Could someone help me derive this problem You have to do 2100J of work to push a
ID: 1727103 • Letter: C
Question
Could someone help me derive this problemYou have to do 2100J of work to push a 90kg crate 2.5m upalong a slope inclined at 20 degrees. What is the coefficient ofstatic friction between the crate and the slope?
I am not quite sure how frictional force is derived, I knowFfr = kFN
You have to do 2100J of work to push a 90kg crate 2.5m upalong a slope inclined at 20 degrees. What is the coefficient ofstatic friction between the crate and the slope?
I am not quite sure how frictional force is derived, I knowFfr = kFN
Explanation / Answer
Work, W = 2100 J Mass, M = 90 kg Distance, S = 2.5 m Angle of inclined plane, = 20 degrees Coefficient of friction = Normal force, N = M g cos Frictional force, f = N = M g cos Total force along the plane in downwarddirection, F = M g sin + f = M g sin + M g cos Component of weight along the plane = M g sin Work, W = F * S = ( M g sin + M g cos ) * S = M g S ( sin + k cos ) ( sin + cos ) = W / M gS = 2100 / ( 90 * 9.8 * 2.5 ) = 0.9523 Coefficient of friction, = (0.9523 - 0.3420 ) / 0.9397 = 0.65