A 5.00 kg block is projected up an inclined plane with an initial speed nu_i = 8

Question

A 5.00 kg block is projected up an inclined plane with an initial speed nu_i = 8.00 m/s. The incline is arranged to make an angle theta = 30.0 degree with respect to the horizontal. The block travels a distance d = 3.00 m up the incline before coming to a stop. For this motion, determine the change in the block's kinetic energy, the change in potential energy of the block-Earth system, and the friction force exerted on the block (assumed to be constant). What is the coefficient of kinetic friction?

Explanation / Answer

A) Change in KE = 0 - 0.5*mvi2 = -160 J

b) change in potential energy = mgLsin30 = 5*9.8*0.5*3 = 73.5 J

c) work done by friction =  Change in KE - change in potential energy

- F*3 = -160+73.5 = -86.5

F = 86.5/3 = 28.8 N

d) umgcos30 =28.8

u = 28.8/[5*9.8*0.866] = 0.679

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