Definition of potential energy

In summary: is not always equal to the change in potential energy between the starting point and the ending point of the path through which it moves.
  • #36
jtbell said:
Something seems to be missing from that equation.
what?please guide
 
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  • #37
Where's the = sign?
 
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  • #38
Wconservative=ΔPE+ΔKE##
 
  • #39
That doesn't agree with the second equation in your post #9.

I advise you to never use Wconservative and ΔPE in the same equation, except this one: Wconservative = -ΔPE, which gives the relationship between them.

In effect, we use ΔPE and the concept of PE in general, as a replacement for Wconservative.
 
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  • #40
Please tell me if there is only conservative force present i.e ##Wconservative##=##Wnet##
in such conditions the below formula is always applicable
##Wconservative##=##-ΔPE##=##ΔKE##
 
  • #41
gracy said:
Please tell me if there is only conservative force present i.e ##Wconservative##=##Wnet##
in such conditions the below formula is always applicable
##Wconservative##=##-ΔPE##=##ΔKE##
If you slide down a hill on a sled with rusty runners do you have the same kinetic energy as when you slide down on one with freshly waxed runners?
 
  • #42
jbriggs444 said:
If you slide down a hill on a sled with rusty runners do you have the same kinetic energy as when you slide down on one with freshly waxed runners?
I THINK, NO.
 
  • #43
gracy said:
I THINK, NO.
I apologize. My snarky comment had ignored your caveats that no non-conservative force was involved. So no rusty runners for your sled. And the equation does hold.
 
  • #44
gracy said:
in such conditions the below formula is always applicable
WconservativeWconservative=−ΔPE-ΔPE=ΔKE
Mark (notice) my word "always"
is your answer still yes?
 
  • #45
Please answer
 
  • #46
jbriggs444 said:
However, the work done by a non-conservative force on an object moving in its field can have nothing whatsoever to do with the change in potential energy between the starting point and the ending point for the simple reason that there is no such thing as potential energy in a non-conservative field.
But
##Wnet##=##Wconservative##+##Wnonconservative##
##Wnet##=##ΔKE##
(WORK ENERGY THEOREM)
##Wconservative##=##-ΔPE##
(definition of potential energy)
hence we can derive from above equations
##Wnnconservative##=##Wnet## - ##Wconservative##
=##ΔKE## - (##-ΔPE##)
=##ΔKE##+##ΔPE##
so there seems to be relation between Work due to non- conservative forces and potential energy atleast mathematically (as we know mathematics is language of physics)
 
  • #47
gracy said:
so there seems to be relation between Work due to non- conservative forces and potential energy atleast mathematically

An object's PE depends only its location. For example, gravitational PE depends only on elevation (height). If we fix the starting and ending locations, then ΔPE is also fixed, and changing Wnon-conservative can change only ΔKE. I think that's the kind of situation that jbriggs444 is thinking about.

However, if we keep the starting location fixed but allow the ending point to vary, then by changing Wnon-conservative we can also change the ending location. If we do more work on a ball while lifting it up in the air, we can raise it up further (increasing its ΔPE).
 
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  • #48
##Wnc##=##ΔPE##+##ΔKE##
Is only applicable when starting and /or ending locations are not fixed,right?
 
  • #49
No, it applies regardless. Can you describe a specific situation in which you think it doesn't apply?
 
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  • #50
jtbell said:
Can you describe a specific situation in which you think it doesn't apply?
No.
 
  • #51
Ok.There is a question.A charge s moved in an electric field of a fixed charge distribution from point A to another point B SLOWLY.The work done by external agent /force in doing so is 100J.Find the work done by external agent/force.
Here we are not told whether this external force is conservative or non conservative.But I think I will suppose it to be non conservative because in our portion /syllabus there are only three types of conservative forces spring force,electric field force and gravitational force.As I a m taking it to be non conservative
I can apply the formula below
##Wnc##=##Wextenal##=##ΔPE##+##ΔKE##
But the question indirectly indicates that there is no change in kinetic energy by the term :slowly"(I have created separate thread on it)
So the equation becomes
##Wnc##=##Wextenal##=##ΔPE##=100J
My answer is correct.I know the other method to solve this also
as explained by @jbriggs444.But I want to know is the above method also correct?
 
  • #52
Gracy, the best thing for you to remember is that, by definition, the work done by a conservative force equals ##-ΔPE##. What follows may only muddy the waters, but you need to see it and if you remember nothing else remember that, by definition, the work done by a conservative force equals ##-ΔPE##.

gracy said:
##Wnonconservative##=##ΔKE##+##ΔPE##

This is a perfectly valid dynamical relation and will serve you well when calculating answers to chapter-end problems and problems on tests and homework for those portions of your introductory physics courses. It is not, however, a valid energy relation. For that reason many textbook authors are abandoning this approach. It follows from a definition of work that cannot be generalized to include energy transfers in thermodynamic effects.

