B How does an object gain potential energy?

[adjurovich]

When we are acting with some force $F = mg$ on an object of mass $m$ vertically in the positive $y$ direction, we are doing work that is equal to the work done by gravity on the same object, but of course opposite in sign. It means that net work is equal to zero, hence no change in kinetic energy according to the work energy theorem. However, does the change in potential energy only result from object’s change in elevation independent of external work - since external work + work done by conservative force (gravity) sum up to zero in this case?

The object will obviously be moving uniformly in the $y$ direction and will change elevation so I understand that it’s very logical to conclude that there must be some potential energy that body will gain, but from where?

 

[kuruman]

. . . it’s very logical to conclude that there must be some potential energy that body will gain, but from where?

From here

When we are acting with some force $F=mg$ . . .

"We" are doing work $W_{we}=+mgh$, gravity is doing work $W_g=-mgh$, the total work done on the object is zero, the change in kinetic energy is zero and the work energy theorem is satisfied. The negative of the work done by the conservative force of gravity is the change in potential energy.

 

[adjurovich]

From here

"We" are doing work $W_{we}=+mgh$, gravity is doing work $W_g=-mgh$, the total work done on the object is zero, the change in kinetic energy is zero and the work energy theorem is satisfied. The negative of the work done by the conservative force of gravity is the change in potential energy.

What confuses me is that fact that external force needs to bring energy to the system. If 50 joules of energy is brought into the system and 50 joules taken by the force of gravity (weight), how can we still have 50 joules of energy in the system? (Potential energy)

 

[kuruman]

You seem to be unclear about what your system is. It could be one of
1. Just the object
2. The object and the Earth

Specifically, if your system is just the object, it cannot have potential energy no matter what you read in textbooks or on the web. Potential energy needs at least two interacting objects as a system and depends on the relative position of the system's components.

When the system is only the object, there are two external forces doing work, gravity and we. As we have seen, there is no net joules transfer in the case.

When the system is the Earth and the object, "we" is the only external force and we act on the component "object". "We" do work that changes the relative position of the system components by moving them farther apart. The work that we, the external force, do on the objcet appears as increase of the potential energy of the two-component system. In this case there is transfer of joules from the external force into the system: The work that we do $mgh$ is equal to the increase $mgh$ of the potential energy of the system.

See how it works?

 

[adjurovich]

You seem to be unclear about what your system is. It could be one of
1. Just the object
2. The object and the Earth

Specifically, if your system is just the object, it cannot have potential energy no matter what you read in textbooks or on the web. Potential energy needs at least two interacting objects as a system and depends on the relative position of the system's components.

When the system is only the object, there are two external forces doing work, gravity and we. As we have seen, there is no net joules transfer in the case.

When the system is the Earth and the object, "we" is the only external force and we act on the component "object". "We" do work that changes the relative position of the system components by moving them farther apart. The work that we, the external force, do on the objcet appears as increase of the potential energy of the two-component system. In this case there is transfer of joules from the external force into the system: The work that we do $mgh$ is equal to the increase $mgh$ of the potential energy of the system.

See how it works?

I understand the first part, but I just wanted to get straight to the point.

But basically what you’re trying to say that even if there is no net work, the object can gain the potential energy?

By the way, should we actually call gravity the external force? As I recall it’s a conservative force?

 

[kuruman]

But basically what you’re trying to say that even if there is no net work, the object can gain the potential energy?

I never said that. I said that the object by itself CANNOT have potential energy,

Specifically, if your system is just the object, it cannot have potential energy

so it cannot gain or lose potential energy. It can only have kinetic energy. It's the two-component system of (Earth + object) that can have potential energy which depends on the distance of the object from the center of the Earth.

By the way, should we actually call gravity the external force? As I recall it’s a conservative force?

You recall correctly, but "external" is not antithetical to "conservative" when referring to forces. Conservative forces can be external or internal and the same goes for non-conservative forces.

In this case, when your system is only the object, all forces acting on it are external because they are generated by entities that are not part of the system. The Earth is not part of the system so the conservative force it exerts on the system is external. Because it is external, joules may cross the boundary of the system as work is done by gravity which results in a change of the kinetic energy of the object.

If you choose the object + Earth two-component system, the conservative force of gravity is internal to the system. One part of the system, the Earth, may do work on another part, the object, and increase the kinetic energy of the object. However, no joules cross the boundary of the Earth+object system. The amount of joules of kinetic energy gained by the object is accompanied by a reduction by the same amount of the gravitational potential energy. That's what mechanical energy conservation is about.

 

[Standard]

You can of course have INTERNAL potential energy.

Then term "Potential" doesn't refer specifically to "gravity".
It refers to the "potentiality", the "possibility" of an energy to be transmitted for real.
Potential and real are the two words that distinguish the two concepts.

Potential energy is classically divided into 2 categories :
1. Macroscopic : The energy is related to some EXTERNAL refrence frame.
2. Microscopic : The energy is related to the INTERNAL reference frame, so you dont have any external reference. In this case, we dont have setted up some reference where we can say " this is the null point", like we can do it with macroscopic energy (we did this improperly, but historicaly it is what we do with macroscopic energy, and it is how it works and because it is usefull we stay with that actually) .
So the ABSOLUTE value of microscopic energy is not known (unless we could start counting from the null point of the "dirac sea" but we can't really do the calculation, we can only produce some theorical estimation) , and we can only know the change of the value of the so called Internal energy U.
https://en.wikipedia.org/wiki/Internal_energy

Per example.
If you climb on a hill on Earth, using or not using some motorisation, you ADD potential energy to the Macroscopic system.
This is because you transfert (internal) potential chemical energy (your body, what you eat, coal or nuclear energy (E=m*c*c +...) coming from the Sun (a source of internal potentential energy) to the potential (external) GRAVITATIONAL energy.

