A 12.0-kg mass is placed on a 25.0 $\degree$ incline and friction keeps it from sliding. The coefficient of static friction in this case is 0.580, and the coefficient of sliding friction is 0.520. The mass is given a shove causing it to slide down the incline. What is the frictional force while the mass is sliding?
A) 55.4 N
B) 107 N
C) 49.7 N
D) 61.2 N

Question

Quizplus · Accepted Answer

The frictional force while the mass is sliding is given by the coefficient of sliding friction times the normal force. The normal force is equal to the weight of the mass times the cosine of the angle of the incline. Therefore, the frictional force is:$$f = \mu_k mg \cos 	heta = (0.520)(12.0 	ext{ kg})(9.81 	ext{ m/s}^2) \cos 25.0^\circ = 55.4 	ext{ N}$$

A 120-Kg Mass Is Placed on a 25 °\degree° Incline and Friction Keeps It from Sliding

A 120-Kg Mass Is Placed on a 25 $\degree$ Incline and Friction Keeps It from Sliding