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

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
3 replies · 1K views
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