 

[adjurovich]

I never said that. I said that the object by itself CANNOT have potential energy,

so it cannot gain or lose potential energy. It can only have kinetic energy. It's the two-component system of (Earth + object) that can have potential energy which depends on the distance of the object from the center of the Earth.

You recall correctly, but "external" is not antithetical to "conservative" when referring to forces. Conservative forces can be external or internal and the same goes for non-conservative forces.

In this case, when your system is only the object, all forces acting on it are external because they are generated by entities that are not part of the system. The Earth is not part of the system so the conservative force it exerts on the system is external. Because it is external, joules may cross the boundary of the system as work is done by gravity which results in a change of the kinetic energy of the object.

If you choose the object + Earth two-component system, the conservative force of gravity is internal to the system. One part of the system, the Earth, may do work on another part, the object, and increase the kinetic energy of the object. However, no joules cross the boundary of the Earth+object system. The amount of joules of kinetic energy gained by the object is accompanied by a reduction by the same amount of the gravitational potential energy. That's what mechanical energy conservation is about.

I am really thankful for such profound analysis, but I think that’s not actually what I asked. My question was relatively simple: it there is no net work, what adds potential energy to the body? Is it just the indirect result of some work (net work is zero, but work is still done to balance the gravity)?

 

[adjurovich]

Per example.
If you climb on a hill on Earth, using or not using some motorisation, you ADD potential energy to the Macroscopic system.

So you are saying that the potential energy is only something object possesses due to elevation but is not the direct result of external work?

When I say direct result, I mean that no true work is done one the body. If $F_{ext}$ is doing work and $F_g = F_{ext}$ is doing work in the opposite direction, energy that is added basically gets taken away? So my real question is: does it really matter that energy is taken away? Is potential energy actually independent of the way object has reached its position, but is only concerned about kinetic energy (velocity) the object could gain if it was left to free fall from the height in the case of gravitational potential energy - I think of it as imaginary/energy that could turn into useful energy that will do work (kinetic)?

 

[Ibix]

it there is no net work, what adds potential energy to the body?

Nothing does. Potential energy does not belong to the body. That was the whole point of @kuruman's post.

 

[Standard]

So you are saying that the potential energy is only something object possesses due to elevation but is not the direct result of external work?

I say potential energy is some energy that is available for further work or for further energy transformation.
The elevation, so gravitational potential energy, is only one of the possibility to have potential energy available.

When I say direct result, I mean that no true work is done on the body. If $F_{ext}$ is doing work and $F_g = F_{ext}$ is doing work in the opposite direction, energy that is added basically gets taken away?

If you add energy from the outside of your system so as to add potential gravitational energy to your system, there is no reason this energy will fade away, even if you are talking about gravitational potential energy.
What you mean is : the thing will fall and finish onto earth, so the gravitational energy will attain 0, but doing so, when the thing reach earth, the thing stop falling because it gives its energy (cinetic energy in this case) to the earth. The earth is pushed back (organised kinetik energy) a little (very very little), some heat is produced (disorganised kinetic energy) and some molecules are broken down (internal energy is take away).
An the heat produces electromagnetic waves that send the energy into space (until you close your system energy will be lost )

So my real question is: does it really matter that energy is taken away? Is potential energy actually independent of the way object has reached its position, but is only concerned about kinetic energy (velocity) the object could gain if it was left to free fall from the height in the case of gravitational potential energy - I think of it as imaginary/energy that could turn into useful energy that will do work (kinetic)?

Energy alone is always "imaginary".
Motion alone is also "imaginary".
Voltage alone is "imaginary"
Etc.
There is no meaning of speaking about energy, motion or whatever (there are exceptions but not many) unless you define some frame of reference.
Therefore you start with some situation S1 and you compare it with the situation S2.
You can not really say how many energy you got with situation S1 or with situation S2.
The only thing you care about is the difference you can observe (some differences can't be observed so you dont care about) between the two situations.
Therefore we can per exemple talk about potential energy , the energy (so the difference) we can retrieve or loose when we observe how energy is transformed from one form to another.
The sum of the energy in a closed systeme is CONSTANT, not null (we dont know the value and we dont need to know the value to do common calculations)

In a closed system : Total of energies of all types in situation S1 = Total of energies of all types in situation S1
Thats all.

 

[adjurovich]

I say potential energy is some energy that is available for further work or for further energy transformation.
The elevation, so gravitational potential energy, is only one of the possibility to have potential energy available.



