Question

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

Answer 1

840 J

Step-by-step explanation:

In the y direction:

∑F = ma

Fn + F sin θ − mg = 0

Fn = mg − F sin θ

In the x direction:

∑F = ma

F cos θ − Fn μ = 0

F cos θ = Fn μ

F cos θ = (mg − F sin θ) μ

F cos θ = mg μ − F μ sin θ

F (cos θ + μ sin θ) = mg μ

F = mg μ / (cos θ + μ sin θ)

F = (49) (9.8) (0.2) / (cos 36.9° + 0.2 sin 36.9°)

F = 104.4 N

You correctly found the force applied by the man, but remember that only the horizontal component of that force is used to do work.

W = F cos θ d

W = (104.4) (cos 36.9°) (10)

W = 835

Rounding to two significant figures, the man does 840 J of work.