A block is placed on a rough wooden plane.It is found that when the plane is tilted 30F° to the horizontal, the block will slide down at constant speed.The coefficient of kinetic friction of the block with the plane is:
A)0.500
B)0.577
C)1.73
D)0.866
E)4.90

Question

Quizplus · Accepted Answer

When the block slides down at a constant speed, it means that the force of friction is equal and opposite to the component of gravitational force parallel to the plane. In this case:
$$F_{friction} = F_{gravity}\sin 30^\circ$$
$$\mu_k F_{normal} = mg\sin 30^\circ$$
where $\mu_k$ is the coefficient of kinetic friction, $F_{normal}$ is the normal force exerted on the block by the plane, and $m$ is the mass of the block. The normal force is equal and opposite to the component of gravitational force perpendicular to the plane:
$$F_{normal} = mg\cos 30^\circ$$
Putting it all together, we get:
$$\mu_k = \frac{mg\sin 30^\circ}{mg\cos 30^\circ} = 	an 30^\circ = \boxed{0.577}$$

A Block Is Placed on a Rough Wooden Plane