If you add energy from the outside of your system so as to add potential gravitational energy to your system, there is no reason this energy will fade away, even if you are talking about gravitational potential energy.
What you mean is : the thing will fall and finish onto earth, so the gravitational energy will attain 0, but doing so, when the thing reach earth, the thing stop falling because it gives its energy (cinetic energy in this case) to the earth. The earth is pushed back (organised kinetik energy) a little (very very little), some heat is produced (disorganised kinetic energy) and some molecules are broken down (internal energy is take away).
An the heat produces electromagnetic waves that send the energy into space (until you close your system energy will be lost )



Energy alone is always "imaginary".
Motion alone is also "imaginary".
Voltage alone is "imaginary"
Etc.
There is no meaning of speaking about energy, motion or whatever (there are exceptions but not many) unless you define some frame of reference.
Therefore you start with some situation S1 and you compare it with the situation S2.
You can not really say how many energy you got with situation S1 or with situation S2.
The only thing you care about is the difference you can observe (some differences can't be observed so you dont care about) between the two situations.
Therefore we can per exemple talk about potential energy , the energy (so the difference) we can retrieve or loose when we observe how energy is transformed from one form to another.
The sum of the energy in a closed systeme is CONSTANT, not null (we dont know the value and we dont need to know the value to do common calculations)

In a closed system : Total of energies of all types in situation S1 = Total of energies of all types in situation S1
Thats all.

I’m thankful for your effort in this reply, but I don’t think this all answers my simple question. My question is rather technical, but not philosophical.

I think I will recompose the question: does the body gain energy even if the net work done on the body is zero?

 

[Vanadium 50]

I think I will recompose the question: does the body gain energy even if the net work done on the body is zero?

Potential energy does not belong to the body.

 

[Chestermiller]

The simplest way to interpret this is to say that the change in potential energy is minus the work done by gravity on the body $-mg\Delta h$: $$W+W_g=W-mg\Delta h=0$$or $$W=-W_g=mg\Delta h=\Delta PE$$where W is the work done on the body by the applied external force F.

 

[adjurovich]

The simplest way to interpret this is to say that the change in potential energy is minus the work done by gravity on the body $-mg\Delta h$: $$W+W_g=W-mg\Delta h=0$$or $$W=-W_g=mg\Delta h=\Delta PE$$where W is the work done on the body by the applied external force F.

I understand what equations say, but I don’t understand how is it possible for something to gain potential energy?

 

[Ibix]

I understand what equations say, but I don’t understand how is it possible for something to gain potential energy?

Perhaps you should read the people telling you the answer: it doesn't. The system gains energy, not the body.

 

[adjurovich]

Perhaps you should read the people telling you the answer: it doesn't. The system gains energy, not the body.

But the same question I’ve asked also applies system? Body alone cannot have potential (gravitational potential) energy. Got it. There has to be at least two bodies, otherwise we couldn’t speak of gravity nor electric force. What happens to the energy “imported into the system” by the work of external force?

 

[Chestermiller]

I understand what equations say, but I don’t understand how is it possible for something to gain potential energy?

It gains potential energy if its elevation decreases since, under this situation, gravity does negative work on the body.

 

[kuruman]

On edit:
This post was stripped of its equations after posting for reasons unknown. See post #27 for a reconstruction.

I am really thankful for such profound analysis, but I think that’s not actually what I asked. My question was relatively simple: it there is no net work, what adds potential energy to the body?

You asked a question that has no answer because it is a non-question. You might as well have asked how many angels can dance on the head of a pin. Until you understand why it is important to have a well-defined system with clearly-set boundaries before you apply the work-energy theorem, you will not be able to get out of the rabbit hole that you fell in. I am not blaming you for this, I blame the people who may have taught it to you and sites like this Wikipedia article in which one reads

​

In physics, potential energy is the energy held by an object because of its position relative to other objects, stresses within itself, its electric charge, or other factors.​

Note the wording "held by an object" that leads one to believe that potential energy joules is something that an object "has" just like kinetic energy joules. Also note that this potential energy depends on the position of the object's position relative to other objects. The statement quoted above is misleading when you read it and start wondering about joules being transferred. You must be clear about the energy transfer being between what and what, in other words what your system is and where the energy comes from.

You are asking the question about the transfer of joules to the object. By asking this question you have implicitly defined the object as the "system". Joules can be transferred to it through two forces doing work, on the object.
Gravity transfers work
"We" transfer work
The net transfer of joules is zero. This comes from the work-energy theorem Here is the sum of the works exerting forces outside the system. In this case Note that potential energy does not enter the equation expressing the work-energy theorem and you cannot just add it in because you think that the object has it.

So how does potential energy enter the picture? First note that I can rewrite equation (1) as I now define the Earth and the object as the system. Assuming that the kinetic energy of the Earth hardly changes as the book is raised, Also the net work done on the system by the external forces is Finally, the definition of gravitational potential energy change is the negative of the work done by gravity If you put all these pieces in equation (2), you get

What happens to the energy “imported into the system” by the work of external force?

Study and compare equations (1) and (3) carefully. On the right hand side of each you have the net external done on (or imported if you prefer) the system.

• When the system is only the object, equation (1), the imported work is the sum of the two works which is zero. The mechanical energy (kinetic + potential) of the system does not change.
• When the system is the object and the Earth, equation (3), the imported work is the work done by "We", . The mechanical energy (kinetic + potential) of the system increases by which, in this case, goes entirely into raising the potential part and not the kinetic.

 

[adjurovich]

You asked a question that has no answer because it is a non-question. You might as well have asked how many angels can dance on the head of a pin. Until you understand why it is important to have a well-defined system with clearly-set boundaries before you apply the work-energy theorem, you will not be able to get out of the rabbit hole that you fell in. I am not blaming you for this, I blame the people who may have taught it to you and sites like this Wikipedia article in which one reads

​

In physics, potential energy is the energy held by an object because of its position relative to other objects, stresses within itself, its electric charge, or other factors.​

Note the wording "held by an object" that leads one to believe that potential energy joules is something that an object "has" just like kinetic energy joules. Also note that this potential energy depends on the position of the object's position relative to other objects. The statement quoted above is misleading when you read it and start wondering about joules being transferred. You must be clear about the energy transfer being between what and what, in other words what your system is and where the energy comes from.

