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I was reading morin's intro to mechanics book, and i reached the point where he was discussing work potential energy equivalence, which i find very confusing right now... basically we are analysing me lifting up a book at constant speed, and there are some cases for the system we can take...

  1. system = book; this is clear to me, I'm doing positive work =$mgh$ and earth is doing negative work $-mgh$

2.system = book + earth; this is also clear, I'm doing positive external work $mgh$ and the potential energy of system increases by $mgh$; all good....

But it is the third case where I have the confusion... This is what morin says...enter image description here

But now my confusion is this: I know there's no net force acting, and everything is constant speed, so the net change in potential energy must be $0$... but how can we analyse the potential energy of a system piece wise; cos the potential energy is a function of the entire system, and we can't like exactly say which part of system contributes this much potential energy, cos its actually a function of the whole system... but even then, if i ignore that, ok the potential energy of earth block system is increased my $mgh$ ; which is simple to analyse, cos the work done by the internal force is $-mg \times h$; so $mgh$ is the increase in potential energy, but how can I calculate the change in my potential energy, how can i say that is $-mgh$? (like obviously i can see that the sum is zero, so it must be that, but morin used the $\iff$ (iff) sign, so it seems there is a way to say clearly that my potential energy decreased my $mgh$... i don't get how to do that at all, like if we go by the definition of $\Delta V = -$Work done by internal force; i don't see how this works at all, like maybe internal forces on me are gravity and blocks normal reaction, but how do these get you $-mgh$...

I am really trying to understand energy well, and would really appreciate any help in doing so... Thanks!

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When they say "your potential energy" they're not just talking about gravitational potential energy. They're also considering the chemical potential energy stored in various chemicals (ATP, IIRC) in your body that needs to be converted to g.p.e. (of the book) when you lift the book.

how can I calculate the change in my potential energy, how can i say that is −$mgh$?

It must be that because that's how much energy was needed to lift the book.

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  • $\begingroup$ Strictly speaking the chemical PE expended must be greater than $mgh$ as the muscles are far from 100% efficient. The rest is lost as heat $\endgroup$
    – RC_23
    Commented May 12 at 17:45
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    $\begingroup$ @RC_23, yes, and the quoted text in the OP also explains that in the next paragraphs. $\endgroup$
    – The Photon
    Commented May 12 at 17:46
  • $\begingroup$ Thank you so much for this answer :) $\endgroup$
    – Aditya_math
    Commented May 13 at 11:37
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He's blurring the line between mechanical energy and other types of energy, in particular here, chemical energy. The energy stored in chemical bonds decreases by at least $mgh$ to conserve energy as you point out. But there's no way that I know of to calculate that number from the microscopic chemistry of the human body. He also ignores the part of chemical energy that becomes thermal energy, which is why I said at least $mgh*. I think he didn't help anyone by blurring that line.

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  • $\begingroup$ Thank you so much, this clarified my main doubt... cos i was worried that there was some simple way of seeing the energy was -mgh that i wasn't seeing, thanks! $\endgroup$
    – Aditya_math
    Commented May 13 at 11:36
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In general you are right: potential energy is a function of the entire system, not a single object. However, in many cases, that potential energy is separable. In many cases, especially with discrete objects, the potential energy function of the system can be written as a sum of smaller functions which only depend on the state of one object, such as the mechanical energy stored in a spring which depends soley on how compressed it is.

Sometimes it doesn't work so well. In the case of gravity, the potential energy is a function of the positions of two objects, not one. For celestial mechanics, this is a big deal. But in simple cases with you and a book, most of the sources of energy actually separate out quite nicely. And when that happens it becomes intuitive to talk about the energy of an object, not just the energy of the system.

As for the potential energy you lose, I think the important lesson to take from the scene is one of conservation of energy. If you see a system where energy is not being conserved, this is a good sign that you should look for sources of energy that you overlooked. In this case, it's the chemical energy in the body.

The extra rambling about efficiency points out that you can't always calculate such energy expenditures. Sometimes there's enough odd losses, like thermal energy, which are very hard to calculate. But it's important to recognize they're there. You didn't get energy for free.

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  • $\begingroup$ Thank you so much, The first parahraph cleared a pretty big doubt of mine... thanks :) $\endgroup$
    – Aditya_math
    Commented May 13 at 11:37

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