Greatest acceleration of the mass

[J89]

Homework Statement

A 2.50 kg mass is pushed against a horizontal spring of force constant 25 N/cm on a frictionless air table. The spring is attached to the tabletop, and the mass is not attached to the spring in any way. When the spring has been compressed enough to store 11.5 J of potential energy in it, the mass is suddenly released from rest. Find..

a) Greatest speed that the mass reaches (already solved)
b) Greatest acceleration of the mass..help!

Homework Equations

[tex]\sum[/tex]=ma
k= 1/2mv^2
W=1/2kx^2

The Attempt at a Solution

For part a) I already solved it, and the answer is 3.03 m/s (set 11.5J=1/2mv^2 and solve for v)

For part b, I know you have to use [tex]\sum[/tex]=ma, but I tried several methods and they all failed :(

Help would be appreciated. Thanks!

 

[LowlyPion]

Don't get carried away.

F = m*a so at what point is the force on the block the greatest?

What is the equation that relates what that force is?

 

[nlightner]

As the mass is released, the spring will push it with a force equal to the force constant (k) times the displacement (x) of the spring. This force will cause the mass to accelerate.

Using the equation F=kx and substituting the given values, we get F=25 N/cm * x. We know that the maximum displacement of the spring is the point at which it has stored 11.5 J of potential energy, so we can set kx=11.5 J and solve for x. This gives us x=0.46 cm.

Now, we can use the equation \sum=ma to find the maximum acceleration of the mass. The net force acting on the mass is the spring force (kx) minus the weight of the mass (mg). So, we have \sum=kx-mg=ma. Substituting the values we know, we get 25 N/cm * 0.46 cm - 2.50 kg * 9.8 m/s^2 = 2.50 kg * a. Solving for a, we get a=2.58 m/s^2.

Therefore, the greatest acceleration of the mass is 2.58 m/s^2.