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Unformatted text preview: work done on the object is: W = F × d = 10 N × 10 m = 100 J . 5. The Power required to do 100 J of work on an object in .5 s is: P = W/ Δ t = 100 J/. 5 s = 200 W. To do this same amount of work in 1 s requires a power of: P = W/ Δ t = 100 J/ 1 s = 100 W. 7. If you do 100 J of work in lifting a bucket, you will increase its potential energy by 100 J. If you lift the bucket twice as high, you will double the original increase in P.E., so the bucket will have 200 J of gravitational P.E. 1...
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This note was uploaded on 10/19/2011 for the course PHY 2020 taught by Professor Staff during the Spring '08 term at University of Florida.
 Spring '08
 Staff
 Physics, Energy, Force, Kinetic Energy, Potential Energy, Work

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