Question

If you use a horizontal force of 30.0 N to slide a 12.0 kg wooden crate across a floor at a constant velocity, what is the coefficient of friction between the crate and the floor?

Answer 1

I found: `mu_k=0.25`

Explanation:

Have a look:

Newton's law is split into two components along `x` and `y`, that, using our data, give:
`30-mu_kR_n=0`
`R_n=mg=12*9.81=117.7~~118N`
So:
`mu_k=30/118=0.25`