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Main Question or Discussion Point
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?
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.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.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?
This is incorrect. When the negative gradient of the PE is zero then the force is zero.Why is the max force found when the negative derivative of potential energy is equal to zero?
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.
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.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.
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.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)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.
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.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)
I would've thought the quoted post below would've clarified what information was and was not from the Hyperphysics website.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 was told by someone that when you set the derivative to zero, you get the max force.
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".I would've thought the quoted post below would've clarified what information was and was not from the Hyperphysics website.
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.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 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.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.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: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.
And this is what I argued against (bold added here):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)
In other words, there isn't any need for there to be a LOCAL MAXIMUM OR LOCAL MINIMUM.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.
If the potential is reallly ½ kx^{2} 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 kx^{2} won't be valid anymore before this happens.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.
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!If the potential is reallly ½ kx^{2} 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 kx^{2} won't be valid anymore before this happens.