Question: A block on an inclined plane has a mass of 10 kg and the angle of inclination is 30 degrees. What is the force of friction if the coefficient of kinetic friction is 0.2?
Options:
Correct Answer: 5.88 N
Solution:
The normal force N = mg * cos(θ) = 10 kg * 9.81 m/s² * cos(30°) ≈ 84.87 N. The force of friction Ff = μk * N = 0.2 * 84.87 N ≈ 16.97 N.