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A mass spring system is in equilibrium. If I pull on the load by $x$ meters, the energy stored in the spring is (this is what is given in my book):

$$E=\frac12kx^2 $$

However, doesn't the load lose gravitational potential energy as it moves down? Where would this energy go? By conservation law, shouldn't the energy equation be: $$E_{stored}= \frac12kx^2 + mgx$$

In short, where does the loss of gravitational potential energy (mgx) of the load get transferred to if it is not stored in the spring?

(Referring to a vertical mass spring system)

The picture in my mind: enter image description here

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  • $\begingroup$ The wording of the problem sounds like they're really just asking for the change in elastic potential energy. Could you post the actual question text to confirm? $\endgroup$
    – BMS
    Dec 14, 2013 at 17:55
  • $\begingroup$ @BMS: it isn't an actual question... I just wanted to understand the formula and also to know where the loss in potential energy (mgx) of the load would go if it isn't stored in the spring... $\endgroup$
    – Eliza
    Dec 14, 2013 at 17:58

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There are some very interesting subtleties here. Let's analyze the situation very carefully.

Let's choose our system to consist of the block, spring, and Earth. By choosing the Earth and block to be in our system, we will have a change in gravitational potential energy.

In the beginning, the (massless) spring hangs vertically with a block of mass $m$ attached at the bottom. We could calculate how much the spring is stretched by equating the gravitational and spring forces ($kx_1=mg$) but we won't need this.

Now, during the pulling process you describe, it's important to note that you are doing positive work on the system, which means that the energy in the system increases. It is tempting to say that the change in energy is zero, but this isn't the case for the system we've chosen.

Let's use the work-energy theorem to answer your question of where the gravitational potential energy "goes." $$\underbrace{W_\text{net, external}}_\text{Positive}=\Delta E_\text{tot}=\underbrace{\Delta U_\text{grav}}_\text{Negative}+\underbrace{\Delta U_\text{elastic}}_\text{Positive}$$

Yes, the gravitational potential energy decreases. Where does it go? Well, the only other term that could (mathematically) compensate for this decrease in gravitational potential energy is the increase in elastic potential energy. But be careful with wording here. The spring is not storing gravitational potential energy; rather, gravitational potential energy was converted to elastic potential energy.

As a side note, since the left-hand side of the equation above is positive, the absolute value of $\Delta U_\text{elastic}$ is greater than that of $\Delta U_\text{grav}$. So, not only was the gravitational potential energy converted to elastic potential energy, the positive work you did on the system also adds to the increase in elastic potential energy.

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    $\begingroup$ so when the spring is pulled by X2 meters by an external agent, positive work done by the external agent AND the loss in gravitational potential energy of the load will be converted to elastic potential energy? $\endgroup$
    – Eliza
    Dec 15, 2013 at 17:56
  • $\begingroup$ That is my interpretation. $\endgroup$
    – BMS
    Dec 15, 2013 at 20:10
  • $\begingroup$ I finally get it. Just one more thing, if this was a horizontal mass-spring system, would the elastic potential energy stored in the spring just equal to the work done by the external agent (i.e. no complications involving gravitational potential energy)? $\endgroup$
    – Eliza
    Dec 16, 2013 at 12:03
  • $\begingroup$ @Eliza if $\Delta K=0$ during the process, yes. If $\Delta K \ne0$, then some of the work "goes into" kinetic energy as well. $\endgroup$
    – BMS
    Dec 16, 2013 at 16:15
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The difference in potential energy is due to different definitions of what $x=0$ means. Since from the perspective of the spring this would be when the spring is not compressed nor stretched (rest length). However in the case of a mass-spring system in a gravity field (assumed to be a constant acceleration, $g$) this position is often chosen to be the equilibrium position, so where the force of the spring is equal to the force of gravity. The difference between these position can be derived with the following equation, $$ k\Delta x=mg. $$ If you would substitute this into the potential energy equation you get: $$ E=\frac{1}{2}k\left(x+\Delta x\right)^2=\frac{1}{2}k\left(x^2+2x\Delta x+\Delta x^2\right)=\frac{1}{2}kx^2+mxg+\frac{1}{2}km^2g^2, $$ where $x$ is relative to the rest length, so the position relative to the equilibrium position would be $x_{eq}=x+\Delta x$.

You can also remove the last therm ($\frac{1}{2}km^2g^2$) since it is independent of $x$, because you are free to choose the position of zero potential energy since you only look at changes in potential energy.

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  • $\begingroup$ The term mgx... where is this change in gravitational potential energy stored in? $\endgroup$
    – Eliza
    Dec 15, 2013 at 3:52
  • $\begingroup$ @Eliza: I am not sure what you mean, but $E=\frac{1}{2}kx_{eq}^2$ would represent the sum of potential energy of the spring and gravity. $\endgroup$
    – fibonatic
    Dec 15, 2013 at 4:33
  • $\begingroup$ Could you please look at my edit. I have made my question clearer using a diagram $\endgroup$
    – Eliza
    Dec 15, 2013 at 4:58
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    $\begingroup$ The work done by the external force would be equal to: $$W=\frac{1}{2}k\left((x_1+x_2)^2-x_1^2\right)-mgx_2$$ But $x_1=\frac{mg}{k}$, which simplifies the work done to $$W=\frac{1}{2}kx_2^2$$ So this is what I mean by $x$ relative to the equilibrium position. $\endgroup$
    – fibonatic
    Dec 15, 2013 at 5:17
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    $\begingroup$ Like I showed in my previous comment, the decrease of gravitational potential energy will reduce the amount of work done, since gravity acts in the same direction as the external force. So you could say that the work done by the external force and gravity together wil be equal to minus (in the other direction) the work done by the spring. So you could say that the spring does store the change of gravitational potential energy. $\endgroup$
    – fibonatic
    Dec 16, 2013 at 12:24
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The energy stored in the spring is the one that is give $\frac{1}{2}{kx^2}$. As you mention, by conservation of energy there also a reduction of potential energy, but that reduction is not energy that it's stored by the spring but the complete change of energy of the whole system.

Take into consideration, that the problem just states what's happening with the spring, for example in a horizontal configuration where there would be no gravitational force applied to the mass/spring system.

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  • $\begingroup$ :I am referring to a vertical mass spring system. Also, what do you mean by the complete change of energy of the whole system? $\endgroup$
    – Eliza
    Dec 14, 2013 at 17:18
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The potential energy turns into kinetic energy. It makes causes a simple harmonic motion. Whenn you release the mass, it means a free fall. It never comes at rest. Dissipative forces neglected. To stop it another force must be applied, which results in loss of energy.

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