Question

A vertical spring with a spring constant of 310 N/m is mounted on the floor. From directly above the spring, which is unstrained, a 0.28-kg block is dropped from rest. It collides with and sticks to the spring, which is compressed by 3.6 cm in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height above the compressed spring was the block dropped?

Answer 1

Answer:

7.3cm above the compressed spring.

Explanation:

We can use the conservation energy theorem to solve this problem:

$U_{1}+K_1=U_e+U_2+K_2\\m*g*h+0=\frac{1}{2}k*x^2+0+0\\\\h=\frac{k*x^2}{2*m*g}\\\\h=\frac{310N/m*(3.6*10^{-2})^2}{2*0.28kg*9.8m/s^2}\\\\h=0.073m$

The block was dropped 7.3cm above the compressed spring.