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When a spring is inelastically deformed, the work done is no longer equal to the elastic potential energy stored in the spring – what accounts for this disparity?

Put another way, the spring is no longer storing all of the energy from mechanical work that has been performed on it as elastic energy, so what else is it storing it as? What is the type of energy 'stored' in irreversible deformation?

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  • $\begingroup$ Your system is in a new state, that is what deformation means. Put another way, your spring's atoms have rearrange their positions in space. I' m not sure if this is what you ask. $\endgroup$ – Constantine Black May 25 '19 at 15:20
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The irreversible deformation causes a change in the internal structure and therefore a change in the total electrostatic energy of your spring (more precisely, the new arrangement modifies the electrostatic energy). As long as your spring doesn't break, the "internal" electrostatic energy should increase with the deformations.

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The spring stretches mainly by torsion (twisting) of the wire of which it is made. At stresses below the elastic limit this occurs by small, temporary changes in the separations of the atoms. Beyond the elastic limit, planes of atoms in the metal start to slip a finite distance over each other. This is an irreversible process. [Energy would have to be supplied to return the planes to their old positions, even if this were practicable. Put simply, once the slip has occurred, the atoms in one plane are perfectly happy with their new neighbours!] The slipping increases the random vibrational energy of the atoms, rather than the elastic potential energy.

[The increase in random thermal energy is only temporary as there will be a flow of heat to the surroundings until the spring's temperature has fallen to that of the surroundings.]

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