Question

If the coefficient of kinetic friction between a 29 kg crate and the floor is 0.38, what horizontal force is required to move the crate at a steady speed across the floor?

Answer 1

Let F be the magnitude of the force needed to keep the box sliding at constant speed, and f the magnitude of kinetic friction. Then by Newton's second law, the net horizontal force on the crate is

F - f = 0

so that

F = f

f is proportional to the magnitude of the normal force n,

f = 0.38n

The net vertical force is

n - mg = 0

which tells us that

n = mg = (29 kg) (9.8 m/s²) ≈ 284 N

Then the required force must have magnitude

F = f = 0.38 (284 N) ≈ 110 N