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?

1 Answer
Nov 4, 2015

I found: mu_k=0.25μk=0.25

Explanation:

Have a look:
enter image source here
Newton's law is split into two components along xx and yy, that, using our data, give:
30-mu_kR_n=030μkRn=0
R_n=mg=12*9.81=117.7~~118NRn=mg=129.81=117.7118N
So:
mu_k=30/118=0.25μk=30118=0.25