Question

A 4 kg bowling ball is drop from rest at a height of 1 m. There is an unstretched spring below. Once the bowling
ball touches the spring, the bowling ball slows down as it compresses the spring. The spring constant is 490 N/m.
Find the velocity of the ball at position D (when the spring has compressed 0.2 m)

Answer 1

Answer:

M g H = 1/2 M v^2 + 1/2 K x^2

H = 1 + .2 = 1.2 m    Left side is original potential energy that is converted to kinetic energy and potential energy     (H given as 1.2 m)

v^2 = 2 g H - K x^2 / M

v^2 = 2 * 9.8 * 1.2 - 490 * .04 / 4 = 23.5 - 4.9 = 18.6 m^2 / s^2

v = 4.31 m/s