Physics Question

Question

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?

Answer 1

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

F = ma = (15.0 kg)(0.271 m/s²) = 4.07 N.

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

F_t = mg sin 36° = (15 kg)(9.81 m/s²)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_kN

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

F_t - µ_kN = ma

F_f = µ_kN = 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