You are asking the question about the transfer of joules to the object. By asking this question you have implicitly defined the object as the "system". Joules can be transferred to it through two forces doing work, on the object.
Gravity transfers work
"We" transfer work
The net transfer of joules is zero. This comes from the work-energy theorem Here is the sum of the works exerting forces outside the system. In this case Note that potential energy does not enter the equation expressing the work-energy theorem and you cannot just add it in because you think that the object has it.

So how does potential energy enter the picture? First note that I can rewrite equation (1) as I now define the Earth and the object as the system. Assuming that the kinetic energy of the Earth hardly changes as the book is raised, Also the net work done on the system by the external forces is Finally, the definition of gravitational potential energy change is the negative of the work done by gravity If you put all these pieces in equation (2), you get

Study and compare equations (1) and (3) carefully. On the right hand side of each you have the net external done on (or imported if you prefer) the system.

• When the system is only the object, equation (1), the imported work is the sum of the two works which is zero. The mechanical energy (kinetic + potential) of the system does not change.
• When the system is the object and the Earth, equation (3), the imported work is the work done by "We", . The mechanical energy (kinetic + potential) of the system increases by which, in this case, goes entirely into raising the potential part and not the kinetic.

Would it be correct to say that potential energy is not the energy stored in object, but rather measure of kinetic energy that would be gained by the object if object was released to free fall (gravity will do work on it and thus lead to change in kinetic energy according to the work energy theorem)?

 

[kuruman]

Would it be correct to say that potential energy is not the energy stored in object, but rather measure of kinetic energy that would be gained by the object if object was released to free fall (gravity will do work on it and thus lead to change in kinetic energy according to the work energy theorem)?

No. The definition of potential energy is In terms of the negative of the work done by a conservative force when the relative position of two objects changes as a result of that force acting on them.

Your statement is specific to a very special case and cannot serve as a definition.

 

[Ibix]

What happens to the energy “imported into the system” by the work of external force?

When you look at the system the net work done by external forces is not zero. The only external force is the force lifting the mass; the gravitational force is internal to the Earth/mass system. So the work you do is stored in the system.

The potential energy is a feature of the Earth, the mass, and the gravitational field collectively. You will struggle to pin it down more precisely than that because, as you found out, if you chop one part of the system out and ask 'is the potential energy here' the answer is 'no'.

You can think of the potential energy as being stored in the field if you like. It's not a bad mental model most of the time

 

[adjurovich]

When you look at the system the net work done by external forces is not zero. The only external force is the force lifting the mass; the gravitational force is internal to the Earth/mass system. So the work you do is stored in the system.

The potential energy is a feature of the Earth, the mass, and the gravitational field collectively. You will struggle to pin it down more precisely than that because, as you found out, if you chop one part of the system out and ask 'is the potential energy here' the answer is 'no'.

You can think of the potential energy as being stored in the field if you like. It's not a bad mental model most of the time

So if I understood it right, if some external force $F_{ext}= mg$ is doing work, energy is being stored in the system, so:​

$W_{ext} = mgh$
​

But work-energy theorem states that the net work of all forces acting on single body is equal to the change of body’s kinetic energy:​

$W_{ext} - W_{int} = \Delta{K}$

From here we get:

$mgh - mgh = \Delta{K} = 0$
​

But since the external force has still done positive work on body, it must increase mechanical energy of the entire system:​

$W_{ext} = \Delta{E}$
​

And since we know that the change in kinetic energy is zero:​

$W_{ext} = (0 + U_2) - (0 + U_1)$

From here we get:

$W_{ext} = \Delta{U}$
​

But if we stop acting with the force equal to gravity, body is no longer in equilibrium and we can apply work-energy theorem (since there is net work of all forces: currently the only one force, gravity):​

$W_{int} = \Delta{K}$
​

Since gravity is acting so that it decreases object’s elevation, it must decrease it’s potential energy:​

$W_{int} = -\Delta{U}$

We can combine these two equations to get:

$\Delta{K} + \Delta{U}$ = 0

And it means:

$K_1 + U_1 = K_2 + U_2$
​

So energy of the system is conserved. Please correct me where I’m wrong.​

 

[kuruman]

When you look at the system the net work done by external forces is not zero.

This statement is meaningless without specifying exactly what the system is. In OP's example of the object being raised at constant speed,

• if the system is only the object, the work done by the external forces on the system is zero;
• if the system is the object + Earth, the work done by the external forces on the system is $+mgh.$

In post #23, OP tries to process post #22. OP uses "system" and "object" interchangeably without having specified beforehand whether the system is just the object or the object + Earth. This is evident in the excerpt

But work-energy theorem states that the net work of all forces acting on single body is equal to the change of body’s kinetic energy:​

$W_{ext} - W_{int} = \Delta{K}$

From here we get:

$mgh - mgh = \Delta{K} = 0$
​

But since the external force has still done positive work on body, it must increase mechanical energy of the entire system:​

$W_{ext} = \Delta{E}$​

Note that OP starts by considering the work-energy theorem applied to the "single body" which implies that OP has chosen the single body as the system. Fine. Let's accept this idea and run with it.

