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Imagine I have a malleable metal rectangular rod (such as an Aluminum rod). Suppose either I stick this rod into a wall or I place it between two extremely heavy blocks so that one end of the rod is fixed and the other end is sticking out. Next, suppose I place a mass $m$ box on the end that is sticking out. The initial potential energy of the box is $U_{0} = mgh_{0}$ where $h_{0}$ is the altitude of the center of mass of the box relative to some reference level. Assuming the box is sufficiently heavy, when I let go of the box, it bends the metal rod, and then the box falls to the floor.

If the metal rod were not in the way, the box would have had energy $KE = mg(h_{0}-h_{1})$ right before hitting the ground. However, because the metal rod was in the way, the metal rod slowed the box's fall to some extent, and thus the box actually has energy $KE < mg(h_{0} - h_{1})$ right before hitting the ground.

My question is, where did the energy go? The metal rod doesn't store any potential energy, it doesn't rebound back to its original shape like a spring, and it doesn't have any apparent kinetic energy, so it's not clear how conservation of energy works here. How do we understand this scenario?

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Heat in the rod from the plastic deformation.

Plastic deformation goes from one stable state of the solid to another stable state.

Therefore an energy barrier must exist between the states. And work is necessary to cross it. This work will be recovered in a large part as phonons (heat).

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My question is, where did the energy go? The metal rod doesn't store any potential energy

This is incorrect. The permanent bending of the rod creates* untold numbers of dislocations (one-dimensional defects) in the material, and these dislocations, being imperfectly bonded, carry an energy penalty.

*E.g., at Frank–Read sources.

As the other answer notes, plenty of heat is generated as well, but it's important to note that permanently bending the rod does not leave it in the same thermodynamic state.

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  • $\begingroup$ Is it certqin, that you create more dislocations than you erase unless you start with a single crystalline rod? $\endgroup$
    – tobalt
    Jan 8, 2022 at 17:16
  • $\begingroup$ For a macroscale object, it is as certain as one can be that far more dislocations are generated than leave the crystal. $\endgroup$
    – Chemomechanics
    Jan 8, 2022 at 17:23
  • $\begingroup$ This is interesting. What happens when you try to straighten the rod back? I assume that would create even more dislocations because you're not actually restoring it to its original state if you do it mechanically (and you'd need to heat it if you really want to recover the original state). Is this accurate or am I off? $\endgroup$
    – Maximal Ideal
    Jan 8, 2022 at 19:26
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    $\begingroup$ Exactly right—this is called cold working. $\endgroup$
    – Chemomechanics
    Jan 8, 2022 at 19:46

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