What does the formula $$ {\bf F} ~=~ \frac{\mathrm{d}W}{\mathrm{d}{\bf x}}$$ actually mean? Here $\mathrm{d}W$ is the work done in a small period, ${\bf F}$ is the force and $\mathrm{d}{\bf x}$ is a symbol for arbitrary displacement.

How and where do we use this formula? Suppose a body moves in the path of a rectangle and starts from the bottom left corner say a then moves right up left and down coming to the same position. So logically one can see that the work is zero but can we incorporate this formula? If so, how? Do we resolve the force?

Basically, can we use this formula to calculate if the energy is conserved or there is no work? If yes, then how?


2 Answers 2


You should think of the formula the other way around, i.e.

$$ \mathrm{d}W = F\mathrm{d}x$$

which means that the infinitesimal work done along an infinitesimal path is just the force $F$ times the length $\mathrm{d}x$ of the path along which the force was exerted.

If we are now given a real path $\gamma : [a,b] \to \mathbb{R}^3$, the total work done along that path is given by the line integral

$$ W = \oint_\gamma \vec F \cdot \mathrm{d}\vec x$$

For a straight path with length $l$, parallel to which a constant force $F$ acts, this reduces to the more familiar

$$ W = F \cdot l$$

Now, it can be shown that, if the integral above depends only on the start- and endpoints of the path $\gamma$, and not on the specific path taken between them, then the force we are looking at is a conservative force, which means that it can be written as the gradient of a scalar potential as

$$ F = - \nabla U$$

and, in one dimension, this reduces to

$$ F = -\frac{\mathrm{d}U}{\mathrm{d}x} $$

so forces which do not depend on the path taken can be written as the derivative of a potential $U$, which we then call the potential energy associated to that force.

  • $\begingroup$ so lets us say there is a particle which travels we dont know where all but if its displacement is zero then the work done is zero,right? $\endgroup$
    – geek101
    Aug 15, 2014 at 14:02
  • $\begingroup$ @user166748: For a conservative force (which are most forces which are not friction, in the usual contexts), that is correct. $\endgroup$
    – ACuriousMind
    Aug 15, 2014 at 14:03
  • $\begingroup$ where is it not correct then? it would be great if you could give me an example. also this line integral is when there is a closed path? $\endgroup$
    – geek101
    Aug 15, 2014 at 14:06
  • $\begingroup$ @user166748: I used the symbol for a closed path, but it is meant for closed and open paths. One example for a force where this is not correct is Stokes friction. It depends on the velocity, and thus on the path taken, and thus the work done will not be zero for every imaginable path (I think). $\endgroup$
    – ACuriousMind
    Aug 15, 2014 at 14:13
  1. This formula is used for conservative forces like gravitational force.
  2. This formula is used when the force is not constant i.e. variable force.
  3. This formula conveys that conservative force is equal to negative potential gradient.
  4. This formula establishes relation between a vector quantity and a scalar quantity(PE).

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