Physics and Power Help (page 2)
Introduction to Physics and Power
In the context of physics, power is the rate at which energy is expended or converted to another form. Mechanically, it is the rate at which work is done. The standard unit of power is the joule per second (J/s), more commonly known as the watt (W). Power is almost always associated with kinetic energy. Sometimes the rate at which potential energy is stored is referred to as power.
In the examples shown by Fig. 8-4, the object acquires potential energy when it is lifted, and this potential energy is converted to kinetic energy as the object falls (if it is allowed to fall). The final burst of sound, shock waves, and perhaps outflying shrapnel is the last of the kinetic energy imparted to the object by lifting. Where does power fit into this scenario?
Fig. 8-4 . Work is done when a force is applied over a specific distance. In this case, the force is applied upward to an object against Earth’s gravity.
A slight variation on this theme can be used to talk about power. This is shown by Fig. 8-5. Suppose that instead of a free end of rope, you have a winch that you can turn to raise the object at the other end of the rope. The object starts out sitting on the floor, and you crank the winch (or use a motor to crank it) for the purpose of lifting the object. This, possibly in conjunction with a complex pulley system, will be necessary if you have a heavy object to lift. Then the pulley had better be strong! The same holds true for the rope. And let’s not forget about the manner in which the pulley is anchored to the ceiling.
Fig. 8-5 . Illustration of power. A winch and pulley can be used to life a heavy object.
Let’s Do It!
It will take energy to lift this object. You can crank the winch, imparting potential energy to the object. If the pulley system is complex, you might reduce the force with which you have to bear down on the winch, but this will increase the number of times you must turn the winch to lift the object a given distance. The rate at which you expend energy cranking the winch can be expressed in watts and constitutes power. The faster you crank the winch for a given object mass, the higher is the expended power. The more massive the object for a given winch crank speed, the higher is the expended power. However, the power does not depend on how high the object is lifted. In theory, you could expend a little power for a long time and lift the mass 100 m, 1 km, or 100 km.
Assume that the winch and pulley system is frictionless and that the rope does not stretch. Suppose that you crank the winch at a constant rotational rate. The power you expend in terms of strain and sweat multiplied by the time spent applying it will equal the potential energy imparted to the object. If P is the power in watts and t is the time in seconds for which the constant power P is applied, then the potential energy imparted to the object E p can be found according to this formula:
E p = Pt
This can be rearranged to
P = E p / t
We know that the potential energy is equal to the mass times the acceleration of gravity times the displacement q . Thus the power can be calculated directly by the following formula:
P = 9.8067 mq/t
Mechanical Power Practice Problem
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 .
Simply use the preceding formula with the acceleration rounded to two significant figures:
P = 9.8 × 200 × 2.50/7.00
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.
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.
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
System Efficency Practice Problem
Consider the scenario of Problem 1 and Fig. 8-6 . If the meter shows 800 W, what is the efficiency in percent?
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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