# Conceptually understanding change in potential energy with 0 net work

• B
• DrBanana
DrBanana
Suppose somehow an object is moving upwards with a speed ##v##, at this point I start applying a force ##F## that is equal to its weight, so the net force on the object is zero. So it will continue moving upwards with its initial speed. Suppose after the height difference is ##h##, I stop applying the force. At this moment, the object has its initial kinetic energy, but now it also has gained some potential energy. If the net force on the object was zero, the net work must also have been zero. But that should have meant no energy flowed to the object, so how did it gain potential energy? I know this question has been asked a lot of times.

The usual explanation goes like this (Work applied by me, potential energy, kinetic energy, work done by field):
##W_{applied}=\Delta U + \Delta K##
##F*h=-W_{field}+ \Delta K##
##F*h=-(-F*h) + \Delta K##
##\Delta K = 0##
Therefore ##W_{applied}= \Delta U##

I know my confusion arises from the fact that I've only been considering the box, but if potential energy is concerned, I should consider the box-earth system. In that case, the work done by the force applied by me comes externally, and gets stored in the system as potential energy. In this case, equality of energy is achieved (that is, energy in = energy supplied). However, I'd like to know the 'internals' of this, as I'm still having a hard time digesting it. I apply a force on the box, but the work done by that force gets treated as force done on the system, why? Does gravity do work on the system, or on the box? If the work done by my force on the system gets stored as potential energy, where does the work done by gravity 'go'?

Edit: I think I've boiled my problem down to this: Is the potential energy gained by the system due to the work done by my force, or due to the negative work done by the gravitational field? How can both be true?

Last edited:
DrBanana said:
If the net force on the object was zero, the net work must also have been zero. But that should have meant no energy flowed to the object
This actually is not the case. The net work being zero just means that the KE did not change. It does not imply that any other forms of energy are unchanged.

Dale said:
This actually is not the case. The net work being zero just means that the KE did not change. It does not imply that any other forms of energy are unchanged.
I was sort of suspecting this. So one can say, the energy getting stored in a system is equal to the energy flowing in; some of this energy might manifest itself in kinetic energy of a part of that system, if it can be shown that net work was done on that part of that system, and the rest is other forms of energy (heat, gravitational/electrical potential etc.)?

Draw a dotted line (wherever you like) around the system. Don't move the line halfway through the analysis.

It is more that net work is different from total work.

The net work is the net force times the displacement of the center of mass. That is equal to the change in KE.

The work of an individual force is the individual force times the displacement of the material at the point of application of the force. It is equal to the transfer of energy from whatever is on the other side of that force.

The total work is the sum of the individual works. It represents the total change of energy.

In your scenario the net force is zero so the net work is zero so the change in KE is zero.

The applied force is ##F## and the displacement of the material at the point of application is ##h##. So the applied work is ##W=Fh##. So energy is going in to the object from the applied force.

The gravitational force is ##-F## and the displacement of the material at the point of application is ##h##. So the work from gravity is ##-Fh=-W##. So energy is going out of the object to the gravitational field, increasing the PE of the gravitational field.

The total work is the sum of the individual works ##W+-W=0##. All of the energy that enters the object from the applied force leaves the object and becomes gravitational PE.

Last edited:
DrBanana
DrBanana said:
Edit: I think I've boiled my problem down to this: Is the potential energy gained by the system due to the work done by my force, or due to the negative work done by the gravitational field? How can both be true?
I assume that by system you mean the two component system of Earth + object. In that case, the total energy of the system is mechanical, i.e. kinetic ##K## which is the sum of the kinetic energies of the two components and potential ##U## which depends on the spatial configuration of the two components. Total energy conservation for the two-component system says $$\Delta K+\Delta U=W_h$$ where ##W_h## is the work done by the external non-conservative force of the hand, ##~W_h=F~h.## The work done by gravity does not enter the picture because it is not external work that crosses the boundary of the system, it just converts kinetic to potential energy internally to the system. In your case the change in kinetic energy is zero and the change in potential energy positive, ##\Delta U = +mgh##. So the total energy conservation equation becomes $$0+mgh=F~h\implies F=mg$$which is expected because the net force must be zero.

