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 08:45:19 · 1 answers · asked by JustWondering 1 in Education & Reference ➔ Homework Help
The frictional force is μR, where R = the normal force acting on the keg.
In this case, R = 56 * 9.8, and μ = .10. F = 548.8 * .10 = 54.88N.
W = Fd = 54.88 N * 1.16 m = 63.66 J.
As for the positive/negative question, you may arbitrarily decide which one is positive, and which is negative. Since they are in opposite directions, what truly matters is that one is positive, and one is negative.
2006-10-27 11:14:57 · answer #1 · answered by ³√carthagebrujah 6 · 0⤊ 0⤋