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Assume a particle travelling along a curve, the work done by any Force field on the particle while moving along a curve is given by the line integral of $\vec{\bf{F}} \cdot \vec{\bf{dr}}$, but shouldn't the path be a straight line regardless of the given path as the work done $(W) = F \cdot s$ (disp between A and B), displacement being the straight line path between the two points?

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    $\begingroup$ What is the question ??? $\endgroup$
    – Frobenius
    Commented Aug 17, 2021 at 6:43
  • $\begingroup$ $dW=\overrightarrow{F}\cdot d\overrightarrow{r}=\overrightarrow{F}\cdot \dfrac{\partial \overrightarrow{r}}{\partial s}ds=\overrightarrow{F}\cdot \overrightarrow{t}ds$ $\endgroup$
    – Eli
    Commented Aug 19, 2021 at 7:51
  • $\begingroup$ The phrasing “shouldn't the path be a straight line regardless of the given path” doesn’t make much sense to me. What do you mean by this? $\endgroup$ Commented Nov 29, 2021 at 1:19

4 Answers 4

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The formula

$$W=\int_{\text{path}}\vec F\cdot \vec{dr}$$

is the formula for the general case, when the force $\vec F$ doesn’t have to be constant. If however $\vec F$ is constant, you can move $\vec F$ out of the integral, and the equation simplifies to

$$W=\int_{\text{path}}\vec F\cdot \vec{dr} = \vec F\cdot \underbrace{\int_{\text{path}}\vec{dr}}_{=\vec s} = \vec F\cdot\vec s.$$

So the formula you propose to use instead of the integral is for the case when the force is constant.

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There are two things I think to consider why it might not always be that way:

  1. There may be multiple forces involved, and we are calculating the work done by one of them. Just for example, if we push something on a track and the pushing is not always tangent, then the track is pushing back. But we are maybe not asked to calculate the work done by the combination of pushing and track. Maybe only asked to determine work done by the pushing. (Just one example). Another might be moving something around in a fluid or the wind, etc. Or pushing down and forward, where the ground provides force to keep it moving horizontal - the work done by us, exclusive of the ground, is integral F dr

  2. Momentum. If we get it moving (or if it is initially moving) then we might be pushing some other direction to the motion.

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    $\begingroup$ I would appreciate if you could explain to me what did you understand to be the question here. $\endgroup$
    – Frobenius
    Commented Aug 17, 2021 at 6:50
  • $\begingroup$ @Frobenius Lol. Yeah hes saying if you push on something, why doesnt it just move exactly the way you push? And then if it does, aren’t the force and displacement always exactly parallel? And then if that’s true, why are we worrying about a dot product along the way? Another way to say it is: how can we have curved paths of different shape than the force profile, where the force is different than: always in the exact direction of the displacement? Maybe that last question doesnt help focus on the first ones $\endgroup$
    – Al Brown
    Commented Aug 17, 2021 at 6:50
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    $\begingroup$ I am sorry, it seems that I am not able to see all that in the question, so let leave it as it is. $\endgroup$
    – Frobenius
    Commented Aug 17, 2021 at 6:58
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    $\begingroup$ @Frobenious yes perhaps im over-interpreting. Certainly at least a little maybe more. Best 🙏🏻👍🏻 $\endgroup$
    – Al Brown
    Commented Aug 17, 2021 at 7:01
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    $\begingroup$ Have a good day and good luck. $\endgroup$
    – Frobenius
    Commented Aug 17, 2021 at 7:03
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If the force field is uniform (with a constant F), then you can use the straight line path. It would be like using components of ds which are parallel or perpendicular to F.

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Indeed the path is a straight line. $\vec{\textbf{dr}}$ is a straight line. But it is infinitesimal. The formula $\vec{\textbf{F}}.\vec{\textbf{dr}}$ is for work done along the path $\vec{\textbf{dr}}$ for the constant force $\vec{\textbf{F}}$. $\vec{\textbf{F}}$ is assumed to be constant for $\vec{\textbf{dr}}$.

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