Question

Starting from rest a 5.5 kg block slides 2.5 m down a rough 30 degree incline. The coefficient of kinetic friction between the block and the incline is uk =.436.

A. Determine the work done by the force of gravity.

B. Determine the work done by the friction force between the block and incline.

C. Determine the work done by the normal force.

Show all work and free body diagrams

Answer 1

given that

mass m = 5.5 kg

sliding distance d = 2.5 m

angle = theta = 30 degrees

coefficient of kinetic friction uK = 0.436

a) here normal force n = mgcos(30) = 5.5*9.81*cos(30) = 46.73 N

work done by gravity wg = mgsin(theta)*d = 5.5*9.8*2.5*sin(30) = 67.375 J

b) workdone by friction

wf = -46.73*2.5*0.436 = -50.94 J

c) work done by normal force = 0

because normal force is perpendicular to the sliding distance