A 15.0-kg stone slides down a snow-covered hill, leaving point A with a speed of

Question

A 15.0-kg stone slides down a snow-covered hill, 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 100 m and then runs into a very long, light spring with force constant 2.00 N/m. The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and 0.80, respectively.
( Make a free-body Diagram)
(a) What is the speed of the stone when it reaches point B?
(b) How far will the stone compress the spring?
(c) Will the stone move again after it has been stopped by the spring?

Explanation / Answer

The potential energy at the start is PE = mgh = 15.0 kg*9.81 m/s2*20.0 m = 2943 J The kinetic energy at the start is KE = mv2/2 = 15.0*(10.0)2/2 = 750 J So, the total energy at the start is PE+KE = 2943+750 = 3693 J When the stone comes to rest, the work done against friction and the spring will total this amount. The work done against the spring is Wspring = 1/2*ks*d2 = 1/2(2.0 N/m)*d2 = d2 N/m The work done against friction is Wfriction = kfdmgd = .2*15.0 kg*9.81 m/s2*d = d*29.43 N This gives rise to a quadratic equation: d2 + 29.43d = 3693, or d2 + 29.43d - 3693 = 0. This is solved in the usual way to get d = 47.81 m is the distance the spring will be compressed. At that distance, the spring is exerting a force of ks*d = (2.0 N/m)*(47.81 m) = 95.62 N The static friction is kfsmg = .8*15.0 kg*9.81 m/s2 = 117.7 N Static friction is higher than the force on the stone, so it will not move again.

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