Question

A vertical spring with a spring constant of 420 N/m is mounted on the floor. From directly above the spring, which is unstrained, a 0.15-kg block is dropped from rest. It collides with and sticks to the spring, which is compressed by 3.7 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:

19.53 cm

Explanation:

The computation of the height is as follows:

Here we applied the conservation of the energy formula

As we know that

P.E of the block = P.E of the spring

 m g h = ( 1 ÷ 2) k x^2

where

m = 0.15

g = 9.81

k = 420

x = 0.037

So now put the values to the above formula

(0.15) (9.81) (h) = 1 ÷2 × 420 × (0.037)^2

1.4715 (h) = 0.28749

h = 0.19537 m

= 19.53 cm