Answer to Question #20839 in Electric Circuits for Leandra

Question

A force of 38.0 N accelerates a 5.0-kg block at 5.7 m/s2 along a horizontal surface; how large is the friction force and what is the coefficient of friction?

Answer 1

ma=F-Ffr
Ffr=F-ma=38-5.0*5.7=9.5 N& --the friction force
mu - the coefficient of friction
mu = Ffr/N
N=mg
mu = Ffr/mg=9.5/(5*10)=0.19