Here's a simple example. You exert a force of ##10.0## N on a block pushing it a distance of ##1.0## m across the surface of a level table top at a steady speed. There is no change in kinetic energy during the process. Neither is there a change in potential energy. You do ##10## J of work on the block. If we say the net work done is zero because the work done by friction is -##10## J, then we satisfy the above quoted relation, but we get in trouble when we try to apply the First Law of Thermodynamics:

##ΔE##int = ##Q## + ##W##

where ##ΔE##int is the change in internal energy of a system, ##Q## is the heat energy transferred to that system, and ##W## is the work done on that system. Heat energy ##Q## is defined as energy transferred to a system as a result of a temperature difference, so if we insulate the block and table from its environment ##Q = 0## and we have

##ΔE##int = ##W##.

There is an increase in the internal energy of the block (it warms up) and the table top (it warms up) totaling ##10## J. (Assuming the block and table are isolated from their environment.)

How much work ##W## is done on the system? We don't want to choose the block alone to be our system because it won't help us verify the validity of the relation ##ΔE##int = ##Q## + ##W##. This is because we cannot say ##ΔE##int = +##10## J. The block's internal increases by less than ##10## J. It was the combined block-table system that gained ##10## J of internal energy. (Both the block and the table top warmed up).

So, let's consider the system to be the block and table combined. Now we can write ##ΔE##int = +##10## J. Now ##W = +10## J, the work you did on the block, and we satisfy the relation ##ΔE##int = ##Q## + ##W##. We do not say the work done by friction is -##10## J. To understand why we have to understand the correct way to define ##W##.

We can define ##W## in a way that makes it consistent with the First Law of Thermodynamics. Some popular introductory physics textbooks started doing this about 20 years ago, for example Halliday and Resnick. Others have followed suite, but it's slow to catch on in the classrooms. An alternative approach is to continue to define ##W## in the same way, but call it something else, such pseudo-work. That approach doesn't seem to be favorable as we do not see it appearing in the textbooks.

Here is the abstract of a paper written by Arnold B. Arons. It appeared in the American Journal of Physics in 1999. Citation: Am. J. Phys. 67, 1063 (1999).

The work-energy theorem, derived from Newton’s second law, applies to the displacement of a particle or the center of mass of an extended body treated as a particle. Because work, as a quantity of energy transferred in accordance with the First Law of Thermodynamics, cannot be calculated in general as an applied force times the displacement of center of mass, the work-energy theorem is not a valid statement about energy transformations when work is done against a frictional force or actions on or by deformable bodies. To use work in conservation of energy calculations, work must be calculated as the sum of the products of forces and their corresponding displacements at locations where the forces are applied at the periphery of the system under consideration. Failure to make this conceptual distinction results in various errors and misleading statements widely prevalent in textbooks, thus reinforcing confusion about energy transformations associated with the action in everyday experience of zero-work forces such as those present in walking, running, jumping, or accelerating a car. Without a thermodynamically valid definition of work, it is also impossible to give a correct description of the connection between mechanical and thermal energy changes and of dissipative effects. The situation can be simply corrected and student understanding of the energy concepts greatly enhanced by introducing and using the concept of internal energy, that is, articulating the First Law of Thermodynamics in a simple, phenomenological form without unnecessary mathematical encumbrances.

So the issue here is the application of the friction force to the underside of the block. The underside of the block is not undergoing a displacement of ##1.0## m. The displacements are microscopic and there is no way, or at least no direct or easy way, to calculate them. Thus we cannot know how much of the ##10## J of energy was received by the block, with the remainder of it being received by the table top.
 
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  • #53
gracy said:
Ok.There is a question.A charge s moved in an electric field of a fixed charge distribution from point A to another point B SLOWLY.The work done by external agent /force in doing so is 100J.Find the work done by external agent/force.
Here we are not told whether this external force is conservative or non conservative.But I think I will suppose it to be non conservative because in our portion /syllabus there are only three types of conservative forces spring force,electric field force and gravitational force.As I a m taking it to be non conservative
I can apply the formula below
##Wnc##=##Wextenal##=##ΔPE##+##ΔKE##
But the question indirectly indicates that there is no change in kinetic energy by the term :slowly"(I have created separate thread on it)
So the equation becomes
##Wnc##=##Wextenal##=##ΔPE##=100J
My answer is correct.I know the other method to solve this also
as explained by @jbriggs444.But I want to know is the above method also correct?

Yes, it is a perfectly valid solution. Note that the work done by the electric field is -##100## J.
 
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  • #54
gracy said:
Ok.There is a question.A charge s moved in an electric field of a fixed charge distribution from point A to another point B SLOWLY.The work done by external agent /force in doing so is 100J.Find the work done by external agent/force.
Here we are not told whether this external force is conservative or non conservative.But I think I will suppose it to be non conservative because in our portion /syllabus there are only three types of conservative forces spring force,electric field force and gravitational force.
It does not matter whether the external force is or is not conservative. It is a force. It does work. That is all that matters. However, the question of whether it is conservative or not can be answered using only the information from the problem statement.

