A 15.0 kg stone slides down a snow-covered hill (the figure ), leaving point A with a speed of 10.0 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 100m and then runs into a very long, light spring with force constant 2.50 N/m. The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and 0.80, respectively.
Here is the picture: http://session.masteringphysics.com/problemAsset/1260235/2/YF-07-34.jpg
A) What is the speed of the stone when it reaches point B?
B) How far will the stone compress the spring?
PE = mgh
KE = .5mv^2
spring = .5kx^2
The Attempt at a Solution
A) I did this:
PE = (15.0)(9.80)(20.0) = 2943 J
KE= .5(15.0)(10.0)^2 = 750 J
2943+750 = 3693 J
3693 = .5(15.0)(v)^2-.5(15.0)(10.0)^2
B) work done by spring = .5(2.50)(x)^2 = 1.25x^2
work done by friction = (0.2)(15.0)(9.8)x = 29.4x
1.25x^2 + 29.43x - 3693 = 0
x= 43.84 or -67.38
x= only 43.84 meters.
Do these answers seem right? thanks!