# Elastic Potential Energy of a Sprin

1. Apr 4, 2012

### testme

1. The problem statement, all variables and given/known data
Two trolleys, of mass 1.2 kg and 4.8 kg, are at rest with a compressed spring between them, held that way by a string tied around both. When the string is cut, the trolleys move apart, if the force constant fo the spring is 2400 N/m, by how much must it have been compressed in order that the 4.8 kg cart move off at 2.0m/s

Cart: 4.8 kg, 2 m/s
Cart 2: 1.2 kg
Spring constant, k: 2400 N/m

2. Relevant equations

Ee = 1/2kx^2
Ek = 1/2mv^2

3. The attempt at a solution
E = E'
1/2kx^2 = 1/2mv^2
kx^2 = mv^2
2400x^2 = 4.8(2)^2
x^2 = 4.8(4)/2400
x = 0.09 m

2. Apr 4, 2012

### HallsofIvy

Staff Emeritus
Why are you not including the kinetic energy of the 4.8 kg cart. The 1.2 kg cart moves too. You can use "conservation of momentum" to find its speed.

3. Apr 4, 2012

### testme

So then would I find the velocity of the other cart like this:

mava + mbvb = mava' + mbvb'
0 = 4.8(2) + 1.2vb'
vb' = 8

then we put that back in so

1/2kx^2 = 1/2mv^2 + 1/2mv^2
2400x^2 = 4.8(2)^2 + 1.2(8)^2
x = 0.2 m

4. Apr 4, 2012

### PhanthomJay

you have left out the KE of the smaller mass. It also must move out a certain velocity, using momentum conservation to find that velocity.

5. Apr 4, 2012

### PhanthomJay

Looks like you and HallsofIvy have already worked this out...correctly.....
I posted after you 'spoke'......