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?
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