To understand your question better, consider the case when the system is only the object. In this case, two external forces act on the system and do work that crosses the system boundary, gravity and hand. Total energy conservation says \begin{align} &\Delta K=W_g+W_h\nonumber \\ & 0=-mgh+F~h\implies F=mg.\nonumber \end{align}The result is the same as it should be. However, note that is the first case we considered the change in potential energy only and that in the second case we considered the work done by gravity only as potential energy is meaningless when you have a single-component system. So the answer to your question "How can both be true?" is "Both cannot be true" because each is true depending on the system you choose. I think this is what @Dullard meant by
Dullard said:
Draw a dotted line (wherever you like) around the system. Don't move the line halfway through the analysis.

DrBanana
kuruman said:
I assume that by system you mean the two component system of Earth + object. In that case, the total energy of the system is mechanical, i.e. kinetic ##K## which is the sum of the kinetic energies of the two components and potential ##U## which depends on the spatial configuration of the two components. Total energy conservation for the two-component system says $$\Delta K+\Delta U=W_h$$ where ##W_h## is the work done by the external non-conservative force of the hand, ##~W_h=F~h.## The work done by gravity does not enter the picture because it is not external work that crosses the boundary of the system, it just converts kinetic to potential energy internally to the system. In your case the change in kinetic energy is zero and the change in potential energy positive, ##\Delta U = +mgh##. So the total energy conservation equation becomes $$0+mgh=F~h\implies F=mg$$which is expected because the net force must be zero.

To understand your question better, consider the case when the system is only the object. In this case, two external forces act on the system and do work that crosses the system boundary, gravity and hand. Total energy conservation says \begin{align} &\Delta K=W_g+W_h\nonumber \\ & 0=-mgh+F~h\implies F=mg.\nonumber \end{align}The result is the same as it should be. However, note that is the first case we considered the change in potential energy only and that in the second case we considered the work done by gravity only as potential energy is meaningless when you have a single-component system. So the answer to your question "How can both be true?" is "Both cannot be true" because each is true depending on the system you choose. I think this is what @Dullard meant by
First of all I thank both you and @Dale; the answers are really satisfying to read.

My takeaway from yours (in the first case) is that while gravity is a force which is doing work, it is a force of different mathematical nature, and every time we encounter it, we should not think of the work done by it, rather the change in potential energy it causes.

DrBanana said:
My takeaway from yours (in the first case) is that while gravity is a force which is doing work, it is a force of different mathematical nature, and every time we encounter it, we should not think of the work done by it, rather the change in potential energy it causes.
You missed my point. First of all, "doing work" is an incomplete and ambiguous statement without prepositions. You need to specify what force is doing work and on what system. The correct form is "Force F does work on system S" or "Work is done by force F on system S." Whichever way you choose to say it, the force that is doing the work is external to the system on which the work is done.

The takeaway message is if you choose your system to include an action-reaction pair, then you cannot think of one item in the pair as doing work on the other. You think of potential energy change that is related to the spatial configuration of the two components. However if you choose only one of the two items as your system, then you can find the work done on it by the other item but you have to give up potential energy change because potential energy needs at least two components in a system.

The mathematical "nature" of gravity is the same as that of any other force. As I showed in post #6, whether you use change in potential energy or work done by gravity is contingent upon what you have chosen to be your system when you write $$\Delta K = \dots$$If your choice for system is just the object, then you put ##\int \mathbf F \cdot d\mathbf s## on the right hand side and call it "work done by the external force ##~\mathbf F.##"

If your choice for system is the object and the Earth, then you put ##~\text{-}\int \mathbf F \cdot d\mathbf s## on the left hand side with ##\Delta K## and call it "change in potential energy ##\Delta U.##"

kuruman said:
if you choose your system to include an action-reaction pair, then you cannot think of one item in the pair as doing work on the other. You think of potential energy change that is related to the spatial configuration of the two components.
I know this wasn't the topic of the post, but what if this action-reaction pair is not a conservative force?

Good question. Part of the mechanical energy is converted into heat, vibrations, sound, etc. more generally internal energy of the system.

In this example, if you consider yourself raising the object as part of the system, then there is negative change in biological energy of the system. You burn calories to raise the object and they need to be replaced by eating.

• Classical Physics
Replies
41
Views
684
• Classical Physics
Replies
6
Views
353
• Classical Physics
Replies
6
Views
888
• Classical Physics
Replies
29
Views
2K
• Classical Physics
Replies
46
Views
2K
• Classical Physics
Replies
31
Views
2K
• Classical Physics
Replies
21
Views
1K
• Classical Physics
Replies
11
Views
2K
• Classical Physics
Replies
2
Views
925
• Classical Physics
Replies
4
Views
1K