How Much Internal Energy Increases in a Damped Oscillating System?

[jimithing]

A light unstressed spring has length d. Two identical particles, each with charge q, are connected to the opposite ends of the spring. The particles are held stationary a distance d apart and then released at the same time. The system then oscillates on a horizontal frictionless table. The spring has a bit of
internal kinetic friction, so the oscillation is damped. The particles eventually stops vibrating when the distance between them is 3d. Find the increase in internal energy that appears in the spring during the oscillations. Assume that the system of the spring and two charges is isolated.

If the system is isolated, we can assume
$$\Delta E_{mec} + \Delta E_{th} + \Delta E_{int} = 0$$
Since friction is negligible, the change in thermal energy can be neglected, so
$$\Delta E_{mec} + \Delta E_{int} = 0$$

Now I realize that $$W = \Delta E_{mec}$$, but where exactly can I start?

 

[Tide]

First you need to find the spring constant which you can determine from the forces acting on the spring in its final configuration. Then apply energy conservation noting that the potential energy of the spring is $\frac {1}{2} k x^2$ and you can figure out the electrical potential before and after.

 

[jimithing]

I got
$$F = \frac{q^2}{4\pi \epsilon_{0}(3d)^2} , F = -k(3d)$$

So

$$k = -\frac{q^2}{4\pi \epsilon_{0}27d^3}$$

Do I now just sub into

$$U = \frac{1}{2}k(3d)^2$$ ?

 

[Tide]

Jimi,

I didn't check the details but it looks like you did the right steps!