watch?v=u6y2RPQw7E0

- What is the difference between positive and negative work?
- What are the standard units for power?
- What is horsepower?
- How many grandfather clocks could you power with the same amount of power as is used by a single light bulb?

The following website has practice problems on work and power.

http://www.angelfire.com/scifi/dschlott/workpp.html

#### Review

- If the circumference of an orbit for a toy on a string is 18 m and the centripetal force is 12 N, how much work does the centripetal force do on the toy when it follows its orbit for one cycle?
- A 50.0 kg woman climbs a flight of stairs 6.00 m high in 15.0 s. How much power does she use?
- Assuming no friction, what is the minimum work needed to push a 1000. kg car 45.0 m up a 12.5° incline?