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# Physics and Power Help (page 2)

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By McGraw-Hill Professional
Updated on Sep 5, 2011

#### Problem 1

Suppose that an object having mass 200 kg is to be lifted 2.50 m in a time of 7.00 s. What is the power required to perform this task? Take the acceleration of gravity to be 9.8 m/s 2 .

#### Solution 1

Simply use the preceding formula with the acceleration rounded to two significant figures:

P = 9.8 × 200 × 2.50/7.00

= 700W

When we say 700 W, we are technically justified in going to only two significant figures. How can we express this here? One way is to call it 7.0 × 10 2 W. Another way is to call it 0.70 kW, where kW stands for kilowatt , the equivalent of exactly 1,000 W. Yet another way is to say it is 700 W ± 5 W, meaning “700 W plus or minus 5 W.” (This is the extent of the accuracy we can claim with two significant figures.) However, most physicists probably would accept our saying simply 700 W. There is a limit to how fussy we can get about these things without driving ourselves crazy.

You will notice that in Problem 1 we did not carry the units through the entire expression, multiplying and dividing them out along with the numbers. We don’t have to do this once we know that a formula works, and we are certain to use units that are all consistent with each other in the context of the formula. In this case we are using all SI base units (meter, kilogram, second), so we know we’ll come out all right in the end.

## Electrical Power

You might decide to spare yourself the tiring labor of cranking a winch to lift heavy objects over and over just to perform experiments to demonstrate the nature of power. Anyhow, it’s hard to measure mechanical power directly, although it can be calculated theoretically as in Problem 1.

You might connect an electric motor to the winch, as shown in Fig. 8-6. If you then connect a wattmeter between the power source and the motor, you can measure the power directly. Of course, this assumes that the motor is 100 percent efficient, along with the other assumptions that the rope does not stretch and the pulley has no friction. All these assumptions are, of course, not representative of the real world. A real pulley does have friction, a real rope will stretch, and a real motor is less than perfectly efficient. As a result, the reading on a wattmeter connected as shown in Fig. 8-6 would be greater than the figure we would get if we used the scheme in Problem 1 to calculate the power.

Fig. 8-6 . Electrical power can be measured directly when a motor is used to drive a winch to lift a heavy object.

## System Efficiency

Suppose that we connect the apparatus of Fig. 8-6 and do the experiment described in Problem 1. The wattmeter might show something like 800 W. In this case we can calculate the efficiency of the whole system by dividing the actual mechanical power (700 W) by the measured input power (800 W). If we call the input power P in and the actual mechanical power P out, then the efficiency Eff is given by

Eff = P out / P in

If you want to calculate the efficiency in percent, Eff % , use this formula

Eff % = 100 P out / P in

#### Problem 2

Consider the scenario of Problem 1 and Fig. 8-6 . If the meter shows 800 W, what is the efficiency in percent?

#### Solution 2

Use the second of the two efficiency formulas presented earlier:

Eff % = 100P/P out /P in

Eff % = 100 × 700/800

= 87.5 percent

If you want to get formal and claim only two significant figures for the 700-W result in Problem 1, then you must round this efficiency figure off to 88 percent.

Practice problems of this concept can be found at: Momentum, Work, Energy, And Power Practice Test

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