A 10 lb block is attached to unstretched spring

[Ryan1059]

A 10-lb block is attached to an unstretched spring of constant k=12lb/in. The coefficients of static and kinetic friction between the block and the plane are 0.60 and 0.40, respectively. If a force F is slowly applied to the block until the tension in the spring reaches 20lb and then suddenly removed, determine (a) the speed of the block as it returns to its initial position, (b) the maximum speed achieved by the block

Homework Equations

F=-k/x
Ff=mew(Fn)
Espring = 1/2kx(squared)
KE = 1/2mv(squared)

this is first post so I don't really know proper format for equations
I solved for Velocity in returning position, I got an answer of 2.32 ft/sec which is correct according to my books answer key.
I really just need help with solving for V max. I realize that I should go about setting up an equation that sets KE equal to X and then solve for KE max, but I don't know how to begin. Please help, any direction would be appreciated.

 

[PhanthomJay]

Hi, Ryan, welcome to Physics Forums!

If you are familiar with calculus, instantaneous acceleration is defined as the slope of the V-t curve, dV/dt, which indicates a maximum velocity occirs when the slope of the V-t curve is zero, that is, when the instantaneous acceleration is 0. Now when the instantaneous acceleration is 0, what can you say about the forces acting on the block at that point?

 

[Ryan1059]

thank you anyways for responding! and putting up with the god awful equations i set up haha. alright well if there is no instantaneous acceleration then that means there are no forces acting on the block. It makes enough sense. I would think that I should take the derivative of some function of velocity, solve for zero which would give me my max velocity, correct? But I don't know how to set up the equation or anything or where to start.

 

[PhanthomJay]

You've got to first correct your first relevant equation , Hooke's Law, which is F = -kx.

Then the point of 0 acceleration in the x direction is the point where there are no NET forces acting in the x direction. There are actually 2 forces acting... which two? Their sum must add to zero... and the max speed can be calculated in the same manner in which you calculated (or I think you calculated, no work was shown) the answer to part a. You don't want to get into any more calculus than you have to.