Solving for Speed of Mass After Spring Compression: .15m

[Lorbersf]

A vertical spring is created by affixing one end of the spring with k=500N/m to the floor. A 2.0kg mass is held .8m above the equlibrium position of the free and of the spring and released from rest. What is the speed of the mass when the spring is compressed .15m?

I need help setting up the equation for conservation of mechanical energy...
Ki+Ui=Kf+Uf >>>>>And then solving.

 

[Leong]

Initially we have the gravity potential energy of the block which is mg(L+h) where L is the length of the spring (when it is relaxed) and h is the height of the block from the top of the spring. at the final state, we have the gravity potential energy for the block , kinetic energy for the block and the potential energy for the spring.

 

[Lorbersf]

i need more help than that please

 

[Leong]

$$mg(L+0.8) = mg(L-0.15)+\frac{1}{2}mv^2+\frac{1}{2}kx^2$$