A vertical spring with a spring constant of 430 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 2.2 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

3.54 cm

Explanation:

From the question,

The potential energy of the block = energy needed to compress the spring

mgh = (ke²)/2 Equation 1

Where m = mass of the block , h = height from which the block was dropped, g = acceleration due to gravity, k = spring constant of the spring e = compression.

Firstly, we make h the subject of the equation

h = ke²/2mg Equation 2

Given: k = 430 N/m, e = 2.2 cm = 0.022 m, m = 0.3 kg, g = 9.8 m/s²

Substitute these values into equation 2

h = 430(0.022²)/2×0.3×9.8

h = 0.20812/5.88

h = 0.0354 m

h = 3.54 cm

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