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Physics 18 Online
OpenStudy (anonymous):

A 10 kg box, sliding to the right across a rough horizontal floor, accelerates at -2 m/s^2 due to the force of friction. Calculate the coefficient of kinetic friction between the box and the floor and the magnitude of the net force acting on the box.

OpenStudy (anonymous):

\[F=uN\] u=coefficient N=normal force

OpenStudy (anonymous):

and also F=ma

OpenStudy (anonymous):

i know the formulas i just dont know how i can get the coefficient of friction without the force of friction

OpenStudy (stormfire1):

\[a=\frac{F}{m}=\frac{u_kmg}{m}=u_kg\]Now you have \[2.0 m/s=u_k(9.8 m/s^2)\]\[u_k=0.2041\]

OpenStudy (stormfire1):

Note that my answer above is right although when I typed it in the 2.0 m/s should have been squared (2.0 m/s ^2=....).

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