How far will the stone compress the spring?

A 13.0kg 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 2.30n/m . The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and 0.80, respectively.

Question:

How far will the stone compress the spring?

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Total energy TE at A = ke + pe = 1/2 mU^2 + mgh. As there are no losses between A and B, TE is the same at point B but all kinetic energy there. U = 12 mps at A. h = ? is the height at A above ground level.

Total energy at the spring = TE – WE = TE – kmgS; where k = .2, m = 13 kg, g is g, and S = 100 m. WE is the work done by friction as the stone slides along the flat between B and the spring.

The spring will store PE = TE – kmgS = 1/2 ndX^2; where dX = ? m the compression you are looking for. n = 2.3 N/m, the spring constant. Solve for dX.

TE – kmgS = 1/2 mU^2 + mgh – kmgS = 1/2 ndX^2

dX^2 = m(U^2 + 2gh – k2gS)/n; everything is given on the RHS but the h, the starting height at point A. I presume that’s in the figure. You can find dX^2 and dX = sqrt(dX^2); you can do the math.

How far will the stone compress the spring?