"Elephant in the way" kinetic energy problem.

[Bob Loblaw]

1. The problem statement, all variables and given/known data
http:/www.imgred.com/http://static.scribd.com/docs/736egzypod07v_files/image004.jpg

I found this on one of those "OMG FUNNY" internet web pages. I am trying to attempt a solution.

2. Relevant equations

Mechanical energy is conserved so Einitial=Efinal. Uinitial+Kinitial=Ufinal+Kfinal.
mgyinitial+1/2mvinitial^2=mgyfinal+1/2mvfinal^2

3. The attempt at a solution

KE+PE = Total Energy. When the 3kg object is released from a 5m. I am unsure on how to procede with this problem. How does the force constant k of the spring factor into the solution?

[Calbear]

You're doing it wrong.
How to do this problem:
U(initial) {which is gravitational potential energy} = mgh, where h is 5m, m is 3kg, g is 9.81m/s^2.
U(final) {which is elastic potential energy} = 1/2kx^2, where k is 100.
WE WANT X.
W=delta E
Conservation of energy in sys mass, plane, earth. W=0
0=1/2kx^2 - mgh
mgh=1/2kx^2
(2mgh)/k=x^2
So, x=Square Root ((2mgh)/k)
So, the answer in this problem is SqRt((2(3kg)(9.81m/s^2)(5m))/100)
~SqRt 3 = 1.73205 meters

Hope that helped, if you care at this point, months later.