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Suppose I have a book kept on the floor. I pick it up, and keep it on the table, at a constant velocity. This means I've to apply a force $mg\hat{y}$ to counteract the force of gravity. The work that I do to pick up the book is $mgh$. Similarly, gravity is also acting on the book with a force $-mg\hat{y}$. Hence, the work done by gravity is $-mgh$.

Thus, if we consider the book to be our system, the net force and by extension, the net work comes out to be $0$. Moreover, the work done by gravity is defined as the negative of the change in potential energy of the book-Earth system. Hence, we can say that the change in potential energy $\Delta U=mgh$.

Now I want to analyze the same situation by considering the book and the Earth to be our system.

The force that I apply, is now the external force, and the internal forces are the equal and opposite forces between the book and the Earth. Remember, in order to raise the book with a constant velocity, I have to apply force on both the book and the Earth in the opposite direction. This will ensure that the center of mass moves at a constant velocity.

Thus, as you can see, the net force on each of the objects is $0$, and so, the net work done must also be $0$, just like the previous case.

let us now calculate the net work done by each of the four forces $F_{us/book},F_{us/earth},F_{book/earth},F_{earth/book}$

Well, as I know, the work done by us, on the book is simply $F_{us/book}. h=mgh$

Similarly, the work done by the Earth, on the book, is given by $F_{earth/book}=-mgh$

Now, I'm confused about two things here. First of all, what is the work done by us, on the earth ? Since the Earth stays in place, I can say this work done is $0$. However, what is the work done by the book on the Earth ? The answer again, should be $0$, but isn't the Earth moving relative to the book ? From the book's perspective, the earth is moving downward, while the gravity is acting upward, towards the book. So, shouldn't the book do a work of $-F_{book/earth}.h=-mgh$ on the Earth ? I'm told this is not true, and I don't seem to understand why. Is there a special choice of origin that we stick to, throughout the entire analysis ? Else, if the Earth does $-mgh$ work on the book, then from the book's perspective, it is stationary, and the Earth is moving, so shouldn't it do $-mgh$ work on the Earth ?

But this would mean that the total work done by the internal forces is $-mgh-mgh=-2mgh$. However, we have established that we do no work on the Earth, so the total external work done by us is $mgh$. If we add up all of these, it doesn't come out to be $0$. It is clear that we are overcounting the work done by the book on the earth and viceversa, but I don't seem to understand why.

Finally, what exactly is the potential energy here ? Is it the negative of the work done of the earth on the book and the book on the earth ? Is the work done by the book on the Earth $0$ , since the Earth doesn't move ? But the Earth doesn't move with respect to us, who is picking up the book, with respect to the book the Earth does move.

Is the change in potential energy of a system the negative of the sum of work done by all the internal conservative forces in the system ? In this case there are two internal forces, and the work done by the earth on the book is $-mgh$ and the work done by the book on the earth is $0$(?). Then we can safely say that total potential energy of the system changes by $mgh$.

However, since the book moves up from the Earth, and the earth moves down, according to the book, why is the work done by the earth on the book non zero, while the work done by the book on the earth is zero. Is there a special reference line with which we measure the change in distance, and if so why ?

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  • $\begingroup$ "The force that I apply, is now the external force, and the internal forces are the equal and opposite forces between the book and the Earth". If you are now considering the book as the system, how can you say the equal and opposite forces between the book and the Earth are internal when the Earth is not part of the system? Aren't your force on the book and the force of the earth on the book both external forces on the system? $\endgroup$
    – Bob D
    Dec 6, 2021 at 17:11
  • $\begingroup$ Yeah, in the first case, I was considering the book as my system, and so both my force and earth's force is external. In the second case, I was considering both the book and the earth as a system as a whole, and myself to be the external agent, who must apply force on both of them to separate them. $\endgroup$
    – RayPalmer
    Dec 6, 2021 at 17:24
  • $\begingroup$ OK but then do you realize when you consider the book only as the system the book has no gravitational potential energy because that energy does not belong to the book?. It belongs to the Earth-book system. $\endgroup$
    – Bob D
    Dec 6, 2021 at 17:41
  • $\begingroup$ Duplicate, eg How does W=0 for the following equation and question on conservative forces? $\endgroup$
    – Farcher
    Dec 6, 2021 at 17:44
  • $\begingroup$ @BobD yes it does, and I wrote so. My confusion is when I consider both the earth and the book to be the system. $\endgroup$
    – RayPalmer
    Dec 6, 2021 at 17:50

2 Answers 2

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Thus, if we consider the book to be our system, the net force and by extension, the net work comes out to be 0. Moreover, the work done by gravity is defined as the negative of the change in potential energy of the book-Earth system. Hence, we can say that the change in potential energy Δ𝑈=𝑚𝑔ℎ

First, in this question let’s completely discard the concept of “net work” which is simply the net force times the displacement of the center of mass. It is typically not useful and is often even counter-productive when looking at conservation of energy.

