Question

An object of mass m = 10 kg is released from a height of 2 m on top of a vertically aligned spring with spring constant k = 1000 N/m. The relaxed length of the spring is 1 m and the lower end is attached on a rigid level surface. Assume that the spring is massless and ideal; and neglect air-resistance. Calculate the maximum compression and extension of the spring after the released object falls on top of the spring.

Part of my question is what is being asked here. How can extension be determined if the falling mass is not attacted wouldn't it just fall and then stop? The other part that I am confused about is what is the height used to find max compression? Is it 2m or 1m?

Answer 1

Given,

mass = m = 10 kg ; h = 2m ; k = 1000 N/m ; L = 1 m

We need to determine the maximum compression and extension of the spring.

The gravitational potential energy of the mass at height H = h + x

E = m g (h + x)

This energy will get tnasferred to spring the moment the mass falls over spring. So,

1/2 k x² = m g (h + x)

Solving for x we get,

kx² - 2 m g x - 2 m g h = 0

1000x² - 2 x 10 x 9.8 x x - 2 x 10 x 9.8 x 2 = 0

1000x² - 196 x - 392 = 0

x² - 0.196 - 0.392 = 0

Solving this quadratic equation we get:

x = 0.73 m

X = 1 m - 0.731 = 0.27 m

Hence, compression = x = 0.73m and extension = 0.27 m