High School Confusion about the Strain Energy when loading/unloading a sample

[KevMilan]

The answer says B, it does makes sense, but why D is regarded as incorrect. If the Elastic Potential energy when loading is the area under the loading curve, so why not with the unloading?

 

[Chestermiller]

D is the work done by the sample while unloading.

 

[KevMilan]

What about the area under the loading curve? Is it the workdone on the sample or by the sample?

 

[Chestermiller]

What about the area under the loading curve? Is it the workdone on the sample or by the sample?

Like I said, it is the work done by the sample. The force exerted by the sample on its surroundings is directed toward the sample, and the displacement at the boundary where the work is being done is directed toward the sample. So the work done by the sample is positive. The work done on the sample is negative.

 

[KevMilan]

For the area under the loading curve, shouldn't it be the workdone ON the sample? Because we are the one putting force/stress on it? Could you clarify? Thankyou.

 

[Chestermiller]

For the area under the loading curve, shouldn't it be the workdone ON the sample?
Because we are the one putting force/stress on it? Could you clarify? Thankyou.

Yes. The force we apply and the displacement are in the same direction.

 

[PeterO]

What about the area under the loading curve? Is it the workdone on the sample or by the sample?

Work done ON the sample, but that area is the sum of the shaded area and the "white" sliver.
Area under top line = work done on the sample, Energy Input
Shaded area = = work done BY the sample when unloaded - Energy returned
The white sliver represents in Energy Lost (Put in, but not returned, ie Option A).

 

[KevMilan]

One more thing, why is it the workdone BY the sample (not ON)? (Is it because we're not putting any force/tension on the sample when unloading?) Any simple explanation would be very helpful!

 

[PeterO]

One more thing, why is it the workdone BY the sample (not ON)? (Is it because we're not putting any force/tension on the sample when unloading?) Any simple explanation would be very helpful!

I suspect you would get similar load/unload curves for a trampoline, and I am happy to consider that the trampoline is doing work on me (work done by the trampoline) when it is flinging me back into the air after I have jumped on it and pushed the mat down (I did work on the Trampoline).

 

[KevMilan]

I see. Thanks a lot man [emoji106]

 

[A.T.]

One more thing, why is it the workdone BY the sample (not ON)? (Is it because we're not putting any force/tension on the sample when unloading?)

We are applying a force on the sample during unloading. But that force is opposite to the displacement so we are doing negative work. If you interpret the dark area as negative, then it represents the work done ON the sample during unloading.

 

[sophiecentaur]

One more thing, why is it the workdone BY the sample (not ON)?

'On' or 'By' is not necessarily obvious and it's actually a bit anthropomorphic, imo.
It may be better to think in terms of the system as a whole and whether energy is being put in or taken out. During displacement from the equilibrium position, Energy is put in (the direction of the arrow on the diagram is an indication of this). As the system returns to equilibrium position, Energy is taken out.