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Consider a deformable bar, fixed at one end and acted upon by a load P (gradually increasing), as shown, through a rigid plate attached to its end. At the end of loading, potential energy is stored in the internal resistive force field (which was developed during loading). If at the end of loading I take any element as shown, it will be acted upon by equal and opposite internal resistive forces.

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Now let's say we gradually decrease the applied load, P. The internal forces on the particle considered will also keep on reducing and the particle will move towards the left. During its 'left' journey some work will be done by the forces $F_1$ and $F_2$ and these works will be equal and opposite in sign so that the net work done on the particle is zero during its entire 'left' journey. The force $F_1$ does a positive work thereby trying to increase the particle's kinetic energy, but at the same time, $F_2$ does a negative work so that the energy (that otherwise would've been the kinetic energy of the particle) is sucked out.

The potential energy stored in the force field is not going to the particle (since net work done on it is zero) so where does it go?

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During its 'left' journey some work will be done by the forces $F_1$ and $F_2$ and these works will be equal and opposite in sign so that the net work done on the particle is zero during its entire 'left' journey.

This isn't correct. Since bar as a whole is shrinking, the element also must shrink. Therefore the distance over which the two forces act is not equal. $F_2$ will operate over a greater distance than $F_1$.

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After the loading process the kinetic energy didn't change, so the work of the net force over the bar (and over each small bar elements) is zero. The same after unloading.

There is an increase of the elastic potential energy after loading, and a decrease after unloading. But as we don't have information about the external force F, nothing more can be said.

The potential energy is conserved if F is the gradient of a potential. For example, a vertical bar with a weight attached. The decrease of gravitational potential energy equals for this case the elastic potential energy after loading.

But it is not conserved for any type of force. For example, I take a weight from the ground and put on a table. The potential energy of the weight increased, and its kinetic energy was zero before and zero after the process.

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During its 'left' journey some work will be done by the forces F1 and F2 and these works will be equal and opposite in sign so that the net work done on the particle is zero during its entire 'left' journey.

No. The work done by $F_1$ and $F_2$ is $F_1(x-\delta L/2)=P(x-\delta L/2)$ and $F_2(-x-\delta L/2)=P(-x-\delta L/2)$, respectively, where the original load $P$ caused translation $x$ and stretching $\delta L$ of the element/particle. (Try sketching a large degree of stretching and release to convince yourself of this.) The total work done while removing $P$ is thus the sum of these or $-P\delta L$, which removes the strain energy originally supplied by $P$.

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  • $\begingroup$ Chemomechanics, the same would apply during elongation as well, isn't it? In that case, some net positive work will be done on the element (considering both displacement and elongation of the element). But if some net +ve work is done on the element during elongation its kinetic energy should increase, but it doesn't. Isn't this violating the work-kinetic energy theorem? $\endgroup$
    – Harshit Rajput
    May 11, 2022 at 9:15
  • $\begingroup$ That theorem applies only to perfectly rigid bodies. You specified that this is a deformable bar that stores potential energy. $\endgroup$
    – Chemomechanics
    May 11, 2022 at 16:33
  • $\begingroup$ Increasingly, I’m thinking that your account is not operated by a single human. The reason is that it’s frequently necessary to explain topics that you yourself have already discussed. $\endgroup$
    – Chemomechanics
    May 11, 2022 at 16:33
  • $\begingroup$ The textbook that I follow first derives the work-energy theorem for a single particle and then for a system of particles. A deformable body is a system of particles so I believe it should be applicable to it as well. The work-energy theorem for a system of particles says that the net work done by external and internal forces on all the particles of the system = change in ke of the entire system of particles. For a rigid body the work done by internal forces cancels out and we just have work done by external forces in the work-energy theorem. $\endgroup$
    – Harshit Rajput
    May 11, 2022 at 17:24
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    $\begingroup$ "A deformable body is a system of particles so I believe it should be applicable to it as well." It is a system of particles with an extra feature: it can store strain energy. If you correspondingly add a strain energy term to the work–kinetic energy relation, everything will work out. $\endgroup$
    – Chemomechanics
    May 11, 2022 at 18:09

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