Question Detail
Question
A 0.5-kg block rests on a horizontal frictionless surface. It is pressed against a spring with k = 625 N/m, compressing it by 10.0 cm. When released, the block slides up a frictionless incline at θ = 30°. What is the maximum distance d the block travels up the incline?
Correct Answer
Option B is the correct answer.
Detailed Explanation
Initial elastic PE = ½ k x² = ½ × 625 × (0.1)² = 3.125 J. At max height on incline, this converts to gravitational PE = m g h, where h = d sin θ. So, m g d sin θ = 3.125 J. Thus, d = 3.125 / (0.5 × 9.8 × sin 30°) = 3.125 / (4.9 × 0.5) = 3.125 / 2.45 = 1.28 m.
Hint
Apply conservation of mechanical energy. The initial spring potential energy is entirely converted to gravitational potential energy at the highest point on the incline.