Relationship between force and potential energy

  • #1
I_laff
41
2
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?
 

Answers and Replies

  • #2
Chandra Prayaga
Science Advisor
652
150
Could you give an example problem or situation when this is done? Where did you see this? It is not correct.
 
  • #4
I_laff
41
2
I was told by someone that when you set the derivative to zero, you get the max force.
 
  • #5
anorlunda
Staff Emeritus
Insights Author
10,871
8,182
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.
Screen%20Shot%202014-08-31%20at%202.33.00%20PM.png
 

Attachments

  • Screen%20Shot%202014-08-31%20at%202.33.00%20PM.png
    Screen%20Shot%202014-08-31%20at%202.33.00%20PM.png
    14.2 KB · Views: 1,092
  • Like
Likes Asymptotic and sophiecentaur
  • #6
I_laff
41
2
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?
 
  • #7
sophiecentaur
Science Advisor
Gold Member
27,591
6,221
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).
 
  • #8
anorlunda
Staff Emeritus
Insights Author
10,871
8,182
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.
 
  • #9
Dale
Mentor
Insights Author
2021 Award
33,291
10,675
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
I_laff
41
2
Thanks sophiecentaur, your answer makes sense.
 
  • #11
ZapperZ
Staff Emeritus
Science Advisor
Education Advisor
Insights Author
35,995
4,711

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
I_laff
41
2
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.
 
  • #13
ZapperZ
Staff Emeritus
Science Advisor
Education Advisor
Insights Author
35,995
4,711
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
willem2
2,090
349
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
ZapperZ
Staff Emeritus
Science Advisor
Education Advisor
Insights Author
35,995
4,711
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
I_laff
41
2
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 was told by someone that when you set the derivative to zero, you get the max force.
 
  • #17
ZapperZ
Staff Emeritus
Science Advisor
Education Advisor
Insights Author
35,995
4,711
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
I_laff
41
2
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
sophiecentaur
Science Advisor
Gold Member
27,591
6,221
I thought we had already decided that this "person" was not a reliable source of information but that Hyperphysics is pretty well trustworthy.
 
  • #20
willem2
2,090
349
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
sophiecentaur
Science Advisor
Gold Member
27,591
6,221
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
jbriggs444
Science Advisor
Homework Helper
11,301
5,883
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
ZapperZ
Staff Emeritus
Science Advisor
Education Advisor
Insights Author
35,995
4,711
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
willem2
2,090
349
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
ZapperZ
Staff Emeritus
Science Advisor
Education Advisor
Insights Author
35,995
4,711
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
sophiecentaur
Science Advisor
Gold Member
27,591
6,221
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
ZapperZ
Staff Emeritus
Science Advisor
Education Advisor
Insights Author
35,995
4,711
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
sophiecentaur
Science Advisor
Gold Member
27,591
6,221
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
ZapperZ
Staff Emeritus
Science Advisor
Education Advisor
Insights Author
35,995
4,711
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
sophiecentaur
Science Advisor
Gold Member
27,591
6,221
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.
 
  • #31
ZapperZ
Staff Emeritus
Science Advisor
Education Advisor
Insights Author
35,995
4,711
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.

The ONLY usage of the word "derivative" was in relation to finding the gradient of "U". Until you brought it up, the derivative with respect to any other variables, such as time, was NEVER an issue. So if anyone here is confused, it might be you.

BTW, unless you have mixed things up, the OP and the person who posted Post #14 that I am disputing are two different people.

Zz.
 
  • #32
sophiecentaur
Science Advisor
Gold Member
27,591
6,221
Until you brought it up, the derivative with respect to any other variables, such as time, was NEVER an issue
If time is not an issue then how would a student be confused in the context of a harmonic oscillator? The maximum displacement in SHM is arbitrary and depends on where it's been let go.- That's my second category (2) of Maximum and it doesn't involve differentiation with respect to displacement. I am not confused about that. I really think that bringing in SHM was not helpful.

Talking to the wrong person is a common problem :smile: but it doesn't make my Physics wrong.
 
