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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 Spring '08
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 Physics, Energy, Force, Kinetic Energy, Potential Energy, Work, 4 m, 5 kg, 2 m, 10 kg

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