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?
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.