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Imagine a book that we lift it with a force that is exactly equal to the force of gravity so the forces cancel out and the book moves with a constant velocity.

Consider the situation after the book has been lifted, and it has come to rest once again. According to the work and kinetic energy laws $\Delta W = \Delta K.$ This seems to hold here, since both are zero. That's okay, but where did the increase in the potential energy come from? Is energy not conserved?

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What do you think happened to the person or object that did the lifting? Did it expend any energy? –  Danu Aug 6 at 16:51
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Also, a force equal and opposite that of gravity cannot cause a constant velocity if the book starts from a different velocity than the final constant velocity. A slight extra push and slow down is required. –  Andres Salas Aug 6 at 16:54
    
you are true Andres Salas but ignore that ,it does not matter in rest or in motion with the question –  Mohamed Osama Aug 7 at 15:45

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You say:

Imagine a book that we lift it with a force that is exactly equal to the force of gravity so the forces cancel out and the book moves with a constant velocity.

so I'm guessing your reasoning is that the net force on the book is zero so the amount of work done on the book is zero. And you are absolutely correct - no work is done on the book and its internal energy doesn't change.

However work is clearly being done because if I'm lifting the book I'm exerting a force of $mg$ over the distance $d$ the book moves, so I do an amount of work $W= mgd$. The question is where this work is going.

The answer is that the work is being done on the Earth-book gravitational system. The work I do is being done on this system and is raising its potential energy. To make this clearer imagine the book is attached to a long spring. If I lift the book I'm doing work, but the work isn't being done on the book but instead the work is being done on the spring. As a result the length of the spring changes and therefore its potential energy changes. The change in the spring potential energy is the same as the work I put in.

Response to comment:

To use the spring analogy again, the source of the work is that your body is also acting like a compressed spring that expands and does work as the book is lifted. When you push upwards with your hand, your feet are pushing downwards on the ground so as you lift the book you are doing work to increase the distance between your hand and your feet.

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But John the problem is you first said " so the amount of work done on the book is zero. And you are absolutely correct - no work is done on the book " but then you said " However work is clearly being done because if I'm lifting the book I'm exerting a force of mg over the distance d the book moves, so I do an amount of work W=mgd. The question is where this work is going." the question is the work by my hand is cancelled by the work of gravity , so why there is an increase in Energy although the work done as a total ( summation because work is scalar ) = zero –  Mohamed Osama Aug 6 at 22:39
    
@MohamedOsama: I've edited my answer to respond to your comment –  John Rennie Aug 7 at 9:56

Let's say the book starts and stops from rest, as I believe you are assuming. The motion within this interval is unimportant, as you'll see.

The increase in gravitational potential energy of the Earth-book system came from your body. You did positive work on the system since your hand force and displacement are in the same direction, resulting in an increase in energy, as inferred from the work-energy theorem $W_\text{net, ext}=\Delta K + \Delta U$.

The Earth is internal to the system, so you don't include its work in $W_\text{net, ext}$. Instead, the effect of the Earth is built into $\Delta U$ in this case.

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the theorem you wrote is derived from the kinetic energy theorem and work , no meaning to use it here and i don't fully understand the first concept –  Mohamed Osama Aug 6 at 22:42

Imagine a book that we lift it with a force that is exactly equal to the force of gravity so the forces cancel out

Ok, so sum of the forces is 0 and the acceleration is zero.

and the book moves with a constant velocity.

Spooky. Was the book moving initially?

...after the book has been lifted, and it has come to rest once again. According to the work and kinetic energy laws ΔW=ΔK This seems to hold here, since both are zero.

Presumably not then; sounds like magic.

Energy is always conserved (globally). If there was a change in height, a force was applied to the book and energy was added to the book-Earth system. If no net force was applied to the book, it didn't change height and no gravitational potential was added to the system.

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If you lift the book with a force that is exactly equal (but opposite) to the force of gravity acting upon it the book won't go anywhere. After all, that's exactly what a table does when you leave the book on it.

To get to book going you need to lift with a force greater in magnitude than that of gravity, resulting in a net force and an associated acceleration. Thus the book gains velocity and hence kinetic energy, which is converted into potential energy as the book rises.

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Yes, but you could put in a very short extra force at the beginning then let the book move at constant speed for an arbitrarily long time. So the PE gain can be much higher than the energy needed for the initial acceleration. –  John Rennie Aug 6 at 17:02
    
@JohnRennie, Yes, my answer addresses the slightly different pre-edit question where we could quibble about what it means to "move with inertia". It is always good to think about the dynamics of the problem, but yes, for a lift over an arbitrary distance at constant velocity that is pretty small potatoes. –  Dave Aug 6 at 17:11

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