Relationship between force and potential energy

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SUMMARY

The relationship between force and potential energy is defined by the first derivative of potential energy with respect to position, indicating that force is the gradient of potential energy. The maximum force occurs where the potential energy changes most rapidly, not where the derivative is zero. Setting the derivative to zero identifies points of local maxima or minima in potential energy, but does not directly yield maximum force. This discussion clarifies misconceptions about the interpretation of derivatives in the context of potential energy graphs.

PREREQUISITES
  • Understanding of calculus, specifically derivatives and their applications.
  • Familiarity with the concepts of force and potential energy in physics.
  • Knowledge of Hooke's law and harmonic oscillators.
  • Ability to analyze graphs of potential energy versus position.
NEXT STEPS
  • Study the relationship between force and potential energy in classical mechanics.
  • Learn about the implications of the second derivative in determining concavity and inflection points.
  • Explore Hooke's law and its applications in real-world scenarios.
  • Investigate the mathematical modeling of potential energy curves and their physical interpretations.
USEFUL FOR

Students of physics, educators teaching mechanics, and anyone interested in the mathematical foundations of force and potential energy relationships.

I_laff
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I am aware that the negative derivative of potential energy is equal to force. Why is the max force found when the negative derivative of potential energy is equal to zero?
 
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Could you give an example problem or situation when this is done? Where did you see this? It is not correct.
 
I was told by someone that when you set the derivative to zero, you get the max force.
 
The answer to your question does not depend on any physics. It is a matter of simple math. Maxima and minima occur at the places where the slope of the curve is zero. Also, the sign of the slope flips as we pass a maximum or minimum.
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So if we set the derivative to zero, we can calculate the minima and maxima of a potential energy graph, how does that help us find the max force?
 
I_laff said:
So if we set the derivative to zero, we can calculate the minima and maxima of a potential energy graph, how does that help us find the max force?
It doesn't. If there is a turning value of Potential Energy with distance, the force is zero. As the hyperphysics link tells you, the highest force is where the Potential Energy is changing fastest with distance.
If you were told otherwise by "someone" then perhaps they are not a reliable source of info (or they misinterpreted the question that you asked them).
 
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I_laff said:
So if we set the derivative to zero, we can calculate the minima and maxima of a potential energy graph, how does that help us find the max force?

No. You don't set the derivative to zero, you find the potential energy where the derivative is zero. That place is a ax or min.
 
I_laff said:
Why is the max force found when the negative derivative of potential energy is equal to zero?
This is incorrect. When the negative gradient of the PE is zero then the force is zero.
 
  • #10
Thanks sophiecentaur, your answer makes sense.
 
  • #11
I_laff said:

Where exactly in the hyperphysics link does it claim that "... max force found when the negative derivative of potential energy is equal to zero ... "? I do not see it, and you're asking us to correct an non-existing error.

Force is the gradient of the potential energy. This means that the quicker the potential energy changes over distance, the higher the absolute value of the force. So it is not the absolute value of the potential energy, but rather the change in the potential energy that corresponds to the force.

Zz.
 
  • #12
ZapperZ said:
Where exactly in the hyperphysics link does it claim that "... max force found when the negative derivative of potential energy is equal to zero ... "? I do not see it, and you're asking us to correct an non-existing error.

Force is the gradient of the potential energy. This means that the quicker the potential energy changes over distance, the higher the absolute value of the force. So it is not the absolute value of the potential energy, but rather the change in the potential energy that corresponds to the force.

Zz.
I didn't say that Hyperphysics stated that the max force is when the derivative was set to 0. I said that I was told that it was, which confused me since it didn't make sense. My only mistake was assuming that the person who told me that the max force could be calculated by setting the derivative to zero was correct.
 
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  • #13
I_laff said:
I didn't say that Hyperphysics stated that the max force is when the derivative was set to 0. I said that I was told that it was, which confused me since it didn't make sense. My only mistake was assuming that the person who told me that the max force could be calculated by setting the derivative to zero was correct.

Look at Posts 1, 2, and 3 and read them again in sequence to see why it appears that you are using the Hyperphysics link to justify what you were told.

Zz.
 
  • #14
I_laff said:
I didn't say that Hyperphysics stated that the max force is when the derivative was set to 0. I said that I was told that it was, which confused me since it didn't make sense. My only mistake was assuming that the person who told me that the max force could be calculated by setting the derivative to zero was correct.
So the maximum occurs where the slope of the potential energy curve is steepest. You can find those points by differentiating the potential energy twice with respect to position and setting that equal to 0. Only in those points can the force be maximal. (there could also be a minimum, or an inflection point)
 
  • #15
willem2 said:
So the maximum occurs where the slope of the potential energy curve is steepest. You can find those points by differentiating the potential energy twice with respect to position and setting that equal to 0. Only in those points can the force be maximal. (there could also be a minimum, or an inflection point)

This is not correct either. 2nd derivative tells you how rapidly the slope is changing, and thus, how rapidly the force changes. It doesn't tell you how large the force is.

You can have an almost vertical straight line on the V vs x graph (i.e. 2nd derivative is zero but not at a max or min or inflection point), and yet, this is where a force can be a maximum.

