An inclined plane of angle 200 has a spring of force consta

An inclined plane of angle = 20.0° has a spring of force constant k = 495 N/m fastened securely at the bottom so that the spring is parallel to the surface as shown in the figure below. A block of mass m = 2.41 kg is placed on the plane at a distance d = 0.297 m from the spring. From this position, the block is projected downward toward the spring with speed v = 0.750 m/s. By what distance is the spring compressed when the block momentarily comes to rest?

Solution

Initial total energy = final potential energy

m g (d + s) sin 20 + 0.5 m v² = 0.5 k s²

[ 2.41 * 9.8 * (0.297 + s) * 0.342 ] + [ 0.5 * 2.41 * 0.75² ] = 0.5 * 495 * s²

(2.399 + 8.08 s ) + 0.678 = 247.5 s²

247.5 s² - 8.08 s - 3.08 = 0

s = 0.129 m