How Is the Total Maximal Elastic Energy Calculated in This System?

[Edothegray]

Homework Statement

Given the scheme
https://www.dropbox.com/s/nm943gleo2nc8ft/Screenshot%202016-08-28%2013.47.33.png?dl=0
Springs' constant k=40 N\m
mass of bulk m=6kg
The wooden bulk is released at rest when all the springs are not stretched
Question: What is the elastic total maximal Energy of the system?
2. Homework Equations
E=0.5*K*(deltaX)^2
A=0.375m
Maximal kinetic energy 11.25Joul

The Attempt at a Solution

If I substitute all the givens in the equation given in section 2 above, I get Etot=4*0.5*40*0.375^2=45/4=11.25
Let me know if it's correct because what I saw in the book answers is 45J, which doesn't match my calculations. Maybe I'm missing something
Thanks for your attention

 

[mfb]

How did you get deltaX and what is A?
Initially the springs are at rest, at which point are they stretched maximally?

I get the same answer as the book (assuming g=10m/s^2).

 

[Edothegray]

Let me see if I got it right:
A-is the point beyond the initial rest point, in which we get equilibrium. So you've got to add the 0.375 of movement towards the bottom so eventually we get 0.75 of stretch of the upper springs down and equally 0.75 shrink of the bottom ones.
Well, you get the 0.375 from equilibrium equation on the bulk:
mg=4k*deltaX
is that what you meant?
Thanks for the response

 

[mfb]

So you've got to add the 0.375 of movement towards the bottom so eventually we get 0.75 of stretch of the upper springs down and equally 0.75 shrink of the bottom ones.

Right. And if you plug that in, you get 45 J.

 

[Edothegray]

gr8! Thank you