Zz.
 
  • #16
ZapperZ said:
Look at Posts 1, 2, and 3 and read them again in sequence to see why it appears that you are using the Hyperphysics link to justify what you were told.

Zz.
I would've thought the quoted post below would've clarified what information was and was not from the Hyperphysics website.

I_laff said:
I was told by someone that when you set the derivative to zero, you get the max force.
 
  • #17
I_laff said:
I would've thought the quoted post below would've clarified what information was and was not from the Hyperphysics website.

I get that you were "told" by someone, but when asked for a source in Post 2, you cited Hyperphysics, as if you were using that page to justified what you were told by this "someone". That's why I asked where specifically on that Hyperphysics page matches what you were told by this "someone".

Many of us are familiar with Hyperphysics. I even used it as an additional educational source for my students. I don't ever recall them making this type of error, and that is why I was particular concerned in trying to find where exactly in there that matches the erroneous info that you were told. If nothing there matches what you were told, why did you cite it without understanding it?

Zz.
 
  • #18
Apologises if I misled you, I will be more careful when citing sources next time so it is obvious what is and is not from the source. I cited hyperphysics because it was from there that I read about the relationship between force and potential. The first reply to my post simply said that what I had stated was 'not correct'. They didn't specify what part of what I said was incorrect, so in reply to this I just posted where I got all my information from. I think my understanding of what hyperphysics was on about was adequate enough for me to cite it. Once again, I never said that hyperphysics stated the incorrect information I was told. You insinuated that this is what I meant from my posts, but that is not the case.
 
  • #19
I thought we had already decided that this "person" was not a reliable source of information but that Hyperphysics is pretty well trustworthy.
 
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  • #20
ZapperZ said:
This is not correct either. 2nd derivative tells you how rapidly the slope is changing, and thus, how rapidly the force changes. It doesn't tell you how large the force is.

You can have an almost vertical straight line on the V vs x graph (i.e. 2nd derivative is zero but not at a max or min or inflection point), and yet, this is where a force can be a maximum.
I don't get this at all. You say you disagree with me, but then you produce a case where the 2nd derivative is 0 and the force is a maximum, but this is not a max or min or inflection point? This appears to be a contradiction.
The force is the first derivative of the potential energy. The only place where the force can be a maximum is where the derivative of the force, or the second derivative of the potential energy is 0.
 
  • #21
willem2 said:
I don't get this at all. You say you disagree with me, but then you produce a case where the 2nd derivative is 0 and the force is a maximum, but this is not a max or min or inflection point? This appears to be a contradiction.
The force is the first derivative of the potential energy. The only place where the force can be a maximum is where the derivative of the force, or the second derivative of the potential energy is 0.
I think your confusion is that you didn't just stop after the relationship was described properly with Maths. There is no point in 'talking around' something as straightforward as the way Hyperphysics states it. Whatever "somebody" said about it, it just adds confusion when that statement is included in the formal definition. If you understand the Maths then just use it and get familiar with the way it applies. The content of the first post is correct so why are we still discussing it, twenty posts later? Ignore your unhelpful, unofficial source.
 
  • #22
willem2 said:
I don't get this at all. You say you disagree with me, but then you produce a case where the 2nd derivative is 0 and the force is a maximum, but this is not a max or min or inflection point? This appears to be a contradiction.
The force is the first derivative of the potential energy. The only place where the force can be a maximum is where the derivative of the force, or the second derivative of the potential energy is 0.
You are correct, @ZapperZ was confused.
 
  • #23
willem2 said:
I don't get this at all. You say you disagree with me, but then you produce a case where the 2nd derivative is 0 and the force is a maximum, but this is not a max or min or inflection point? This appears to be a contradiction.
The force is the first derivative of the potential energy. The only place where the force can be a maximum is where the derivative of the force, or the second derivative of the potential energy is 0.

This is what you said:

willem2 said:
ou can find those points by differentiating the potential energy twice with respect to position and setting that equal to 0. Only in those points can the force be maximal. (there could also be a minimum, or an inflection point)

And this is what I argued against (bold added here):

ZapperZ said:
You can have an almost vertical straight line on the V vs x graph (i.e. 2nd derivative is zero but not at a max or min or inflection point), and yet, this is where a force can be a maximum.

In other words, there isn't any need for there to be a LOCAL MAXIMUM OR LOCAL MINIMUM.

I'll give you another example since there straight-line curve didn't sink in with you: U = ½ kx2

What is the 2nd derivative of that? Is it zero?

Yet, there IS a maximum force for this potential when we do a simple Hooke's law experiment.

Zz.
 
  • #24
ZapperZ said:
I'll give you another example since there straight-line curve didn't sink in with you: U = ½ kx2

What is the 2nd derivative of that? Is it zero?

Yet, there IS a maximum force for this potential when we do a simple Hooke's law experiment.

If the potential is reallly ½ kx2 for all x, there is no place where the second derivative is 0, but there is also no place where the force has a maximum. Of course real springs will break at some point, but U = ½ kx2 won't be valid anymore before this happens.
 
