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I am confused about changes in gravitational potential energy, net external force, and work-energy theorem.

For a single particle system, I understand that a net external force results in a change in kinetic energy and thus work is done on the system (particle).

What I am confused about is the work done on a system and the increase in gravitational potential energy.

If I define the system to include a book + Earth, (where I am external to the system) and I lift the book at a constant speed, then the net force on the system is not zero - correct? However, the net force on the book is zero. So, by Newton's Second Law, the system should accelerate and there should be a change in kinetic energy and thus work will be done on the system.

However, most books state that there is no change in kinetic energy but a change in potential energy instead. How can I apply an external force to the system increasing the energy of the system but the system not accelerate? If the amount of kinetic energy remains constant then is work done on the system?

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  • $\begingroup$ Although the applied force on the book and the gravitational force add to zero, the book is part of a system that includes the Earth. So if I am applying a force to the book which is part of the system then isn't the system affected by the force - not just the book? So shouldn't the system accelerate? $\endgroup$ – ConfusedinPhysics Jan 31 '18 at 22:19
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Work does not always have to involve changes in kinetic energy. I believe your confusion stems from a misapplication of the Work-Energy Theorem. Work is only equal to change in kinetic energy if no potential fields are present (i.e. the Work-Energy Theorem only applies for free, rigid bodies). Gravity is one of these fields, so it is possible to do work without altering kinetic energy.

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  • $\begingroup$ If the net force on the system is zero how is the energy of the system increasing. I thought that if the net force on the system is zero then there is no change in the amount of energy in the system. $\endgroup$ – ConfusedinPhysics Jan 31 '18 at 22:21
  • $\begingroup$ @ConfusedinPhysics Edited to maybe address the source of the problem better. $\endgroup$ – probably_someone Jan 31 '18 at 22:24
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Imagine the Earth and the book as your system.

You need to apply an external force on the book and an equal and opposite external force on the Earth to "lift" the book if you do not want to accelerate the centre of mass of your system.

Taking the the book as a system it has two equal and opposite forces on it, the gravitational attraction of the book by the Earth and the applied external force, so the net work done on the book is zero and the kinetic energy of the book does not change.
The same is true of taking the Earth as a system and it too has no work done on it so the kinetic energy of the Earth does not change.

Gravitational potential energy is something the book and Earth system possess jointly and the external forces do work on the book & Earth system increasing the separation between them thus increasing the gravitational potential energy of the book and Earth system.

Evaluation of the work done by the external forces is usually restricted to the work done by the external force on the book because the distance the book moves is much, much greater than the distance the Earth moves as the Earth is so more massive than the book.

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  • $\begingroup$ So to make sure I understand this: the net force on the book+Earth system is zero but external forces are doing work on the system increasing the potential energy?? If the net external force on the system is zero, I do not understand how the total energy of the Earth+book system is increased as their distance of separation increases. This is what is really not making sense. $\endgroup$ – ConfusedinPhysics Feb 2 '18 at 15:42
  • $\begingroup$ @ConfusedinPhysics You, who is not part of the system, push the book up (external force) and also push the Earth down (external force) thus increasing the separation between the book and the Earth. $\endgroup$ – Farcher Feb 2 '18 at 16:22
  • $\begingroup$ OK - but just to clarify, the net force on the system is zero. However, since the book is moved a greater distance than the Earth, the work done by me on the book is not equal to the work done by me on the Earth, and thus the energy added to the system is equal to the product of the weight of the book and the distance over which it is moved. So work can be done on a system, increasing only the potential energy while the kinetic energy remains unchanged. $\endgroup$ – ConfusedinPhysics Feb 2 '18 at 16:52
  • $\begingroup$ Yes and the work done on the Earth is neglected. $\endgroup$ – Farcher Feb 2 '18 at 18:51
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Imagine you apply a force to your book upwards equal to it's weight just a moment after it dropped from the shelf and already has a small velocity downward.

It will keep falling but with no acceleration. So even after applying the force the book is loosing its potential energy. The force has just cancelled the gravity and removed the acceleration.

In example you have when you apply a force upward, with no horizontal components three things happen. - F < W

The book will fall but with less than g acceleration. And it's kinetic energy will increase due to Earth gravity work.

  • F = W

The book will keep it's velocity or stay still if it was not moving. No change in kinetic energy.

  • F > W

The book will climb with acceleration with initial speed of what it has at the time of applying the force. And it's kinetic energy will increase.

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