Finding Force from the Potential Energy

In summary, the conversation discusses how to estimate the force in Newtons for a given graph of -du/dx = F (in x direction). Various methods are proposed, including using a cubic equation and taking the derivative, or "eyeballing" the slope at specific points. The accuracy of these methods is debated, with the conclusion that a more accurate method would be to estimate the slope of the particle and take its negative. However, the conversation also notes that this is still just an approximation.
  • #1
joedozzi
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Homework Statement



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Homework Equations



-du/dx = F (in x direction)

The Attempt at a Solution



Would I have to use system equations to estimate a cubic equation, and just take the negative derivative of it? Cause it turns out to be a mess and I end up getting a equation not a Force in Newtons. Any ideas?
 
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  • #2
joedozzi said:
Would I have to use system equations to estimate a cubic equation, and just take the negative derivative of it? Cause it turns out to be a mess and I end up getting a equation not a Force in Newtons.
Ummm. :rolleyes: I'm not sure I'm following you there.
Any ideas?
What's the slope of the line, at any given point on the curve? (Think "rise over run" if it helps.)

How does the slope of the line relate to -du/dx?
 
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  • #3
The slope of the line at any given point, is the derivative of u(x) at that point. But seeing how the slope is constantly changing how would I estimate the Force.

And what I was saying is that it looks like the graph can be modeled by a cubic equation. If I take 4 points off the graph, I can use the points to estimate a cubic equation. Then I take the derivative of this, times it by negative 1, and this will give the force equation. However, this is not a number rather an equation, so I can't estimate the amount of Force in Newtons. Unless I am totally off track here.
 
  • #4
joedozzi said:
The slope of the line at any given point, is the derivative of u(x) at that point. But seeing how the slope is constantly changing how would I estimate the Force.
If the problem statement is actually asking for a specific number, I'm guessing that it's asking for the negative of the derivative of u(x) evaluated at x = 4 [m], where particle A happens to placed in the figure.

But on the other hand, if you wanted to, you could estimate the negative of the derivative of u at all points, from evaluated at x = 0 to 9 [m], and then end up with a plot of F(x).

(It's not 100% clear to me if the problem is asking for a particular number or a plot.)
And what I was saying is that it looks like the graph can be modeled by a cubic equation. If I take 4 points off the graph, I can use the points to estimate a cubic equation. Then I take the derivative of this, times it by negative 1, and this will give the force equation. However, this is not a number rather an equation, so I can't estimate the amount of Force in Newtons. Unless I am totally off track here.
Yes, you could model it by a cubic equation, or even more accurately by a polynomial of a larger order. But that's probably overkill for this problem, I'm guessing. The problem statement did say "estimate" the Force, after all.

If you could just "eyeball" -du/dx at points x = 0.5, 2.5, 4.0, 6.0, 8.0 and 8.75 [m], it's enough to reproduce a good plot by connecting the points. Or, if the problem is just asking for a number, just eyeball the negative slope around x = 4 [m].
 
  • #5
Ok Thanks so Much :) . I think it asking for it A, or else It would be asking for too much.

I have to hand this question in. I already found a cubic equation and found the force to be 1.47N... but when I use a tangent line to find the force I get 1.66N. Which one do you think would be more accurate?
 
  • #6
joedozzi said:
Ok Thanks so Much :) . I think it asking for it A, or else It would be asking for too much.

I have to hand this question in. I already found a cubic equation and found the force to be 1.47N... but when I use a tangent line to find the force I get 1.66N. Which one do you think would be more accurate?
Neither, actually.

When I "eyeball" the particle around x ≈ 4 m (where it's almost linear), it looks to me that when x changes from 3.75 m to 4.25 m, u changes from 4 J to 3 J.

(I haven't taken the negative of the slope yet. My point above is that I "eyeball" a different, nice, even number.)
[Edit: of course that's just an approximation.]
 
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FAQ: Finding Force from the Potential Energy

What is potential energy?

Potential energy is the energy that an object has due to its position or state. It is stored energy that can be converted into other forms, such as kinetic energy, when the object moves or changes position.

What is force?

Force is a push or pull that can cause an object to accelerate or change its motion. It is a vector quantity, meaning it has both magnitude and direction.

How are potential energy and force related?

Potential energy and force are related through the principle of conservation of energy. When an object moves from a higher potential energy to a lower potential energy, the difference in potential energy is converted into kinetic energy. This change in kinetic energy is caused by a force acting on the object.

How can potential energy be used to find force?

Potential energy can be used to find force by using the formula: force = -∂U/∂x, where U is the potential energy and x is the position. This formula calculates the force required to move an object from one position to another.

What are some real-life applications of finding force from potential energy?

One common application is in determining the force required to lift an object against gravity. Another application is in analyzing the forces involved in simple machines, such as levers and pulleys. Additionally, potential energy and force are important concepts in understanding the behavior of springs, pendulums, and other systems in physics.

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