I was wondering if potential energy for spring is always positive.
$$PE=\frac{1}{2}kx^2$$ where $k$ is spring constant and $x$ is the change in length.

Thank you.

  • $\begingroup$ -1. No research effort. Why do you think the answer might not be the obvious one? $\endgroup$ – sammy gerbil Oct 11 '17 at 13:06

We tend to define the potential energy stored in the spring to be zero when the spring is relaxed. As Schrodinger's Cat says in his answer, you can add an arbitrary constant to potential energy without changing the physics, but it would be somewhat eccentric to do this.

Anyhow, to displace the spring from its relaxed state you have to do work on it, and that applies whether you are compressing the spring or stretching it. Because you are doing work on the spring, i.e. transferring energy to it, you are increasing the potential energy stored in it.

So if you graph the potential energy stored in the spring as a function of the displacement $x$ you'd get a quadratic with its minimum at $x=0$. Making the sensible definition that the PE is zero when $x=0$ the potential energy can never be negative.

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    $\begingroup$ If others have been confused too: This is not related to Schrödinger's thought experiment including a cat - it's just the user name of an other answer's author. :) $\endgroup$ – Cedric Reichenbach Oct 9 '17 at 13:40
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    $\begingroup$ Ah yes :-) I've now added some links for clarification. $\endgroup$ – John Rennie Oct 9 '17 at 15:17
  • $\begingroup$ To avoid the confusion, it's better to write the author's name without "beautifying" it: its true version doesn't have space nor apostrophe. $\endgroup$ – Ruslan Oct 10 '17 at 5:29

Actually, the formula, you mentioned, should be for change in potential energy. $$\mathrm{Change \, in \, P.E.}= \frac{1}{2}kx^2 \color{red}{\not}\Rightarrow \mathrm{P.E.}= \frac{1}{2}kx^2$$

You can always add some negative constant to your P.E. and make it negative.

$$\mathrm{P.E.}= V_0+\frac{1}{2}kx^2$$ where $V_0$ is just a real constant.

This was just the mathematical part.

For a proper physical interpretation, observe that the force corresponding to this potential is given by $$\vec F=-\vec \nabla V=-kx \, \hat x$$ which indicates a simple harmonic motion (oscillatory motion like that of a pendulum, spring etc.)

So, for any motion of the spring from its equilibrium position (elongation or compression), external energy needs to be supplied. This energy compresses or elongates the spring, which results in storing of potential energy inside the spring. So, the motion involves net increase of potential energy whether it is being compressed or elongated.

Hence, in that sense, change in potential energy is always positive, in this case, as John Rennie has mentioned in his post.

  • $\begingroup$ One should be careful to distinguish $\Delta x^2$ and $(\Delta x)^2$... $\endgroup$ – gj255 Oct 9 '17 at 17:37
  • $\begingroup$ @gj255 I couldnt understand what you wanted to tell me... I have clearly expressed my notation... it is quite standard notation. $\endgroup$ – SchrodingersCat Oct 9 '17 at 17:53
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    $\begingroup$ Your expression for the change in PE is misleading. If $x = 1$ initially and $x=2$ finally then $\Delta x = 1$ but the change in potential energy is not $k/2$. $\endgroup$ – gj255 Oct 9 '17 at 20:17
  • $\begingroup$ @gj255 The OP chooses $x$ to denote change in length. I used $\Delta x$ to denote change in length. Perhaps that was what the confusion was about. Edited my answer. $\endgroup$ – SchrodingersCat Oct 10 '17 at 3:52

It does not need to be positive or negative, because generally what counts when using energy to describe a system, is the exchange of this quantity.

In classical mechanics it is usually defined as you presented it, but sometimes in quantum systems, the level of zero energy is chosen where the kinetic and potential energy balance each other, and in those cases the elastic potential has a negative sign.

Furthermore, the choice is also made to comply with other conventions, as @SchrodingersCat points out with the relation between force and gradient of the potential.


Depends what you mean with "spring".

If you always mean the corkscrew spring from elastic materials, then yes, every deviation from the stationary position increases the potential energy.

But if you mean a general spring-like material like a rubber band, then the answer is "no". If you use an relaxed rubber band on a level and draw, potential energy of the rubber band increases, but if you push it, potential energy remains the same (I don't count the welling up of the band). Unsymmetric potentials always exist if a material is anisotropic (reacts different for the same amount of force for different directions). The energy could even be negative if the molecules rearrange (e.g. triggered by a phase change, a switch in the molecular configuration, as example a shape-memory alloy).


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