# Negative Potential Energies

1. May 13, 2014

### tmobilerocks

Hi All,

I just have a question regarding potential energies. Say I have a block above a spring, and I define the potential energy of gravity and the spring to be zero at the uncompressed point. I then drop the block, trying to find the maximum compression of the spring using conservation of energy. I also define my coordinate system as the down direction as being positive. If I use conservation of energy equations, do I need to manually put in minus signs? For example in this problem:

ΔK + ΔUs + ΔUg = 0
mgh - mgho + .5ky2 = 0 (General equation)

Coordinate system down as positive
mgh + mgho - .5ky2 = 0

Coordinate system up as positive
-mgh - mgho + 5ky2 = 0

2. May 13, 2014

### Feodalherren

Put them in manually as opposed to what? I'm not quite sure I understand the question but whether you reverse a coordinate system or not you need to stay consistent. If you're saying up is negative then any movement in that direction needs a negative sign.

The sign conventions in the energy equations come from the derivation - the integration - so you can't just swap signs. Unless you're going to re-derive your equations you need to stick with the convention that was used when the equation was derived.
For the spring this is important because a spring's Force is defined as F= -kx and this is used to derive the equation. You're not free to chose a sign convention of your liking when you're dealing with this equation unless you're willing to do the integration by yourself. The negative sign signifies that the force exerted by
the spring is always directed opposite the displacement from equilibrium.

3. May 13, 2014

### tmobilerocks

I drop a block on a spring compressing it. Let's say 0 potential energy is defined at spring equilibrium and down is positive direction. The spring is now compressed. The change in potential energy would be mgh + mgh0 right? (Needed to manually add positive sign to initial potential energy, up direction is now negative)

4. May 13, 2014

### Nathanael

If mgh0 is the initial potential energy then the change in potential energy would be mgh0 - mgh regardless of whether up or down is defined as positive.

The definition of up or down being positive is irrelevant to the potential energy, potential energy is a scalar.

5. May 13, 2014

### tmobilerocks

The block is below the reference point; h is negative. If you do mgh - mgh0 it becomes positive or mgh + mgh0

6. May 13, 2014

### BvU

Hello T,
Can I contribute an elementary question about post 1, or rather: two questions:
1. About Ug: I get the impression that you think g = 9.8 m/s2, no matter which way the positive direction is defined. g is pointing to the center of gravity of the earth, so if you choose "up is positive", you will have to fill in (manually) g = -9.8 m/s2. Am I right ?

2. About Us: the y you introduce is equal to h - h0, I suppose ? The energy is 0.5 k y2, In either direction, and for either definition of positive, y2 is the same as (-y)2. Your change in sign for that term is in error. Do you agree ?

There is enough challenging thought left over: is pulling up over 1 cm the same as pushing down 1 cm ?

7. May 13, 2014

### tmobilerocks

In a way. The specific problem deals with a block released above a spring and compressing on it. I define Ug = Us= 0 as the uncompressed point and down as the positive direction. Up is therefore negative, so the block starts out with negative potential energy U = mgh (h is negative). Drawing a energy diagram, the spring potential energy must be negative for conservation of energy to work.

8. May 13, 2014

### Staff: Mentor

Spring potential energy is always positive (or zero). Regardless of your choice of sign convention, the general equation for energy conservation will be: ΔUg + ΔUs = 0. Since the spring starts out uncompressed, ΔUs will be positive.