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

7.3cm above the compressed spring.

Step-by-step 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=(1)/(2)k*x^2+0+0\\\\h=(k*x^2)/(2*m*g)\\\\h=(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.