Now the sentence before the last equation where OP concludes that the external force that does positive work on the body, a.k.a. the system, increases the "mechanical energy of the entire system." What entire system? The only system whose mechanical energy changes is object + Earth. It seems that OP started by considering the object as the system and changed it in the middle of the argument. It's like shifting the origin in the middle of solving an equation of motion.

Since my first post on this thread I have been trying to get OP to understand the need to be clear about the choice of system and to stick with that choice through to the end. OP could use some help in that direction from others who wish to post here.

 

[Herman Trivilino]

I understand the first part, but I just wanted to get straight to the point.

Potential energy is a property of the Earth-object system. It is a common textbook error to attribute the potential energy to the object only.

 

[adjurovich]

This statement is meaningless without specifying exactly what the system is. In OP's example of the object being raised at constant speed,

• if the system is only the object, the work done by the external forces on the system is zero;
• if the system is the object + Earth, the work done by the external forces on the system is $+mgh.$

In post #23, OP tries to process post #22. OP uses "system" and "object" interchangeably without having specified beforehand whether the system is just the object or the object + Earth. This is evident in the excerpt

Note that OP starts by considering the work-energy theorem applied to the "single body" which implies that OP has chosen the single body as the system. Fine. Let's accept this idea and run with it.

Now the sentence before the last equation where OP concludes that the external force that does positive work on the body, a.k.a. the system, increases the "mechanical energy of the entire system." What entire system? The only system whose mechanical energy changes is object + Earth. It seems that OP started by considering the object as the system and changed it in the middle of the argument. It's like shifting the origin in the middle of solving an equation of motion.

Since my first post on this thread I have been trying to get OP to understand the need to be clear about the choice of system and to stick with that choice through to the end. OP could use some help in that direction from others who wish to post here.

System is composed Earth and some body. I will change it later in the post #23. However it doesn't seem very precise to me to say work is being done on the system. System is composed of two bodies in this case, but could be also composed of thirty three. By doing it this way, how would we have any information about where energy is going? Is the work being done on earth or just some box? (for example).

Nevertheless, assuming that the system is Earth and the body, which parts aren't correct in my derivation?

 

[kuruman]

Note: This post is reconstructed post #19 where all the LaTeX equations were stripped clean at some unknown point in time for reasons unknown. I am reposting this because #19 makes no sense without equations.

I am really thankful for such profound analysis, but I think that’s not actually what I asked. My question was relatively simple: it there is no net work, what adds potential energy to the body?

You asked a question that has no answer because it is a non-question. You might as well have asked how many angels can dance on the head of a pin. Until you understand why it is important to have a well-defined system with clearly-set boundaries before you apply the work-energy theorem, you will not be able to get out of the rabbit hole that you fell in. I am not blaming you for this, I blame the people who may have taught it to you and sites like this Wikipedia article in which one reads

​

In physics, potential energy is the energy held by an object because of its position relative to other objects, stresses within itself, its electric charge, or other factors.​

Note the wording "held by an object" that leads one to believe that potential energy joules is something that an object "has" just like kinetic energy joules. Also note that this potential energy depends on the position of the object's position relative to other objects. The statement quoted above is misleading when you read it and start wondering about joules being transferred. You must be clear about the energy transfer being between what and what, in other words what your system is and where the energy comes from.

You are asking the question about the transfer of joules to the object. By asking this question you have implicitly defined the object as the "system". Joules can be transferred to it through two forces doing work, on the object.
Gravity transfers work
"We" transfer work
The net transfer of joules is zero. This comes from the work-energy theorem
$$\Delta K_{\text{system}}=W_{\text{Net}}^{\text{external}}$$ Here $W_{\text{Net}}^{\text{external}}=W_{\text{g}}+W_{\text{We}}$ is the sum of the works done by forces outside the system. In this case $$\Delta K_{\text{system}}=\Delta K_{\text{object}}=W_{\text{g}}+W_{\text{We}} \tag{1} $$ Note that potential energy does not enter the equation expressing the work-energy theorem and you cannot just add it in because you think that the object has it.

So how does potential energy enter the picture? First note that I can rewrite equation (1) as $$ \Delta K_{\text{object}}+(-W_{\text{g}})= W_{\text{We}}\tag{2}$$ I now define the Earth and the object as the system. Assuming that the kinetic energy of the Earth hardly changes as the book is raised, $\Delta K_{\text{system}}\approx \Delta K_{\text{object}}.$ Also the net work done on the system by the external forces is $W_{\text{We}}$ Finally, the definition of gravitational potential energy change is the negative of the work done by gravity, $\Delta U_{\text{g}}.$ If you put all these pieces in equation (2), you get $$ \Delta K_{\text{object}}+(\Delta U_{\text{g}})= W_{\text{We}}\tag{3}$$

What happens to the energy “imported into the system” by the work of external force?

Study and compare equations (1) and (3) carefully. On the right hand side of each you have the net external done on (or imported if you prefer) the system.

• When the system is only the object, equation (1), the imported work is the sum of the two works which is zero. The mechanical energy (kinetic + potential) of the system does not change.
• When the system is the object and the Earth, equation (3), the imported work is the work done by "We", . The mechanical energy (kinetic + potential) of the system increases by which, in this case, goes entirely into raising the potential part and not the kinetic.

 

[kuruman]

Nevertheless, assuming that the system is Earth and the body, which parts aren't correct in my derivation?

