# Where is the energy lost in a spring?

Thinking about springs, and their extensions, I recently came to a confusion which I hope this wonderful community can help me solve.

The question is this. When the block is initially attached to the spring, the spring has some extension $x_0$. Now the spring gets extended to some extension $x=\frac{mg}k$ by an external force maintaining equilibrium at all the points such that $KE=0$ at the bottom.

As my reference is the line shown in the figure, the initial potential energy $U$ is 0 due to both gravity and spring potential energy($x=0$).

Now as the block comes down, the spring potential energy is: $U_(spring)=\frac12kx^2$. Final extension is $\frac{mg}k$. So spring potential energy is $\frac{m^2g^2}{2k}$ But the decrease in gravitational potential energy is $mgx$ which equals $\frac{m^2g^2}k$.

This means that potential energy has decreased. Intitially, $U_{net}=0$ but finally $U_{net}=-\frac{m^2g^2}{2k}$.

Where if any, did this energy get compensated(to ensure COE is still true)?

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You're missing a somewhat subtle point in your analysis. The block on the left in your diagram, where the spring is at its equilibrium position, is moving, so it has kinetic energy (which you're currently ignoring). I'll leave it to you to sort out what the speed needs to be and check that CoE holds.

It needs to be moving because, if it were not, then there must be some force (your hand?) holding it in place. This would count as an external force, and when such a force acts, CoE does not hold.

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The "missing" energy you're referring to actually left the block-spring system when the external force was interacting with the block. One way to think about it is the following.

The work-kinetic energy theorem tells us that, since the KE of the block doesn't change during the lowering process, the net work done on the block is 0: $$W_{net}=\Delta \mathrm{KE}=0,$$ but the net work done on the block may be written as $$W_{net}=W_{grav}+W_{spring}+W_{external}.$$ Gravity, the spring, and the external agent (i.e. my hand) are all doing work on the block, and the sum of these must be 0, since KE didn't change. Gravity does positive work, since displacement is in the same direction as the force of gravity. The spring and the external agent do negative work, since their forces are applied in the opposite direction of the displacement: $$W_{grav}=mgx=\frac{m^2g^2}{k}.$$ The work done by the spring is equal to minus the amount of potential energy it gains (in other words, the spring does negative work on the block, the block does positive work on the spring), so $$W_{spring}=-\frac{1}{2}kx^2=-\frac{m^2g^2}{2k},$$ and, from the work kinetic energy theorem, we then have $$0=W_{grav}+W_{spring}+W_{external}\implies W_{external}=-\frac{m^2g^2}{k}+\frac{m^2g^2}{2k}=-\frac{m^2g^2}{2k}.$$ So the external agent does negative work on the spring (or the spring does positive work on the external agent), which exactly accounts for the energy that you noticed that was "missing".

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Thank you for your answer, but, when this external agent does this work. Where does this work go? Does it increase the temperature of the system? What happens to this energy? –  nobody_nowhere Feb 12 at 20:10
That's a good question. It depends on the nature of the external agent. If it's my arm, the missing energy goes to heat in my bicep. You could design something where the block pushes down on a lever (other end of lever lifting a mass), with fulcrum position being adjusted to account for the changing force required to hold up the block (as the spring starts to hold more weight)--in that case the energy would go to grav. potential in the other block at other end of lever. You could even rig something up to light up a light bulb as the block lowers, so energy would go to light/heat in that case. –  Mike Bell Feb 12 at 20:20
What it boils down to though, as @Kyle said, is that if you are considering the block-spring combination as your system, then by lowering the block slowly, an external agent is introduced, and so the energy of the block-spring system is not required to be conserved. (The energy of the block-spring-external agent system will be conserved, instead.) So, after the external agent is gone, and you analyze the energy of the block-spring system, you see that the total energy has changed. Some of that is now in the external agent, in one form or another. –  Mike Bell Feb 12 at 20:34

The energy goes into the thermal energy of the spring, the air around the block, or the block itself.

If you perform the experiment that you talk about you will find that the block bounces up and down for a while. If there is no energy loss, the block will bounce back to the exact height from which you dropped it and will keep doing this forever. In doing this the energy of the system is transferred back and forth between the kinetic energy of the block and the potential energy of the block in the gravitational field. In real life all materials have losses which convert kinetic energy into thermal energy. This process sucks energy out of the system, and it eventually reaches the equilibrium state which you describe.

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Well, I'm applying an external force to slowly move the block the equilibrium position in a reversible way(the block is in equilibrium at every moment)...Just like the definition of Potential Difference, the block is always moved at equilibrium. –  nobody_nowhere Feb 12 at 19:56