Question

A 2-kg particle is acted upon by a spring with force constant k = 100 N/m while being influenced by a friction. The static friction force and the kinetic friction force are both 50 N in magnitude. The object is released from rest at x = 5.0 m, with the spring stretched by 5.0 m.

What is the speed of the object when it reaches x = 0 for the first time?

Answer 1

the easiest way to solve is problem is by using energy. You start the problem with energy stored as spring potential energy:\[1/2kx^2\] where x=5m and k=100N/m
as the spring retracts and reaches it's equilibrium position (x=0) that spring energy is converted into kinetic energy \[1/2mv^2\] where m = 2kg
you must also remember though, that some of the energy is lost due to the frictional force on the block, an amount equal to the frictional force times the distance traveled\[F _{f}d\]
where d=5m and Ff = 50N
this gives us
\[1/2kx^2 = 1/2mv^2-F _{f}d\] or \[1/2(100)(5^2) = 1/2(2)(v^2) - 50(5)\]
solving for V we get \[V^2=1500 \rightarrow V=\pm \sqrt(1500)\]
depending on where you put your x-axis, it will either be the positive or the negative solution.

Answer 2

actually i want to make a correction, i accidentally got my signs wrong, the correct equation is\[1/2kx^2=1/2mv^2+F _{f}d\] because your spring potential is going INTO both the work due to friction and the work due to kinetic. therefore your answer would actually be \[v=\pm \sqrt(1000)\]