Finding the Spring Constant of a mattress

[012anonymousx]

Homework Statement

A mattress manufacturer estimates that 20 springs are required to comfortable support a 100kg person. When supporting the person, the 20 springs are compressed 0.035m. Calculate the spring constant for one sprint.

Homework Equations

The answer is 1400N/M at the back of my textbook. How does one solve it?
Why does my solution not work?

The Attempt at a Solution

(20)Ee = Eg //The gravitational potential energy is transferred to the 20 springs.
20(0.5)(k)(x^2) = mgh
10(k)(0.035^2) = (100)(9.8)(0.035)
k = 2800N/M
So close yet so far...

 

[haruspex]

You have used conservation of energy. Think about the dynamics of that. Person let go from rest on uncompressed springs... what will happen?
You don't need to think about energy here. Just treat it as a statics question.

 

[012anonymousx]

Ah,
F = kx
k = F/x
k = mg/x
k = (100)(9.8) / 0.035 x 20
k = 1400

I still think my solution should have worked. I don't understand conceptually why energy is not conserved.

 

[haruspex]

I still think my solution should have worked. I don't understand conceptually why energy is not conserved.

As I said, you treated it as though the person was suspended horizontally, just touching the bed, then released. Suppose energy is conserved - what will happen? The person will fall, compressing the springs, but 'overshoot'. When the force from the springs matches the person's weight, the person will stop accelerating, but already has momentum and KE, so will continue to descend. When descent ceases, the springs will be applying a larger force and the person will bounce back up. If energy is conserved they will come all the way back up to starting position before descending again, and bounce forever.
The question required the point at which there was no acceleration, somewhere between the extremes of the bounce.