# What Is The Speed Of The Stone When It Reaches Point B? How Far Will It Compress The Spring? Will The Stone Move Again After It Has Been Stopped By The Spring?

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.
thanked the writer. 