  • #33
ZapperZ
Staff Emeritus
Science Advisor
Education Advisor
Insights Author
35,995
4,711
If time is not an issue then how would a student be confused in the context of a harmonic oscillator? The maximum displacement in SHM is arbitrary and depends on where it's been let go.- That's my second category (2) of Maximum and it doesn't involve differentiation with respect to displacement. I am not confused about that. I really think that bringing in SHM was not helpful.

Talking to the wrong person is a common problem :smile: but it doesn't make my Physics wrong.

No, but it shows that you are equally confused in many of these conversations.

1. The harmonic oscillator has the potential form of U = 1/2 kx2. Notice, I'm not bring in ANY time factor here.

2. Post #14 says that all one has to do to find the "maximum force" is to find where the 2nd derivative (with respect to "x", if there is any confusion here) is zero.

3. Student performs the 2nd derivative of U = 1/2 kx2. He/she gets "k" as an answer, and wonders why he/she can't find the "zero" for this result. Does this mean that there are no values of "x" for the force to be a maximum?

The rule doesn't work when there are no local maxima/minima in the 1st derivative.

Now, where exactly in what I said above is wrong?

Zz.
 
  • #34
sophiecentaur
Science Advisor
Gold Member
27,591
6,221
The harmonic oscillator has the potential form of U = 1/2 kx2.
Just one turning value. A minimum at x=0 and, of course, a finite maximum displacement but the derivative increases all the time. Would you call that "a maximum" at the arbitrary extremes of displacement?
Does this mean that there are no values of "x" for the force to be a maximum?
In the displacement / force law there is no 'maximum' because there is no turning value up there. I think your confused student would just sit and think about it for a minute and realise the difference. Just because the oscillator only explores a part of the Hooke's law doesn't mean that it's found a maximum. If he/she is interested enough to find this a problem all that's necessary is to increase the displacement and prove that the force still increases. Hooke's Law doesn't predict a turning value anywhere until the spring breaks. I guess that would be a valid maximum.

No, but it shows that you are equally confused in many of these conversations.
Now now! That should be in a PM, I think. :wink:
 
  • #35
ZapperZ
Staff Emeritus
Science Advisor
Education Advisor
Insights Author
35,995
4,711
Just one turning value. A minimum at x=0 and, of course, a finite maximum displacement but the derivative increases all the time. Would you call that "a maximum" at the arbitrary extremes of displacement?

There's no "time". The force is linear with displacement. There is no "maximum" as in a "peak" in the curve, and thus, no zeros in the second derivative with x. But there is a maximum force that is applied by the spring, and that occurs at maximum displacement. This can't be found simply by blind mathematics. It can only be found by understanding the physics, i.e. the system has a maximum displacement, and at that maximum displacement, one gets maximum force.

In the displacement / force law there is no 'maximum' because there is no turning value up there. I think your confused student would just sit and think about it for a minute and realise the difference. Just because the oscillator only explores a part of the Hooke's law doesn't mean that it's found a maximum. If he/she is interested enough to find this a problem all that's necessary is to increase the displacement and prove that the force still increases. Hooke's Law doesn't predict a turning value anywhere until the spring breaks. I guess that would be a valid maximum.

But that is exactly what I've been arguing about! This can't be done simply by blind mathematics of "find the zeroes in the 2nd derivative"! It requires the PHYSICS and knowing that the force is linear with displacement, and that there is a range of values for the displacement. The mathematics does NOT tell you this.

Early on in this thread, I gave the simplest answer that I believe matched this thread level, which is simply to look at the slope of the U(x) curve, and find where the slope is maximum. That corresponds to the maximum absolute value of the force. This works no matter if there are local maximum in the force curve or not. If it is a straight line, then the maximum slope is the same everywhere. This simplest approach works and it is what WE ALL know here, and was clearly described in the Hyperphysics link. I do not know why we just did not stick with this and added on the complications of the 2nd derivative that doesn't even work all the time in picking up the location of maximum force. To me, this thread has been sufficiently answered PRIOR to Post #14.

Zz.
 

Suggested for: Relationship between force and potential energy

Replies
55
Views
733
Replies
12
Views
367
Replies
5
Views
182
Replies
9
Views
832
Replies
8
Views
257
Replies
10
Views
395
  • Last Post
Replies
24
Views
282
Replies
2
Views
2K
Replies
3
Views
450
Top