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

• mvpshaq32
In summary, the stone slides down a snow-covered hill with no friction, then travels 100 m on a rough horizontal region before encountering a spring with a force constant of 2.10 N/m. The coefficients of kinetic and static friction are 0.20 and 0.80, respectively. The stone has a velocity of 23.2 m/s at the bottom of the hill and compresses the spring by 28.9 m, but this may not be the final compression as it may come to rest at a different point.

## 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

## 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.

hi mvpshaq32!

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mvpshaq32 said:
… 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?