A 13.0 kg stone slides down a snow-covered hill (the figure ), leaving point A w

Question

A 13.0 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 2.00 N/m. The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and 0.80, respectively. At point b V2 =23.2. How far will the stone compress the spring?

Explanation / Answer

We can solve this with energy. Initially the energy is all Kinetic.
E=(1/2)mv2=(1/2)(13)(23.15)2

E=3484 joules

Next we need to figure out how much energy goes to friction, this is going to be equal to the work done by friction. (use kinetic coefficient because the stone is moving)

W(x)=Fd=mgd=.2(13)(9.8)x

W(x)=25.48x

Now we need to figure out the energy that goes into the spring.

PE=(1/2)kx2

PE(x)=(x-100)2 (its x-100 because the spring doesn't act til 100m)

Now we just set E=W+PE

3484=25.48x+(x-100)2

x2-174.52x+6516=0

x=120.4m (the other solution, 54m, is incorrect because our equation only work when x>100)

Hope that helps

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