Question

A 19.5 kg block is dragged over a rough, horizontal surface by a constant force of 110 N
acting at an angle of angle 31.1◦ above the horizontal. The block is displaced 6.74 m,and the coefficient of kinetic friction is 0.156.What is the net work done on the block?
Answer in units of J.

Answer 1

This look s like one I answered already, posted by you.

Answer 2

I did it the way you said and it said it was wrong.

Answer 3

The answer is 433.719 joules. You find the sum of the x forces by taking 110*cos31.1 - 19.5*9.81*.156 which gives you 64.35 Newtons. You put that answer in the Work formula. Work equals force * distance so take 64.35*6.74 meters and you get your answer, 433.719 Joules.

Answer 4

Nate, the friction force is equal to mu*N, N is equal to mg-Fsin(31.1) not mg.

Answer 5

But you do not need the frictional force to find the total work. Just the dragging force and the displacement.