  • #25
willem2 said:
If the potential is reallly ½ kx2 for all x, there is no place where the second derivative is 0, but there is also no place where the force has a maximum. Of course real springs will break at some point, but U = ½ kx2 won't be valid anymore before this happens.

But if you do a Hooke's law experiment, that IS the potential! This is the common harmonic oscillator potential! We even used this in quantum mechanics!

It means that the force has NO LOCAL MINIMUM OR MAXIMUM. But if you stand in front of the class and claim that no where in the oscillation is there a maximum force, you will be making a false statement! There ARE two points where the force on the spring-mass system is maximum, and those are the two ends/turning points when the spring is extended the most! On the graph, these are the maximum limits of "x".

This was why I insisted that one looks at the actual force curve, rather than blindly applying mathematical rules to find such "maximum force"! One can find maximum force even if the system has no local max/min... IF one pays attention to the physics!

Zz.
 
  • #26
ZapperZ said:
One can find maximum force even if the system has no local max/min...
The use of the term "maximum force" can mean one of two things. 1. The Force / Displacement Law has a local maximum or 2. You chose to limit the range of forces or displacements in your experiment.
Why is this verbal jiggery pokery being involved with a perfectly good mathematical description? If you want to be realllllly precise then include some more formal definitions and caveats such a that the function is continuous and differentiable within the limits being considered etc. etc. but would that really help?
Mods please help and put this thread out of its misery.
 
  • #27
sophiecentaur said:
The use of the term "maximum force" can mean one of two things. 1. The Force / Displacement Law has a local maximum or 2. You chose to limit the range of forces or displacements in your experiment.
Why is this verbal jiggery pokery being involved with a perfectly good mathematical description? If you want to be realllllly precise then include some more formal definitions and caveats such a that the function is continuous and differentiable within the limits being considered etc. etc. but would that really help?
Mods please help and put this thread out of its misery.

I don't understand your complain. This IS a physics question and not purely a mathematics question, is it not? After all, this was attached to the concept of "potential energy".

I gave a very simple example that did not conform to what was claimed, i.e. using the ONLY the 2nd derivative of the potential energy to find the "maximum force". I showed a specific example where using blind mathematics, you cannot get such a thing from something that is a COMMON example in intro physics. And yet, in the physical scenario, there ARE points where one does get "maximum forces".

Someone who reads Post #14, and then learn about harmonic oscillator, will get VERY confused.

Zz.
 
  • #28
ZapperZ said:
Someone who reads Post #14, and then learn about harmonic oscillator, will get VERY confused.
That is out of context but not wrong. The restoring force is proportional to the displacement away from the rest position. Any given oscillator falls into category 2 in my post. The maximum displacement is arbitrary and is chosen of the particular experiment and is nothing to do with the Force Law. Confusion could be caused but only because of lack of description. The PE variation with time is sinusoidal and the maximum of force corresponds to the maximum of PE. But this is just rehearsing what we know already and the sheer number of these posts is contributing more to confusion than any statement in any particular post.
As with many questions and solutions to many problems, the initial statement about the physical situation has to be clear and bombproof. Introducing SHM is a red herring because it introduces Time into the situation and the term "maximum" can either imply dy/dx = 0 or dy/dt = 0. There's your potential confusion.
 
  • #29
sophiecentaur said:
That is out of context but not wrong. The restoring force is proportional to the displacement away from the rest position. Any given oscillator falls into category 2 in my post. The maximum displacement is arbitrary and is chosen of the particular experiment and is nothing to do with the Force Law. Confusion could be caused but only because of lack of description. The PE variation with time is sinusoidal and the maximum of force corresponds to the maximum of PE. But this is just rehearsing what we know already and the sheer number of these posts is contributing more to confusion than any statement in any particular post.
As with many questions and solutions to many problems, the initial statement about the physical situation has to be clear and bombproof. Introducing SHM is a red herring because it introduces Time into the situation and the term "maximum" can either imply dy/dx = 0 or dy/dt = 0. There's your potential confusion.

I did not introduce time. You did. The PE variation is also in "x", the extension of the spring. This is the ONLY scenario that I have used, which is the only parameter that the OP brought up. This is not a question on the dynamics of the system. So if anyone is pushing this out beyond the original confines, it isn't me.

Post #14 implied that there is a general rule of determining maximum force, simply by applying the 2nd derivative of U. I disputed that by giving a very simple and common example and showed why that general rule does not work on something that every general physics student has seen. I do not understand why this is a problem!

Zz.
 
  • #30
ZapperZ said:
I did not introduce time. You did.
Iirc, you introduced the possible problems when students come across the Harmonic Oscillator. Is there an earlier mention of time? If you use the term "derivative" then the independent variable needs to be stated. Up till your post the dx was all we were concerned with and it would be best to avoid mention (implied or otherwise) of dynamics.
You were right to challenge the use of the second derivative as it doesn't help but the OP was almost driven by the thread into over-thinking the issue.
Haha - it didn't help that the original statement in the first post was wrong.
I will get on my soap box and shout the praises of Worked Examples from The Book. Doing a couple of those can usually put people right, far better than any amount of chat.
 

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