A 12.0-kg mass is placed on a 29.0° incline and friction keeps it from sliding.The coefficient of static friction in this case is 0.660,and the coefficient of sliding friction is 0.546.The mass is given a shove causing it to slide down the incline.What is the frictional force while the mass is sliding?
A)57.0 N
B)102.9 N
C)56.2 N
D)67.9 N
E)10.5 N

Question

A 12.0-kg mass is placed on a 29.0° incline and friction keeps it from sliding.The coefficient of static friction in this case is 0.660,and the coefficient of sliding friction is 0.546.The mass is given a shove causing it to slide down the incline.What is the frictional force while the mass is sliding?
A)57.0 N
B)102.9 N
C)56.2 N
D)67.9 N
E)​10.5 N

Quizplus · Accepted Answer

The frictional force while the mass is sliding can be calculated using the formula for kinetic (sliding) friction: $F_{\text{friction}} = \mu_k \cdot N$, where $\mu_k$ is the coefficient of kinetic friction and $N$ is the normal force. The normal force can be found from the component of the gravitational force perpendicular to the incline, which is $mg \cos(\theta)$, where $m$ is the mass, $g$ is the acceleration due to gravity (9.8 m/s$^2$), and $\theta$ is the angle of the incline. Thus, $N = 12.0 \cdot 9.8 \cdot \cos(29.0^\circ)$. Plugging in the values, $N \approx 12.0 \cdot 9.8 \cdot 0.874 = 103.5$ N. Then, the frictional force is $F_{\text{friction}} = 0.546 \cdot 103.5 \approx 56.5$ N, which rounds to $56.2$ N due to the precision of the given options.

A 120-Kg Mass Is Placed on a 29