# Ideal Spring, Spring Constant, Equilibrium Length

1. Oct 5, 2008

### Spartan Erik

1. The problem statement, all variables and given/known data

A 2.0 kg mass is attached to one end of an ideal spring with a spring constant of 500 N/m and a 4.0 kg mass is attached to the other end. The masses are placed on a horizontal frictionless surface and the spring is compressed 10 cm from its equilibrium length. The spring is then released with the masses at rest and the masses begin to move apart. When the spring has reached its equilibrium length, what is the speed of the 2.0 kg mass?

0.67, 1.0, 1.3, 2.1, none of these (all in m/s)

2. Relevant equations

Hooke's law: F = -kx
W(spring): 1/2kx(initial)^2 - 1/2kx(final)^2
W(applied): -W(spring) if stationary before/after displacement

3. The attempt at a solution

I don't know how I can utilize this data and relate it to velocity

2. Oct 5, 2008

### LowlyPion

OK. You kind of have the right idea.

The work that the spring produces has gone into what? (Does it rhyme with Kinetic Energy by any chance?)

Also you have two masses that are being acted on by the same force, even if in opposite directions.
F = m1a1 = m2a2

Since you know that m1 = 2 and m2 = 4, then you know that m2 = 2*m1

What does that mean then for a1 and a2? If that is true for a1 and a2 then it must also be true for v1 and v2 at the moment that PE is converted wholly to KE.

That leads to Work = total KE = ... you should be able to get the rest.