A man exerts a constant force to pull a 49-kg box across a floor at constant speed. He exerts this force by attaching a rope to the box and pulling so that the rope makes a constant angle of 36.9∘ above the horizontal. The coefficient of kinetic friction for the box-floor interface is μk = 0.20.
What is the work done by the man as he moves the box 10 m?
The velocity is constant so Fnet=0 and
F horizontal=F friction
F horizontal=F man × cos36.9
F friction=0.2(9.8×49 - F up)
F up = F man × sin36.9
So
Fm(cos36.9)=0.2(9.8×49 - Fm(sin36.9))
Solve for Fm: 104.4 N
Multiply by 10 to get 1044 J...but that's wrong. Please help.