# Answer to Question #20839 in Electric Circuits for Leandra

Question #20839

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?

Expert's answer

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

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

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