A vertical spring with a spring constant of 400 N/m is mounted on the floor. From directly above the spring, which is unstrained, a 0.30 kg block is dropped from rest. It collides with and sticks to the spring, which is compressed by 2.7 cm in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height (in cm) above the compressed spring was the block dropped?

Answer 1

Gravitational potential energy loss (at maximum compression) = Spring potential energy gained
(There is zero kinetic energy when the drop begins and when maximum compression is attained.)

M g H = (1/2) k *(0.027)^2
k is the spring constant, 400 N/m
H is the distance you want to know.
g = 9.8 m/s^2
M = 0.30 kg

Solve for H