Displacement of wood block originally attached to spring

[usamo42j]

Homework Statement

A 180-g wood block is firmly attached to a light horizontal spring, Fig. 6-26. The block can slide along a table where the coefficient of friction is 0.30. A force of 20 N compresses the string 18 cm. If the spring is released from this position, how far beyond its equilibrium will it stretch on its first string?

Homework Equations

F=kx
Elastic PE = 1/2*kx^2
F_fr=μ_k*F_N

The Attempt at a Solution

We have 20 = 0.18k where k is in N/m. So k= 1000/9. Then this means the elastic potential energy is 1/2*20*(0.18)^2=1.8 J. Work due to friction is F_fr*displacement=0.3*0.18*9.8*displacement=.5292*displacement=1.8, and solving yields ≈3.4 m. Not correct, apparently.

Can anyone show me where I went wrong?

 

[Doc Al]

You set all of the initial elastic potential energy equal to the work done by friction. But since it's stretching past equilibrium, the final position (of momentary rest) will also have some elastic potential energy.