A difficult mechanic problem with spring

[cfkorea]

I hope someone's help.
If someone has some idea of following thing, please write your idea here.
$$Problem$$: There is a spring(spring constant: $$k$$) fixed on the floor with massless board fixed upside of it. Peter put a box of $$m(kg)$$ on the board and make spring in a state of equilibrium.Peter want to make the box off from the board. What is the least value of $$x$$, the distance Peter should press the spring, to make box off from the board?
I hope a proof using forces, such as gravity, spring force, and $$N$$ or using Energy, such as $$mgh, \frac{1}{2} kx^2$$, etc..
Can anyone help me?

 

[ONJ's Noble Steed]

Direct the x-axis upward. Define the zero point as the point from which the massless board must be launched in order for the box to stop touching the board. Define 'a' as the point of equilibrium. Define b as the location of the board if one takes the box off of it.
Then:
(b-a)k=mg, Equilibrium condition.
(kb^2)/2=mgb, Conservation of energy. This equation says that the box must reach point b, but it need not go higher since we're trying to find the least value of initial displacement that will cause the box to just stop touching the board.
Solving the two equations we get: a=mg/k. This is what we needed, 'a' represents the amount of displacement from the equilibrium position needed for the box to stop touching the board for an instant.
Anyone care to check this?

 

[lightgrav]

Your answer is correct.

(We're supposed to see how far they've gotten already)
(... at least seen how cfkorea had defined variables... .)