# Cons. of energy problem

1. Oct 20, 2008

### musicfairy

A spring of length D and constant k is hung from the ceiling. A nonstretchable string of length L is attached to the bottom of the spring. An object of mass m attached to the free end of the string and dropped from the ceiling. What is the lowest the object would reach?

I'm supposed to solve this strictly using energy. No amplitude or anything like that.

Here I go:

E = Ug + Us
Ug = (D + L)mg
Us = .5kx2

That all I know about U. How do I find how far the spring stretches past its equilibrium position, when it stretches the farthest away from the point of origin?

This is my guess:

Since it's all potential energy at the time of release, the spring is unstretched:

mg(D + L) = .5kx2

x = sqrt((2*mg(D + L))/k)

So the lowest point would be D + L + x

So can someone explain to me all the conceptual errors I made and how those concepts are supposed to apply in this situation? Thanks.

2. Oct 20, 2008

### Redbelly98

Staff Emeritus
For gravitational potential, here are 2 things to think about:

At what height or level is the grav. potential equal to zero?

And

What is the height and grav. potential when the mass is at its lowest point?

3. Oct 20, 2008

### musicfairy

I assume that at the lowest there is no gravitational potential energy. That's why I set the two equal to each other.

Do you mean that gravitational potential energy should be mg(D + L +x) since while the spring stretches down it still has some gravitational potential energy?

Am I on track of getting a correct equation? I really need one, and fast. Anyone please help.

4. Oct 20, 2008

### Redbelly98

Staff Emeritus
Yes, good.

Yes, you're on the right track. Gravitational and spring potential energies are all you need to solve this.

5. Oct 20, 2008

### musicfairy

Ok, so if that's true then my equation should be

mg(D + L + x) = .5kx2

And then I'll plug in numbers and solve for x somehow.

Is this right?