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8.2: Power

Difficulty Level: At Grade Created by: CK-12

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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.

$Power=\frac{Work}{Time}$

In SI units, power is measured in Joules per second, which is given a special name: the watt , $W$ .

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: $mg = (70.0 \ \text{kg})(9.80 \ \text{m/s}^2) = 686 \ \text{N}$

$W = Fd = (686 \ \text{N})(4.5 \ \text{m}) = 3090 \ \text{N m} = 3090 \ \text{J}$

$P=\frac{W}{t}=\frac{3090 \ \text{J}}{4.0 \ \text{s}}=770 \ \text{J/s}=770 \ \text{W}$

$P = 770 \ \text{W} = 1.03 \ \text{hp}$

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

$P=\frac{W}{t}=\frac{Fd}{t}=F \frac{d}{t}=Fv$

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.

$W = Fd = (13720 \ \text{N})(3.86 \ \text{m}) = 53,000 \ \text{J}$

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.

$P = Fv = (13720 \ \text{N})(3.86 \ \text{m/s}) = 53,000 \ \text{W}$

Summary

• Power is defined as the rate at which work is done or the rate at which energy is transformed.
• $\text{Power}=\frac{\text{Work}}{\text{Time}}$
• $\text{Power} = \text{Force} \times \text{velocity}$

Practice

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

1. What is the difference between positive and negative work?
2. What are the standard units for power?
3. What is horsepower?
4. 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.

Review

1. 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?
2. A 50.0 kg woman climbs a flight of stairs 6.00 m high in 15.0 s.  How much power does she use?
3. Assuming no friction, what is the minimum work needed to push a 1000. kg car 45.0 m up a 12.5° incline?

Feb 05, 2013