A 11.0kg stone slides down a snow-covered hill (the figure (Figure 1) ), leaving

Question

A 11.0kg stone slides down a snow-covered hill (the figure (Figure 1) ), 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 2.50N/m . The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and 0.80, respectively.

Picture: http://i49.tinypic.com/a9qq1z.jpg

How far will the stone compress the spring?

Explanation / Answer

Initial total energy of the stone at A=550 +2156=2706 J At the starting of the horizontal region,it's velocity let be v. Then 0.5*11*v^2=2706 V=22.18 m/s Deceleration due to friction=1.96 So velocity when after travels 100 m=9.997 So let the spring be compressed by x. Then 0.2*11*9.8*x+0.5*2.5*x^2=0.5*11*v^2 Solving for x,we get x=13.984m

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