How far will the stone compress the spring?

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.


How far will the stone compress the spring?…

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?

Leave a Comment