# Spring deflection problem using energy equations

ago01
Homework Statement:
Given a box of mass 0.8kg how far is the spring compressed if the block is placed on the spring and does not oscillate? In other words, what is the compression of the spring at the equilibrium point of the block- spring system? The spring constant is 22 N/m.
Relevant Equations:
mg, -kx
At the equilibrium position all forces are equal to 0 (they balance). So, naturally we:

mg - kd = 0

d = mg/k

d = 0.356 m

This is the correct answer, I believe. But I want to solve it using the energy equations because I am really trying to understand energy's connection to the rest of kinematics.

At the equilibrium position all forces are 0. But there is potential energy stored in the spring and the box, and to be at equilibrium these both must cancel. Let d be the deflection and the ground be the 0 potential plane. Place the bottom of the spring on the 0 potential plane.

So,

mgd = (1/2)kd^2

d = (2mg)/k

d= 0.712m

Which of course is a factor of 2 larger than the actual answer. I am confused, as I am lead to believe this would be solvable both ways (one using forces, one using energy). I have asked this to several people and they seem to immediately go to the d = mg/k solution, which is fine.

Is this solvable using conservation of energy or am I spinning my wheels for no reason?