A 15.0 kg box is released on a 36.0° incline and accelerates down the incline at 0.271 m/s2.

Find the friction force impeding its motion.

How large is the coefficient of friction?

A 15.0 kg box is released on a 36.0° incline and accelerates down the incline at 0.271 m/s2.

Find the friction force impeding its motion.

How large is the coefficient of friction?

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First of all, by Newton's 2nd Law the net tangent force to the incline is:

F = ma = (15.0 kg)(0.271 m/s^{2}) = 4.07 N.

The tangential and normal components of the force of gravity are:

F_{t} = mg sin 36° = (15 kg)(9.81 m/s^{2})sin 36° = 86.5 N

N = mg cos 36° = (15 kg)(9.81 m/s2)cos 36° = 119 N

The net force is also given by F = F_{t} - µ_{k}N

To solve for friction force, equate the formula above with ma.

F_{t} - µ_{k}N = ma

F_{f }= µ_{k}N = F_{t} - ma = 86.5 N - 4.07 N = 82.4 N

Finally, the coefficient of kinetic friction is:

µ_{k} = F_{f}/N = (82.4 N) / (119 N) = 0.692

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