A 3-kg block slides at constant velocity down a 32 $\circ$ inclined plane.What is the coefficient of friction between the block and the plane?
A)0.31
B)1.60
C)0.0
D)0.62
E)1.24

Question

Quizplus · Accepted Answer

Since the block slides at constant velocity, the net force acting on it is zero. This means the component of gravitational force acting down the incline is exactly balanced by the frictional force. The gravitational force component along the incline is $mg\sin(\theta)$ and the frictional force is $f = \mu mg\cos(\theta)$, where $\mu$ is the coefficient of friction, $m$ is the mass, $g$ is the acceleration due to gravity, and $\theta$ is the angle of the incline. Setting these equal to each other because of the constant velocity (no acceleration), we get $mg\sin(\theta) = \mu mg\cos(\theta)$. Solving for $\mu$ gives $\mu = \tan(\theta)$. For $\theta = 32^\circ$, $\mu = \tan(32^\circ) \approx 0.62$.

A 3-Kg Block Slides at Constant Velocity Down a 32 ∘\circ∘

A 3-Kg Block Slides at Constant Velocity Down a 32^$\circ$