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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

Answer:

I found: #mu_k=0.25#

Explanation:

Have a look:
enter image source here
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#