# Elastic potential energy of spring

1. Sep 11, 2007

### psycovic23

1. The problem statement, all variables and given/known data
A spring of spring constant k is suspended from the ceiling with a mass M attached to its lower end. The spring has negligible mass. Show that the sum of the elastic potential energy of the spring and the gravitational potential energy of the mass is a minimum at the equilibrium position.

2. Relevant equations

3. The attempt at a solution

So I think the basic setup would be:
(Assuming the datum is at the position when the spring doesn't have anything on it)

$$\frac{1}{2}kx^{2} - mgh = E(x)$$
But the graph shows that energy decreases until a minimum and then increases. If x=0 is the equilibrium point (and thus minimum energy), why does that function continue to decrease? Any help would be appreciated.

2. Sep 11, 2007

### genneth

x=0 is not the equilibrium point...

3. Sep 11, 2007

### fikus

the point of equilibrium: kx - mg = 0 --> x=mg/k

energy $$E(x)= 1/2kx^2 - mgx$$
Minimum is where derivative is zero and that is exactly the fist equation.