Question Detail
Question
A 120 N concrete block on a 30° incline has a coefficient of kinetic friction μk = 0.5. What is the magnitude of the kinetic friction force if the block is moving?
Correct Answer
Option B is the correct answer.
Detailed Explanation
The friction force is μk * F_N. The normal force F_N on an incline is mg cos θ = 120 cos 30°. cos 30° = √3/2 ≈ 0.866. So, F_N ≈ 120 * 0.866 = 103.92 N. Friction = 0.5 * 103.92 N = 51.96 N.
Hint
First calculate the normal force. On an inclined plane, the normal force is the component of weight perpendicular to the surface, which is mg cos θ.