Question

A certain bedspring is observed to compress by 1.00 cm when a certain person is at rest atop the spring. What would be the spring’s compression if the same person was dropped onto the spring from a height of 0.250 m above the top of the spring? Assume the person is dropped from rest. 

Answer 1

Let the spring have a spring constant of k, x is the compression.
Restoration force = F = - k x,    weight of person = m g
Potential energy of the person = mg h (h = x)
Potential energy of the spring = 1/2 k x²

When a person sits on the bed at rest, the Decrease in PE of person = Increase in PE of spring.

       m g x = 1/2 k x²
       k  =  2 m g / x  = 2 m g / 0.01meter = 200 mg  N/m

When a person is dropped from a height of 0.250 meters:

       m g (0.250+ x)  = 1/2 k x²
       mg (0.25+x) = 1/2 * 200 * mg * x²
       0.25 + x = 100 x²
       x = (1 +- √101 )/200 = 0.0552 m

Compression of the spring is 5.52 cm