Work/KE/PE Problem: Solving for Stone Compression Distance

[mvpshaq32]

Homework Statement

A 12.0 kg stone slides down a snow-covered hill (the figure ), leaving point A with a speed of 12.0 m/s. There is no friction on the hill between points A and B, but there is friction on the level ground at the bottom of the hill, between B and the wall. After entering the rough horizontal region, the stone travels 100 m and then runs into a very long, light spring with force constant 2.10 N/m. The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and 0.80, respectively.

How far will the stone compress the string?
http://session.masteringphysics.com/problemAsset/1000077656/3/YF-07-34.jpg

Homework Equations

The Attempt at a Solution

I found the velocity at the bottom of the hill, which equals 23.2 m/s. Then I set up the equation 0.5(12)(23.2^2)=0.5kx^2 + 0.2mgL to solve for x, giving 28.9 m, but apparently this is wrong.

 

[tiny-tim]

hi mvpshaq32!

(try using the X² icon just above the Reply box )

… The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and 0.80, respectively.

How far will the stone compress the string?
…
I found the velocity at the bottom of the hill, which equals 23.2 m/s. Then I set up the equation 0.5(12)(23.2^2)=0.5kx^2 + 0.2mgL to solve for x, giving 28.9 m, but apparently this is wrong.

hmm … for the maximum compression of the spring, that looks correct to me

maybe, since they give you the coefficient of static friction, they want the compression at which the stone finally comes to rest?