How Does Charge Affect Spring Compression in a Frictionless System?

[velvetchere]

two small blocks are resting (and touching) side-by-side on a frictionless tabletop. the block on the left is connected to a wall to the left of it by a spring having stiffness constant k=75 N/m. the one on the right is connected to a wall on to the right of it by a spring of k=50 N/m. when the two blocks are touching, the springs are at their natural lenghts. a charge of Q=45 microcoulombs is placed on the left block, and a charge of 16 is placed on the right block, causing them to move apart and compressing their respecitve springs. if the blocks are allowed to slowly move apart until they stop, determine the amount each spring will be compressed at this point.

would conservation of energy be applied to this problem?

 

[Euclid]

If I am understand the question correctly, I believe you merely have to equate the electrostatic force of repulsion with the spring force.
I don't think energy is conserved in this situation (what causes the blocks to stop?).

 

[velvetchere]

thanks for your reply. ok, so I'm assuming that we're basically treating each block as a particle, and that the electrostatic force would equal
(ke)(q_1)(q_2) divided by r^2. and i suppose r is essentially the distance that the springs compress?

i'm a little iffy on the spring aspect of the problem, being how we're given two k values.

 

[daniel_i_l]

You have 2 unknowns (the compression of each spring) and 2 equations, the eqilibrium of the forces in the end, and energy conservation. Solve them and you'll get your answer.

 

[Euclid]

The two equations you need are

$$k_1x_1 = \frac{1}{4\pi \epsilon} \frac{q_1q_2}{(x_2-x_1)^2}$$

and

$$k_2x_2 = \frac{1}{4\pi \epsilon} \frac{q_1q_2}{(x_2-x_1)^2}$$

Energy conservation is not applicable to this problem. There must be some external force doing work on the system, as evidenced by the phrase "blocks are allowed to slowly move apart until they stop".