Now, the key to such problems is to carefully define the system. In this case, the system is the book. Note that in $mgh$ the $h$ is not a property of the book. It is a relationship between the book and the earth. Since it is not part of the book the PE $\Delta U$ does not belong to the book.

So the correct energy analysis for the book as the system is that the contact force does positive work on the book and gravity does negative work on the book, so the system consisting only of the book has no change in energy.

Usually we consider the earth as being so massive that it doesn’t move. In that case, treating the potential energy as “belonging” to the book does not cause any problems. But it is technically wrong and can cause confusion like yours.

Now I want to analyze the same situation by considering the book and the Earth to be our system.

Now, in this case $h$ is a property of the system. It is a “degree of freedom” that can be used to change the configuration of the system and store energy. For energy, we are only interested in the external forces. The internal forces just shuffle energy around inside the system, only external forces change the energy of the system.

The work done by the contact force on the book is positive, thus increasing the energy of the system by $\Delta U= mgh$. However, since the Earth doesn’t move, the contact force on the earth does no work. So the external work done is equal to the change in energy, so energy is conserved.

Now, if you do choose to analyze the work done by the internal forces, you will find that it is negative. Since that is an external force, that is not energy lost, it is just energy that is converted to internal potential energy rather than kinetic energy.

From the book's perspective, the earth is moving downward, while the gravity is acting upward, towards the book. So, shouldn't the book do a work of −𝐹𝑏𝑜𝑜𝑘/𝑒𝑎𝑟𝑡ℎ.ℎ=−𝑚𝑔ℎ on the Earth ? I'm told this is not true, and I don't seem to understand why. Is there a special choice of origin that we stick to, throughout the entire analysis ?

You can pick any inertial reference frame. There is no special inertial frame, but you do need to stick to that frame throughout the analysis. You cannot calculate part of the energy in one frame and the rest of the energy in another frame. This is because the energy will be different in different frames (but it will be conserved in all frames).

However, even though you can, in principle, use any inertial frame, the math is much simpler in the frame where Earth is at rest. If you want to do the analysis in a frame where the Earth moves then you must consider the conservation of momentum, and you must consider the earth to have a finite mass.

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  • $\begingroup$ The line that I need to stick to a particular frame, and can't change it in the middle of the calculation, sounds rather obvious now, and a bit uncomfortable to me, at the same time. It begs the question, who exactly is measuring the work done by a certain force. Imagine, two bodies attracting each other with gravity $F$. I want to know the work done by both of these bodies on each other. If I want to find the work done by a certain body, do I have to consider it at rest and the other moving ? $\endgroup$
    – RayPalmer
    Dec 6, 2021 at 20:49
  • $\begingroup$ If I now want to find the work done by the other body, do I change my frame to be at rest with the other body, or do I keep my current position and notice the other body as it is ? $\endgroup$
    – RayPalmer
    Dec 6, 2021 at 20:50
  • $\begingroup$ So, the work done by an object exerting some force, depends on the reference frame of the observer and not the rest frame of the object itself, right ? That is what I did wrong ? $\endgroup$
    – RayPalmer
    Dec 6, 2021 at 20:51
  • $\begingroup$ @RayPalmer yes, that last comment is right. The rest frame of the object is not important. It needs to be any inertial frame, and you need to be consistent and not change frames part way through. That is all $\endgroup$
    – Dale
    Dec 6, 2021 at 21:12
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I think that it all boils down to the following statement,

So, shouldn't the book do a work of $−F_{\rm book/earth}.h=−mgh$ on the Earth?

which is incorrect.

Tie the reference frame to the centre of mass of the book and Earth system and note how far the Earth (and any forces acting on the Earth) moves when the book (and any forces acting on the book) moves a distance $h$. It is not $h$ but something which is very much smaller than $h$ because the mass of the Earth is so much greater than that of the book.

Now transport the frame of reference so that it is tied to the book which is moving a constant velocity relative to the centre of mass of the book and Earth system and note that the distance any force acting on the Earth moves does not change from that measured in a frame tied to the centre of mass.

Just because the Earth moves a distance $h^+$ away from the book does not mean that a force acting on the Earth moves that distance.

...

Imagine me sitting on a non-frictionless table and observing you pushing a block across the table at constant velocity.
The work you have done is the force that you applied times the distance moved by the force relative to the table.

You repeat the pushing of the block whist I walk past you and the table at constant velocity. I observe you pushing with the same force and the force moves the same distance relative to the table so I conclude that you have done the same amount of work as before.

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  • $\begingroup$ If I tie my frame of reference to the book, then in this frame of reference the book is at rest and the earth must go away in the opposite direction right ? If we are setting the earth to be the reference frame, than the book moves $h$ away. However, if we are with the book, in the rest frame of the book, then isn't it the earth which moves $h$ away ? $\endgroup$
    – RayPalmer
    Dec 7, 2021 at 0:06
  • $\begingroup$ See the addition to my answer. $\endgroup$
    – Farcher
    Dec 7, 2021 at 8:00

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