The part that says

And since we know that the change in kinetic energy is zero:​

$W_{ext} = (0 + U_2) - (0 + U_1)$

From here we get:

$W_{ext} = \Delta{U}$​

When the system is object + Earth, the external force comes from "We". "We" exert a non-conservative force which does work $W_{\text{We}}=F_{We}h$. This work, being done by a non-conservative force, has nothing to do with potential energy change. It is energy added to the object + Earth system. How it is distributed within the system depends on the details of the displacement of the object.

Here are three examples. Say that "We" do work and add 50 joules to the object + Earth system.

• If the object moves vertically up at constant speed, 50 joules are added to the system's energy in the form of potential energy of the system. There is no change in the kinetic energy of the system.
• If the object moves horizontally, 50 joules are added to the system's energy in the form of kinetic energy of the object. There is no change in the potential energy of the system.
• If the object moves vertically up at increasing speed, 50 joules are added to the system's energy in the form of potential energy of the system and kinetic energy of the object.

In all three cases, the change in potential energy is the negative of the work done by gravity. It has nothing to do with the 50 joules that we put in but a lot to do with where the object starts and where it ends up in space.

 

[adjurovich]

The part that says

When the system is object + Earth, the external force comes from "We". "We" exert a non-conservative force which does work $W_{\text{We}}=F_{We}h$. This work, being done by a non-conservative force, has nothing to do with potential energy change. It is energy added to the object + Earth system. How it is distributed within the system depends on the details of the displacement of the object.

Here are three examples. Say that "We" do work and add 50 joules to the object + Earth system.

• If the object moves vertically up at constant speed, 50 joules are added to the system's energy in the form of potential energy of the system. There is no change in the kinetic energy of the system.
• If the object moves horizontally, 50 joules are added to the system's energy in the form of kinetic energy of the object. There is no change in the potential energy of the system.
• If the object moves vertically up at increasing speed, 50 joules are added to the system's energy in the form of potential energy of the system and kinetic energy of the object.

In all three cases, the change in potential energy is the negative of the work done by gravity. It has nothing to do with the 50 joules that we put in but a lot to do with where the object starts and where it ends up in space.

I am thankful for the time people invest in their answers here, but I feel like we are not going in any direction. As far as I am concerned, Laws of Conservation are nothing close to Quantum Field Theory, but this starts to feel like Quantum Field Theory for some reason. Something very simple is getting endlessly complicated somehow.

Also I would like to note that in my high school textbook, potential energy is studied before systems are even mentioned and at high school classes I was taught that the potential energy is basically the energy body has due to its position.

What my reasoning was is that if two forces do work on the same object in opposite directions, energy is basically “disappearing” and that seems to be very incorrect. But my intuition stops here. Where does the energy go if there’s the net work is zero?

Edit:
Do you have some literature recommendation (high school or less advanced college-level) that explains laws of conservation in detail, because I want to get deep understanding of this topic?

 

[Herman Trivilino]

System is composed Earth and some body. [...] However it doesn't seem very precise to me to say work is being done on the system.

In the scenario you asked about, work is done on the object by you, which equals the increase in the system's potential energy.

No one is saying work is done on the system.

 

[adjurovich]

In the scenario you asked about, work is done on the object by you, which equals the increase in the system's potential energy.

No one is saying work is done on the system.

I think I already said that in the post #23.

$W_{ext} = \Delta{U}$​

 

[kuruman]

Also I would like to note that in my high school textbook, potential energy is studied before systems are even mentioned and at high school classes I was taught that the potential energy is basically the energy body has due to its position.

You were taught incorrectly. That incorrect teaching is what led you to this question

What my reasoning was is that if two forces do work on the same object in opposite directions, energy is basically “disappearing” and that seems to be very incorrect. But my intuition stops here. Where does the energy go if there’s the net work is zero?

I have given you the correct way to consider this situation but I seem not to accept or understand what I am saying and go in that direction. Hence your statement

I am thankful for the time people invest in their answers here, but I feel like we are not going in any direction.

You are welcome. We are doing this as a service to you. The bottom line of all that has been said is that being clear about the boundaries of the system under consideration is good practice that protects against confusing yourself and others. It is a simple plea for clarity. Defining clearly what you are talking about is a simple idea that has nothing to do with quantum field theory.

I have said all I have to say on this thread.

 

[A.T.]

I was taught that the potential energy is basically the energy body has due to its position.

This is a commonly used approximation/shorthand, for cases where one body is much more massive than the other. Here, the PE of the system will almost exclusively go into the KE of the smaller body when they are released. Therefore, the system's PE is being often associated with the smaller body.

Unfortunately this shorthand leads to confusion, once you consider work done.

 

[DrBanana]

I asked the same question here: https://www.physicsforums.com/threa...-in-potential-energy-with-0-net-work.1062154/

Plus there is a three part series on the pitfalls of the work-kinetic energy theorem on the Insights blog, that might illuminate some things as well.

 

[Herman Trivilino]

Something very simple is getting endlessly complicated somehow.

Yes. It's the notion that mechanical work can be generalized to a complete understanding of energy transfers. It cannot. The concept of internal energy must first be introduced. Later it can be followed by heat and the 1st Law of Thermodynamics.

This was one of the great intellectual accomplishments of the 19th century and led to what we now call the conservation of energy.

Also I would like to note that in my high school textbook, potential energy is studied before systems are even mentioned and at high school classes I was taught that the potential energy is basically the energy body has due to its position.

