Conservation of Energy and Springs. Block dropped onto a spring

[coheedcoheed]

1. Homework Statement [/b
A 2.7 kg block is dropped from rest from a height of 4.5 m above the top of the spring. When the block is momentarily at rest, the spring is compressed by 25.0 cm. What is the speed of the block when the compression of the spring is 15.0 cm?

Homework Equations

PE=mgh
KE=.5mv^2
U(s)=.5kx^2

The Attempt at a Solution

Ok so what I did at first was use PE= U(s) to find the spring constant, which i got to be 4026 N/m (I don't know if this is right, correct me if I'm wrong), after I found that, i used mgh + .5mv^2 = .5kx^2, where h was .1m, and x was .15 since it was compressed .15m. but its wrong. please help!

 

[rootX]

1) If the v = 7.52 m/s, I would ask how you found k? I* punched few numbers and had different k (~3000)
2)Why you are taking h to be 0.1 m? in mgh + .5mv^2 = .5kx^2