Work done by a vector field (Force field) on a particle travelling along a curve 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?
 A: 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.
A: There are two things I think to consider why it might not always be that way:

*

*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


*Momentum. If we get it moving (or if it is initially moving) then we might be pushing some other direction to the motion.
A: 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.
A: 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 $\vec{\textbf{dr}}$.