And that's almost correct. The energy a system has due to the relative position of its constituents. There are other textbooks that do this correctly, but they are in the minority.

What my reasoning was is that if two forces do work on the same object in opposite directions, energy is basically “disappearing” and that seems to be very incorrect.

It is incorrect because you are using mechanical work to draw conclusions about the more general concept of energy.

But my intuition stops here. Where does the energy go if there’s the net work is zero?

It goes into increasing the potential energy of the system. It comes from the person doing the lifting.

Do you have some literature recommendation (high school or less advanced college-level) that explains laws of conservation in detail, because I want to get deep understanding of this topic?

Development of energy concepts in introductory physics courses
Arnold B. Arons
Citation: Am. J. Phys. 67, 1063 (1999); doi: 10.1119/1.19182
View online: http://dx.doi.org/10.1119/1.19182
View Table of Contents: http://ajp.aapt.org/resource/1/AJPIAS/v67/i12
Published by the American Association of Physics Teachers

 

[adjurovich]

Yes. It's the notion that mechanical work can be generalized to a complete understanding of energy transfers. It cannot. The concept of internal energy must first be introduced. Later it can be followed by heat and the 1st Law of Thermodynamics.

This was one of the great intellectual accomplishments of the 19th century and led to what we now call the conservation of energy.

And that's almost correct. The energy a system has due to the relative position of its constituents. There are other textbooks that do this correctly, but they are in the minority.

It is incorrect because you are using mechanical work to draw conclusions about the more general concept of energy.

It goes into increasing the potential energy of the system. It comes from the person doing the lifting.

Development of energy concepts in introductory physics courses
Arnold B. Arons
Citation: Am. J. Phys. 67, 1063 (1999); doi: 10.1119/1.19182
View online: http://dx.doi.org/10.1119/1.19182
View Table of Contents: http://ajp.aapt.org/resource/1/AJPIAS/v67/i12
Published by the American Association of Physics Teachers

I think I almost understood it. The last question I will ask is: if the energy coming from external work is stored in the system as potential energy. What happens to the energy from work of gravity?

 

[Herman Trivilino]

The work is not external. It's internal to the Earth-object system.

 

[adjurovich]

The work is not external. It's internal to the Earth-object system.

Let’s be a little more precise: work done by gravity or work done by some force $F$?

 

[Herman Trivilino]

Let’s be a little more precise: work done by gravity or work done by some force $F$?

I answered the question you asked:

What happens to the energy from work of gravity?

As I told you before, the energy expended by the agent separating the object from Earth is stored as potential energy in the object-Earth system.

 

[adjurovich]

I answered the question you asked:

Thanks a lot!! I get it now.

 

[A.T.]

if the energy coming from external work is stored in the system as potential energy. What happens to the energy from work of gravity?

Work is transfer of mechanical energy. If an external force does positive work on one of the bodies then it transfers energy to that body. Some of that energy might go into KE of that body, and some into the PE of the system. The later transfer is represented by the negative work by gravity on that body.

It might help to replace the gravitational field with a spring. Here it's more clear where the PE is stored, while the whole energy transfer scheme is the same.

 

[PeroK]

I think there is a general problem here of starting with the equations and then trying to interpret them. Whereas, the meaning of an equation is inherent in the definitions and assumptions that led to the equation in the first place.

In the case of $F =ma$ it is often forgotten that $F$ is the total external force and that this equation is not valid independent of the definition of the three quantities $F, m$ and $a$.

The same applies to the work energy theorem. That equation holds under a precise set of hypotheses. And if it appears to fail, then one or more of the hypotheses must not apply.

 

[cianfa72]

Here are three examples. Say that "We" do work and add 50 joules to the object + Earth system.

• If the object moves vertically up at constant speed, 50 joules are added to the system's energy in the form of potential energy of the system. There is no change in the kinetic energy of the system.
• If the object moves horizontally, 50 joules are added to the system's energy in the form of kinetic energy of the object. There is no change in the potential energy of the system.
• If the object moves vertically up at increasing speed, 50 joules are added to the system's energy in the form of potential energy of the system and kinetic energy of the object.

In all three cases, the change in potential energy is the negative of the work done by gravity. It has nothing to do with the 50 joules that we put in but a lot to do with where the object starts and where it ends up in space.

I came late to this thread: let's assume "Earth + book" as the system. As pointed out by @kuruman in the post before the referenced one, when "We" lift the book raising its kinetic energy, the Earth kinetic energy changes as well. However one can disregard the latter (reasonably assuming that the location of system's center of mass doesn't change) hence all the system's kinetic energy actually goes into the book's kinetic energy.

 

[kuruman]

I came late to this thread: let's assume "Earth + book" as the system. As pointed out by @kuruman in the post before the referenced one, when "We" lift the book raising its kinetic energy, the Earth kinetic energy changes as well. However one can disregard the latter (reasonably assuming that the location of system's center of mass doesn't change) hence all the system's kinetic energy actually goes into the book's kinetic energy.

Good point. One should keep in mind that work done on a single object by "something outside it" is always accompanied by equal and opposite work done by the single object object on the "something outside it." This is guaranteed by Newton's third law.

 

[jbriggs444]

Good point. One should keep in mind that work done on a single object by "something outside it" is always accompanied by equal and opposite work done by the single object object on the "something outside it." This is guaranteed by Newton's third law.

This is correct if we add the proviso that there is no relative motion between the thing inside the system which is subject to the external force and the thing outside the system on which the third law partner force acts.

