A 15.0kg stone slides down a snow-covered hill (the figure (Figure 1) ), leaving point A with a speed of 11.0m/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 2.10N/m . The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and 0.80, respectively.
What is the speed of the stone when it reaches point B?
22.65 m/s (this is correct)
How far will the stone compress the spring?
Will the stone move again after it has been stopped by the spring?
picture: Mastering Physics
PE = mgh
KE = .5mv^2
spring energy = .5kx^2
The Attempt at a Solution
I set up the equation
0.5(15)(22.65^2)=0.5(2.1)x^2 + 0.2(15)(9.8)(100)
I solved for x to equal 29.4, but this is wrong