What does this mean ##U=\int F\times d\vec{r}## in the Work-Energy Theorem?

Click For Summary

Discussion Overview

The discussion revolves around the interpretation of the equation ##U=\int \vec{F}\times d\vec{r}## in the context of the Work-Energy Theorem. Participants explore the meaning of this equation, particularly focusing on the vector nature of force and displacement, and the implications of using different types of vector multiplication.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about the meaning of the equation and seeks clarification on its interpretation.
  • Another participant explains the concept of work as the product of force and distance, suggesting a more complex path requires integration over segments.
  • A participant points out that the use of "x" in the equation implies a cross product, which may not be appropriate, and suggests that a dot product should be used instead.
  • Another participant confirms that the correct operation is the dot product, emphasizing the distinction between the inner product and the cross product in this context.

Areas of Agreement / Disagreement

There is disagreement regarding the correct interpretation of the vector multiplication in the equation, with some participants advocating for the dot product while others note the initial use of the cross product. The discussion remains unresolved as participants have differing views on the appropriate mathematical operations.

Contextual Notes

Participants have not reached a consensus on the correct interpretation of the vector operations involved in the equation, highlighting potential misunderstandings about the notation used.

n3pix
Messages
15
Reaction score
1
Hello,

I'm newly discovering the world of the Energy.

My question is about the equation ##U=\int \vec{F}\times d\vec{r}=-\int \vec{F}_{s}\times d\vec{r}##.

Can you tell me what does this equation means?

Thanks!
 
Physics news on Phys.org
I assume you are familiar with the expression work=force * distance. U=F*r (letting r stand for distance)

But now consider a more complicated path, not a straight line distance.

We can divide the curved line into a number of nearly straight segments dx. The work for each segment dU=Fdr.

Then ##U=\sum{F dr}=\int F dr##

More generally both F and dr can be vectors, not just scalars as in your OP.
 
In your equation F and dr are vectors. When multiplying vectors an “x” means something very specific and not just “times”. I am sure you meant to use a dot for the dot product and not a “x” for the cross product which makes no sense here.
 
  • Like
Likes   Reactions: vanhees71, Delta2, n3pix and 1 other person
Note, by the way, that it's ##\vec F.d\vec r##, not ##\vec F\times d\vec r## - it's the inner product of the force vector and the displacement vector, not the cross product.

Edit - beaten to it by mere seconds, I see...
 
  • Like
Likes   Reactions: vanhees71 and Cutter Ketch

Similar threads

  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 31 ·
2
Replies
31
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 54 ·
2
Replies
54
Views
7K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 27 ·
Replies
27
Views
5K