A 4kg block is lowered down a 37° incline a distance of 5m from point A to point B. A horizontal force (F = 10 N) is applied to the block between A and B as shown in the figure. The kinetic energy of the block at A is 10 J and at B it is 20 J. How much work is done on the block by the force of friction between A and B?

Answer 1

W = +mgsin37(5) + Wf – Fcos37(5) = (4)(9.8)(5)(.6) + Wf – 10(5)(.8) = 77.6 + Wf
W = K = 20 – 10 = 10 J

10 = 77.6 + Wf
Wf = (-77.6+10) = -68 J