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.

 

[rwisz]

Hi there,

First, let's define some variables for the problem:
m = mass of the block (10 lbs)
k = spring constant (12 lb/in)
F = force applied to the block
u_s = coefficient of static friction (0.60)
u_k = coefficient of kinetic friction (0.40)
v_r = speed of the block as it returns to its initial position
v_max = maximum speed achieved by the block

To solve for v_max, we need to consider the forces acting on the block during its motion. Initially, when the block is attached to the unstretched spring, there is no force acting on it. As the force F is slowly applied, the spring will stretch and the block will start to move. At some point, the force F will be equal to the maximum tension in the spring (20 lbs) and the block will stop moving. Let's call this point A.

At point A, the forces acting on the block are:
1) Spring force, Fs = -kx (since the spring is stretched)
2) Friction force, Ff = u_k * mg (since the block is in motion)
3) Applied force, F = 20 lbs (since this is the maximum tension in the spring)

Now, when the force F is suddenly removed, the spring will contract and the block will start to move in the opposite direction. Let's call this point B. At point B, the forces acting on the block are:
1) Spring force, Fs = kx (since the spring is compressed)
2) Friction force, Ff = u_s * mg (since the block is now at rest)
3) No applied force, F = 0 lbs

Using the equations of motion, we can write:
For point A:
F - u_k * mg - kx = 0
For point B:
Fs - u_s * mg = 0

Substituting the values of F and Fs from the equations above and solving for x, we get:
x = 1.25 in

Now, we can use the equation for kinetic energy to solve for v_max:
KE = 1/2 * m * v_max^2
At point A, the KE is:
KE_A = 1/2 * m * v_max^2 = 1/2 * 10 * 0^2 = 0

At point B, the KE is: