I'm pretty much not so much introduced to calculus (I am grade 11 of India and they teach integration part of basic calculus by the end of grade 12) so I would be glad if the answer will be much more mathematically algebraic though it can't be fully algebraic, I understand. I was kind of confused, sometimes we say $$W= \vec{F}\cdot \vec{s}$$ and sometimes when we are calculating work done along a curve $C$ by a variable force we use $$W=\int_C \vec{F} \cdot d\vec{s}$$ which calculates work along a path.

So, is work in general about path not about displacement i.e. if for a constant force magnitude along a curve, we take integral of only force and multiply it by distance travelled, will the answer be same.

up vote 2 down vote accepted

Yes. Work is path dependent. Work is defined by the integral: $$W=\int_C \vec F \cdot d\vec s$$ Where $C$ is the trajectory from x($t_1$) to x($t_2$). Trajectory refers to the path the particle takes. It's confusing because the integral includes the displacement vector $\vec s$ which is path independent.

$W=\vec F \cdot \vec s $ is used only for when the force is constant in magnitude and direction along the path. It's usually the equation students use when they first learn about work as it's easier to use when the students aren't skilled in integration.

If you think about moving a rock; you are more tired after moving a rock from A to B and back to A than you are after moving it from A to B and stopping. Even though your displacement in the first scenario is zero, you did more work.

  • It's not correct to say that "$W = \vec{F} \cdot \vec{s}$ is used only for straight-line trajectories". There are straight-line trajectories for which this is not true, and there are curved trajectories for which this is true. The correct statement is that $W = \vec{F} \cdot \vec{s}$ if the force $\vec{F}$ is constant (in magnitude and direction) along the path. – Michael Seifert Mar 11 at 15:49
  • That is correct, will edit. – Niamh O'SS Mar 11 at 15:51

If the force is constant ( i.e. not changing direction and magnitude) then we could just use $Work=Force.Displacement$ .

This would be the same Work if you integrate $\int F.dr $ as the force is constant and it will come out of the dot product and the integration. And you will only be left with the integral of $\int dr$ which is just the displacement.

If the force is not constant (i.e. changing direction or magnitude) then we cannot use the above formula as the work at each instant is not the same. Here you have to integrate. F will not come out of the integration.

  • So, work in general is path dependent? I mean in the former case displacement came int0 scene because in that case displacement and distance were same. But in the line integral formula we calculate work for the path i.e. distance travelled.So work in general is not about displacement but about distance. Am I correct? – Moti Rattan Gupta Mar 11 at 14:04
  • Take for example friction, If you push a block 10m to the right friction does -12J work(say). And then you push it back 10m to the same position friction again does -12J work. So the total work done by friction is path dependent. – SmarthBansal Mar 11 at 14:08

To answer this question we must consider the type of force which does the work. The force is conservative, it is path independent and for non conservative force it is path dependent. For example gravitational work is a conservative work and frictional force is a non conservative force.Hope this clear your doubt

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