In that case, the relevant displacements are equal. So the work done: $F_\text{action} \Delta s + F_\text{reaction} \Delta s = 0$

If there can be relative motion (e.g. kinetic friction, gravity, electrostatic force) then we can find energy being absorbed into or emerging from the force pair.

 

[cianfa72]

This is correct if we add the proviso that there is no relative motion between the thing inside the system which is subject to the external force and the thing outside the system on which the third law partner force acts.

In that case, the relevant displacements are equal. So the work done: $F_\text{action} \Delta s + F_\text{reaction} \Delta s = 0$

Just to give an example in the realm of Newtonian physics: let's take an isolated binary system made of two bodies A and B rotating around their common Center of Mass (CoM). Now suppose an external thing suddenly interacts with B doing "external" work on the system (where system = binary system). Generally speaking, the system will change both its potential energy and kinetic energy of its components (A and B) w.r.t. their common CoM.

Assuming A much more massive than B, the displacement of A w.r.t. CoM can be neglected, so basically the change in system's kinetic energy is accounted to B kinetic energy (w.r.t. CoM) alone.

If there can be relative motion (e.g. kinetic friction, gravity, electrostatic force) then we can find energy being absorbed into or emerging from the force pair.

Sorry, might you be more explicit about this point?

 

[kuruman]

This is correct if we add the proviso that there is no relative motion between the thing inside the system which is subject to the external force and the thing outside the system on which the third law partner force acts.

I don't think that a proviso is needed. By "single object" I meant a system with no internal structure.

 

[jbriggs444]

Sorry, might you be more explicit about this point?

It goes back to the definition of mechanical work.

The work done by a force is the dot product of the applied force and the displacement of the material at the point of application of that force: $$W = F \cdot \Delta s$$Newton's third law says that if there is an applied force of $F$ then there will be a third law partner force $-F$ applied on some other body. If we apply the definition of work again, we need to worry about the displacement of the material on that other body.

@kuruman would have us believe that because $F_2 = -F_1$, it follows that $F_1 \cdot {\Delta s}_1 + F_2 \cdot {\Delta s}_2 = 0$

But that is clearly not the case if $\Delta s_1 \ne \Delta s_2$.

I don't think that a proviso is needed. By "single object" I meant a system with no internal structure.

Let us try a concrete example.

We have a car. Model it like a block of wood with no internal structure. It is skidding to a stop. Our system of interest is the block of wood. There is an external force of friction between road and block.

@kuruman would assert that the work done by road on block is equal and opposite to the work done by block on road. Let us test that assertion.

We will adopt the frame of reference of the road. It does not matter. Any inertial frame will deliver a similar result.

Let us say that the block has mass $m$, velocity $v$ and kinetic energy $\frac{1}{2}mv^2$. The coefficient of kinetic friction is $\mu$. The local acceleration of gravity is $g$. The frictional force between road and block is $F = -\mu m g$.

The block will skid to a stop over a distance $\Delta s$ which will be given by $\frac{ \frac{1}{2} mv^2} {\mu m g}$.

The work done by road on block is $F \Delta s = -\frac{1}{2}mv^2$. [Force and displacement are oppositely directed. The block is losing KE. This should indeed be negative]

Let us consider the work done by block on road. The road does not move. The work done is zero.

The two works are not equal and opposite. There is a deficit of $\frac{1}{2}mv^2$. This is an invariant. It is the amount of kinetic energy dissipated via kinetic friction.


Let us revisit our choice of inertial frame...

We can adopt a frame where the block begins at rest and accelerates to match speeds with the moving road. We will find that the work done by road on car is $\frac {1}{2}mv^2$. We will find that the work done by car on road is $-mv^2$. Again, there is an invariant deficit of $\frac{1}{2}mv^2$.

We can adopt a frame where both block and road are moving at $\frac{v}{2}$ toward each other. This time, the work done by road on car will be 0 (it begins and ends with the same speed) and the work done by car on road will be $-\frac{1}{2}mv^2$ (half the velocity of the previous case, so half the work done). Again, the invariant deficit is $\frac{1}{2}mv^2$

 

[A.T.]

One should keep in mind that work done on a single object by "something outside it" is always accompanied by equal and opposite work done by the single object object on the "something outside it." This is guaranteed by Newton's third law.

Newton's 3rd is about conserving momentum: equal but opposite impulse.

It doesn't imply conservation of mechanical energy: the work doesn't have to be equal but opposite.

Macroscopic mechanical energy can be converted to/from other forms of energy.

 

[kuruman]

We have a car. Model it like a block of wood with no internal structure. It is skidding to a stop. Our system of interest is the block of wood. There is an external force of friction between road and block.

You can also model a cow as a sphere, but is it realistic? By my reckoning, a car skidding to a stop has internal structure and hence internal energy. The heat dissipated by friction raises the internal (thermal) energy of the molecules in the brake pads, the tires, etc. You are limiting your argument to mechanical work when the law that rules here is the first law of thermodynamics. If the system is the car, then when it is skidding to a stop both mechanical work and heat cross the system's boundary. What happens to the heat entering the system if it doesn't go into raising the system's internal energy?

@kuruman would assert that the work done by road on block is equal and opposite to the work done by block on road.

I would not assert that. I would get around the issue by considering the system to consist of the car and the Earth. I would figure out all the energy transformations and use total energy conservation to analyze the situation. I have already written about this here. See example II.2 which includes friction and is the closest example to a skidding car.