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 October 17th, 2017, 09:07 PM #1 Senior Member     Joined: Nov 2015 From: United States of America Posts: 181 Thanks: 21 Math Focus: Calculus and Physics 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
October 17th, 2017, 10:35 PM   #2
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 Originally Posted by SenatorArmstrong 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.

October 17th, 2017, 10:38 PM   #3
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 Originally Posted by romsek 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!

 October 27th, 2017, 06:12 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 That's pretty much the definition of "conservative" force field!

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