Work/KE/PE problem

  • Thread starter mvpshaq32
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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


Homework Equations





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.
 

Answers and Replies

  • #2
tiny-tim
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hi mvpshaq32! :smile:

(try using the X2 icon just above the Reply box :wink:)
… 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 :redface:

maybe, since they give you the coefficient of static friction, they want the compression at which the stone finally comes to rest?
 

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