We are told that the electrostatic force on the object comes from a fixed charge distribution. It follows that the electrostatic force is conservative.

We are told that the charge is moved slowly. It follows that the unknown external force is approximately equal and opposite to the electrostatic force.

A force is conservative if and only if the work it does along a path between two end points depends only on the endpoints and not upon the path. The work done by a force is the path integral of force times incremental distance. Factoring a -1 out of the integral, it is immediately obvious that the work done by the external force is equal and opposite to the work done by the electrostatic force. If the one integral is independent of path then the other must be as well. If the one force is conservative then the other must be (at least approximately) conservative as well.

 
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  • #55
gracy said:
But
##Wnet##=##Wconservative##+##Wnonconservative##
##Wnet##=##ΔKE##
(WORK ENERGY THEOREM)
##Wconservative##=##-ΔPE##
(definition of potential energy)
hence we can derive from above equations
##Wnnconservative##=##Wnet## - ##Wconservative##
=##ΔKE## - (##-ΔPE##)
=##ΔKE##+##ΔPE##
so there seems to be relation between Work due to non- conservative forces and potential energy atleast mathematically (as we know mathematics is language of physics)
Hi Gracy,

This is pretty much a rehash of the presentation on the video you sent me, so,within the framework of the definitions used in the video, it is correct. However, your speculation that there might be a relation between the work done by non-conservative forces and the potential energy is an incorrect interpretation. All the equation is saying is that the work done by non-conservative forces is equal to the net work done by all forces (the change in kinetic energy) minus the work done by conservative forces (minus the change in potential energy). The work done by non-conservative forces is called the "change in total energy" in the video.

This relates to what I wrote in post #21. In that example, F is the non-conservative force (the net force minus the conservative force), qE is the conservative force, and the net force is F+qE. The work resulting from F (the non-conservative work) is equal to the change in total energy PE+KE.

Chet
 
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  • #56
jbriggs444 said:
Factoring a -1 out of the integral,
What is a1 here?
 
  • #57
gracy said:
What is a1 here?
Minus 1. Written as -1.
 
  • #58
jbriggs444 said:
If the one integral is independent of path then the other must be as well. If the one force is conservative then the other must be (at least approximately) conservative as well.
##∫FdS##=-##∫FdS##
You meant if F of right hand side is conservative F of left hand side has to be conservative as well?
 
  • #59
gracy said:
##∫FdS##=-##∫FdS##
You meant if F of right hand side is conservative F of left hand side has to be conservative as well?
It is poor form to use the same letter in the same equation to denote two different things.

I mean ##∫-FdS = -∫FdS##

If F is conservative then -F is also conservative.
 
  • #60
jbriggs444 said:
Factoring a -1 out of the integral, it is immediately obvious that the work done by the external force is equal and opposite to the work done by the electrostatic force.

Yes. The total work done is zero, so the two amounts of work that sum to zero must be equal but opposite.

If the one integral is independent of path then the other must be as well.

You lost me here. The work done by the electrostatic force is path independent. The work done by the other need not be. It could be, for example, that had a different path been taken the total work done might not be zero.
 
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  • #61
gracy said:
Here we are not told whether this external force is conservative or non conservative.But I think I will suppose it to be non conservative because in our portion /syllabus there are only three types of conservative forces spring force,electric field force and gravitational force.

In the formula

Wnc = ΔPE + ΔKE

you account for the work done in one of two ways. Either as a Wnc or as a ΔPE. The work done by the external force in this question must be accounted for as a Wnc because even if it were a conservative force you have no way of accounting for it as a ΔPE.

If the force is conservative it has a potential energy function, if it isn't it doesn't. Here's another way to look at it.

Wnc = ΔPE + ΔKE
Wnc - ΔPE = ΔKE
Wnc + Wc = ΔKE
Wnet = ΔKE
 
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  • #62
Mister T said:
The work done by the electrostatic force is path independent. The work done by the other need not be. It could be, for example, that had a different path been taken the total work done might not be zero.

Agreed. That is a reasonable objection. If, for instance, we were to push the charge through the field with our hands, we would not normally describe the force from our hands as "conservative". However, the problem statement does not restrict the set of paths that could be taken. It does require slow movement on any path that actually is taken. So the force applied by our hands must indeed always be equal and opposite to the electrostatic force at every place (and time) where it is applied.

Perhaps I can rephrase as "if the external force is supplied by a force that can be described by a static field then that field must be conservative".
 
  • #63
jbriggs444 said:
So the force applied by our hands must indeed always be equal and opposite to the electrostatic force at every place (and time) where it is applied.

Good point. I hadn't thought about that. So in this case the work done by the force applied by the hand actually must be path independent. That, however, doesn't mean that the force is in general path independent. It might be the case, for example, that along some other paths that end in different locations the work done by this particular force might be path dependent. Thus it doesn't qualify as a force for which a potential energy function exists.

Interesting.
 

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