Conservation of energy involving spring and friction

[IniquiTrance]

Homework Statement

A 0.62 kg wood block is firmly attached to a very light horizontal spring (k = 180 N/m) . This block-spring system, when compressed .05 m and released, stretches out 0.023 m beyond the equilibrium position before stopping and turning back.

What is the coefficient of kinetic friction between the block and the table?

The Attempt at a Solution

This is how far I've gotten:

At x = -.05, there is 0 kinetic energy, and potential energy = 1/2 kx².

So:

1/2 kx² = 1/2 mv₂² + \mumg(energy lost due to friction)

But this has 2 unknown variables.

Any help is greatly appreciated.

 

[IniquiTrance]

Nevermind, figured it out.

1/2 kx_initial² = 1/2 kx²_final + \mumg(abs((x_final - x_initial))