A 10.0-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.What is the frictional force in this situation?
A)41.4 N
B)88.8 N
C)46.2 N
D)51.5 N

Question

Quizplus · Accepted Answer

The frictional force is equal to the component of the gravitational force parallel to the incline, which is m*g*sin(θ). For a 10.0-kg mass on a 25.0° incline, this is 10.0 * 9.8 * sin(25°) = 41.4 N. Since the object is not sliding, this is the static frictional force.

A 100-Kg Mass Is Placed on a 25 °\degree° Incline and Friction Keeps It from Sliding

A 100-Kg Mass Is Placed on a 25 $\degree$ Incline and Friction Keeps It from Sliding