Talk:PlanetPhysics/Example of Mechanical Power

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%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: example of mechanical power %%% Primary Category Code: 40. %%% Filename: ExampleOfMechanicalPower.tex %%% Version: 1 %%% Owner: bloftin %%% Author(s): bloftin %%% PlanetPhysics is released under the GNU Free Documentation License. %%% You should have received a file called fdl.txt along with this file. %%% If not, please write to gnu@gnu.org. \documentclass[12pt]{article} \pagestyle{empty} \setlength{\paperwidth}{8.5in} \setlength{\paperheight}{11in}

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{\bf Problem 1.} If a 200 pound man runs up a flight of stairs 20 feet high in seven seconds, at what rate is he working? Express the result in foot-pounds per second, horsepower, and watts.

{\bf Answer.}

First find the \htmladdnormallink{work}{http://planetphysics.us/encyclopedia/Work.html} done in climbing from one floor to the next, which is equal to the potential \htmladdnormallink{energy}{http://planetphysics.us/encyclopedia/CosmologicalConstant.html} gained.

$$Work = Wh = (200 \, pounds)(20 \, feet) = 4,000 \, foot-pounds$$

Since the rate of doing this work, that is, the \htmladdnormallink{power}{http://planetphysics.us/encyclopedia/Power.html}, is the work divided by the time, we find the quotient obtained from (4,000) foot-pounds) / (7 seconds) or 571 foot-pounds per second. This is equivalent to 571/550 or 1.037 horsepower or in terms of watts, (1.037)(746) = 774 watts. It is perfectly possible for a man to work at the rate of a horsepower for a short time; it is also possible for a horse to work at the rate of many horsepower for a short time; but it takes a first-class horse to work at the rate of one horsepower for an entire day.

{\bf Problem 2.} A horse pulls a plow at the rate of two feet per second and exerts a forward force on the plow of 250 pounds. At what rate does the horse work? Express the result in foot-pounds per second, in horsepower, and in watts. How much work does the horse do in five hours?

{\bf Answer.}

The rate of doing work is the power, one \htmladdnormallink{formula}{http://planetphysics.us/encyclopedia/Formula.html} for which is the product of the force and the \htmladdnormallink{speed}{http://planetphysics.us/encyclopedia/Velocity.html}. Since the force is 250 pounds and the speed is 2.00 feet/second, the required power is (250 pounds)(2.00 feet/second) or 500 foot-pounds/second. Since there are 550 foot-pounds/second per horsepower, this power is 500/550 or 0.909 horsepower. It will be noticed that the numerator together with its units is 550 foot-pounds per second per horsepower, or 550 foot-pounds/horsepower-second. When we divide the numerator by the denominator, all the units cancel exept horsepower, which being in the denominator of the denominator can be transferred to the numerator and survives in the result. By multiplying 0.909 horsepower by the conversion factor 746 watts/horsepower, the horsepower cancels giving 678 watts. Since power is the ratio of work to time, it follows that work is the product of power and time. we may therefore multiply any one of our three answers by an amount of time corresponding to five hours and get the work done by the horse. Five hours is 18,000 seconds. When we multiply 500 foot-pounds/second by 18,000 seconds, the seconds cancel and we have 9,000,000 foot-pounds. if we multiply by 0.909 horsepower by 5.00 hours, we have 4.54 horsepower-hours. If we multiply 678 watts by 5.00 hours, we have 3,390 watt-hours. Since 1,000 watts equals one kilowatt, this is equivalent to 3.39 kilowatt-hours.


 * These examples are from the Public \htmladdnormallink{domain}{http://planetphysics.us/encyclopedia/Bijective.html} work of [Frye].

\begin{thebibliography}{9} \bibitem{Frye} Frye, Royal M., {\em Applied Physics}. Prentice-Hall, Inc., New York, 1947. \end{thebibliography}

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