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October 17th, 2017, 09:07 PM   #1
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Work in a force field

Consider a force field $\vec F(x,y)$. If I want to calculate the work done from $(a,b) \rightarrow (c,d)$ on the cartesian plane, and I pick two different paths for my line integral $$\oint_C \vec F \dot \,d\vec s$$

Will I always get the same work for any path taken?

I have been messing with a similar problem today and I keep getting different values for work as I try different paths between the same points in the force field. Doesn't seem right
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October 17th, 2017, 10:35 PM   #2
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Originally Posted by SenatorArmstrong View Post
Consider a force field $\vec F(x,y)$. If I want to calculate the work done from $(a,b) \rightarrow (c,d)$ on the cartesian plane, and I pick two different paths for my line integral $$\oint_C \vec F \dot \,d\vec s$$

Will I always get the same work for any path taken?

I have been messing with a similar problem today and I keep getting different values for work as I try different paths between the same points in the force field. Doesn't seem right
excellent question

If the force field is conservative, i.e. if it is the gradient of a scalar potential function, then the work done moving a particle through the field between two points will be independent of the path taken.

https://en.wikipedia.org/wiki/Conservative_vector_field

If the field is not conservative then the work done will vary with the path taken.
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October 17th, 2017, 10:38 PM   #3
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excellent question

If the force field is conservative, i.e. if it is the gradient of a scalar potential function, then the work done moving a particle through the field between two points will be independent of the path taken.

https://en.wikipedia.org/wiki/Conservative_vector_field

If the field is not conservative then the work done will vary with the path taken.
Ahh that makes perfect sense. Thanks!
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October 27th, 2017, 06:12 AM   #4
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That's pretty much the definition of "conservative" force field!
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