Question Detail
Question
A 5 kg body slides 6 m down a rough plane inclined at 30°, then onto a frictionless horizontal surface where it compresses a spring with k = 300 N/m. μ_k = 0.1. What is the maximum compression of the spring?
Correct Answer
Option D is the correct answer.
Detailed Explanation
Initial PE = mgh = 5 × 9.8 × (6 sin 30°) = 49 × 3 = 147 J. Friction force f_k = μ_k mg cos θ = 0.1 × 5 × 9.8 × 0.866 = 4.24 N. Work done by friction on incline = f_k × d = 4.24 × 6 = 25.4 J. Energy remaining to compress spring = 147 - 25.4 = 121.6 J. Then ½ k x² = 121.6, so x² = (2 × 121.6)/300 = 243.2/300 = 0.8107, x = √0.8107 = 0.90 m.
Hint
The initial potential energy is partially dissipated by friction on the incline. The remaining energy becomes spring potential energy.