A spring being compressed between two masses

[dahoom102]

Homework Statement

The two blocks I and II shown above have masses m and 2m respectively. Block II has an ideal massless
spring attached to one side. When block I is placed on the spring as shown. the spring is compressed a distance
D at equilibrium. Express your answer to all parts of the question in terms of the given quantities and physical
constants.

a. Determine the spring constant of the spring

Relevant Equations

kx=mg

mgh=1/2kx^2

i attempted this problem by using conservation of energy,
mgh=1/2kx^2
mgD=1/2kD^2
2mg=kD
k=2mg/D

why is it wrong ? btw the correct answer used mg = kx which is mg/D

 

[vela]

The term equilibrium means there's no net force on the mass. The condition of equilibrium doesn't really say anything about the relationship between the gravitational and elastic potential energies.

For this specific problem, you seem to be assuming the mass is placed on the uncompressed spring and then released, and when it has fallen to the equilibrium point, the initial gravitational potential energy has been turned into elastic potential energy and the mass is at rest. In reality, you'd find the mass is still moving.

 

[dahoom102]

Thank you! That's quite helpful.