A 56.00 kg keg of beer slides upright down a 2.75 m long plank leading from the back of a truck 1.16 m high to the ground. The coefficient of kinetic friction, muk= 0.10.
a) Determine the amount of work done on the keg by gravity.
Fgrav = ma = 56(9.8)
Wgrav = ma*d = 56(9.8)(1.16) = 636.6 J
b) What is the total work done by friction on the keg?
I'm not sure how to go about doing the second part. I tried 548.8 N * 2.75 m = -150.92 J, but it didn't work. In this situation, is the work due to gravity positive, or am I wrong in my equation?
2006-10-26
05:56:16
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1 answers
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asked by
JustWondering
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