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Imagine there is a train in front of me and I'm on a train following behind pushing with a force of 100 Newtons for one minute. I can't imagine getting twice as tired if the train is moving a 40km/h instead of 20km/h. The only solution I can see is that some of the work the train is doing must be getting included as well. This creates an obvious question - exactly what work is the work that is included here?

Notes:

This question is similar to this one, but is different because in my question I am actually doing steady work by continuing to push on the front train.

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    $\begingroup$ This seems to essentially be a duplicate of physics.stackexchange.com/q/1984/50583 $\endgroup$ – ACuriousMind Mar 29 '20 at 13:08
  • $\begingroup$ @ACuriousMind Not quite because of the second part $\endgroup$ – Casebash Mar 29 '20 at 13:09
  • $\begingroup$ Please ask only 1 question per post. $\endgroup$ – Qmechanic Mar 29 '20 at 13:30
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    $\begingroup$ 1. If the first part is answered by the duplicate please edit your question to remove it. 2. It is not clear to me how the second part is meaningfully different - in the first case your tiredness is not related to any mechanical work done at all (since there is no mechanical work done!). Why would you expect it to be different in the second case? $\endgroup$ – ACuriousMind Mar 29 '20 at 13:30
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    $\begingroup$ @ACuriousMind Edited. In the second case you are doing work $\endgroup$ – Casebash Mar 29 '20 at 13:34
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Let's suppose I push my desk into the wall. It doesn't move at all, but I get tired.

The sensation of getting tired has almost nothing to do with how much external work you do. A human is a very inefficient machine and tiredness is a very non-specific measurement. Whenever you are formulating a question in those terms it is likely not relevant to the underlying physics.

The easiest “fix” is to ask about a spring instead of a human. The amount of energy stored in a spring is easy to calculate exactly and such questions can be answered clearly on the basis of the underlying physics.

A spring can push your desk into a wall for as long as you want without getting tired. It’s energy depends only on the distance pushed and not on the time.

imagine there is a train in front of me and I'm on a train following behind pushing with a force of 100 Newtons for one minute. I can't imagine getting twice as tired if the train is moving a 40km/h instead of 20km/h.

Similarly, we will replace you with a spring compressed to provide 100 N force. As long as the distance between the train cars is constant then the 100 N will be constant and the energy in the spring will be constant.

As the trains move at 20 kph the spring will deliver a certain amount of mechanical power (work per unit time) to the front train. The rear train will deliver that same amount of power to the spring. The internal energy of the spring does not change and all energy that enters from the rear train leaves to the front train. Because the spring’s internal energy is constant the spring does not “get tired”.

As the trains move at 40 kph the situation is similar. Twice as much power (work/time) is indeed delivered to the front train from the spring and twice as much power is delivered to the spring from the rear train. The spring’s internal energy is again constant.

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It doesn't move at all, but I get tired.

The fact that you get tired is about you and not about the table. The work we are talking about is the work absorbed by/from the table. And there is no such work.

Our human bodies just happen to expend energy within our muscles in order to generate a force. If you instead tie and rubber band around the objects so a constant force pushes the table into the wall, then no energy is expended in that rubber band if the table doesn't move.

In short, it is not about what is pushing (which may or may not spend energy in order to generate the push) but about what is being pushed. That is how we define work.

imagine there is a train in front of me and I'm on a train following behind pushing with a force of 100 Newtons for one minute.

The displacement in the work formula is relative. If you stand on one moving train and push on another moving train, and both trains have the same speed and keep a constant distance, then the distance relative to your "ground" (the first train) remains zero and you do no work.

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  • $\begingroup$ I was assuming that both trains were actually accelerating $\endgroup$ – Casebash Mar 29 '20 at 13:37
  • $\begingroup$ @Casebash Why would that make a difference? If both accelerate equally, their relative distance still stays the same. Of the front train accelerates faster, again it is only the displacement as seen from the back train you should use to calculate any work you do. $\endgroup$ – Steeven Mar 29 '20 at 13:46
  • $\begingroup$ But isn't the first train accelerating in part/moving because you are pushing on it? So shouldn't you be performing work on it? $\endgroup$ – Casebash Mar 29 '20 at 13:55

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