I have derived 3 expressions for the power that a force, $\vec{F}$, transfers to a body. (A) and (B) are consistent, but (C) isn't. Where have I gone wrong with (C)? EDIT: THE MISTAKE HAS NOW BEEN SPOTTED. (Explained below.)

(A.) The infinitesimal work done on a body by a variable force, $\vec{F}$, on a body as it moves over an infinitesimal displacement, $\delta \vec{r}$, in an infinitesimal time period, $\delta t$, is given by $$\delta W = \vec{F} \cdot \delta \vec{r}.$$

(This is only exactly true for when the path the body follows over the time interval, $\delta t$, is a straight path, and when the force, $\vec{F}$, is constant over this time interval and over its displacement. But as $\delta t \to 0$, this becomes a better and more accurate approximation.)

And dividing both sides through by $\delta t$, and applying the linearity of the dot product: $$\frac{\delta W} {\delta t} = \frac{\vec{F} \cdot \delta \vec{r}} {\delta t} = \vec{F} \cdot \frac{\delta \vec{r}} {\delta t}$$

And as $\delta t \to 0$, $\frac{\delta W} {\delta t} \to P$, the instantaneous power the force delivers to the body, and $\frac{\delta \vec{r}} {\delta t} \to \vec{v}$, the body's instantaneous velocity. $$\therefore P = \vec{F} \cdot \vec{v}$$

(B.) Generally, the work done on a body by a force, $\vec{F}$, is $$W = \int \vec{F} \cdot d\vec{r} = \int \vec{F} \cdot \frac{d\vec{r}}{dt}\, dt = \int \vec{F} \cdot \vec{v}\, dt$$ (I have had to assume the 2nd equality holds. I've seen an explanation as to why $\int f \frac{dx}{dt}\, dt = \int f \, dx$ but I haven't seen why it holds even when we have a vector $d\vec{x}$ and it is dotted with the integrand.)

The time derivative of the work the force does on the body is the power it transfers to it, $$P = \frac{dW}{dt} = \frac{d}{dt} \left( \int \vec{F} \cdot \vec{v}\, dt \right) = \vec{F} \cdot \vec{v}$$

The last equality holds because of the fundamental theorem of calculus.

(C.) In the specific case where a body follows a straight-line path, the work done by a force, $\vec{F}$, on it, over a displacement, $\vec{s}$, is given by $$W = \vec{F} \cdot \vec{s}$$ And the time derivative of the work that the force does is the power it transfers to the body: $$P = \frac{dW}{dt} = \frac{d}{dt} \left( \vec{F} \cdot \vec{s} \right) = \frac{d\vec{F}}{dt} \cdot \vec{s} + \vec{F} \cdot \frac{d\vec{s}}{dt} = \frac{d\vec{F}}{dt} \cdot \vec{s} + \vec{F} \cdot \vec{v}$$ In the 2nd equality we have applied the product rule for differentiating the dot product of two vectors. And notice that the time derivative of displacement from some fixed initial position is just the body's velocity, $\vec{v}$.

However, my problem is with the $\frac{d\vec{F}}{dt} \cdot \vec{s}$ term. It doesn't feel right because there is no reason why the instantaneous power a force transfers to a body could necessarily depend on it's displacement from some arbitrary chosen initial position. I cannot explain why this term would be zero in order for it to agree with the other results: the force, $\vec{F}$ doesn't necessarily need to be constant; the displacement from the initial position obviously won't always be zero and there is no reason for the vectors $\frac{d\vec{F}}{dt}$ and $\vec{s}$ to be necessarily perpendicular!

So I think I've made an error in (C). Can anyone see where I've gone wrong?


THE MISTAKE HAS BEEN SPOTTED NOW! As Puk has pointed out, for (C) to be correct - for $W = \vec{F} \cdot \vec{s}$ to really give you the work done by a force, $\vec{F}$, on a body following a straight-line path while it is displaced by $\vec{s}$, you need $\vec{F}$ to be constant!

(Not because otherwise the body can't follow a straight line path - even if $\vec{F}$ varies there can be other variable forces acting on the body to ensure there is no component of net force on it in any direction - other than parallel to the path - which will ensure it won't deviate from the straight-line path!) The reason $\vec{F}$ has to be constant is because this is how work done is defined: it is the dot product of a force acting on a body that stays constant for some time and the displacement it moves over that time. Think about it - because if $\vec{F}$ didn't have to be constant then how would you know which time to evaluate $\vec{F}$, in $W = \vec{F} \cdot \vec{s}$? There would be no reason to choose any specific time (e.g. at the start or end or middle of the 'journey' - or at any other time!) to evaluate $\vec{F}$ because it isn't mentioned in the definition!

Having $\vec(F)$ as constant resolves the inconsistency because then $\frac{d\vec{F}}{dt} = 0$ and the $\frac{d\vec{F}}{dt} \cdot \vec{s}$ term disappears.

  • $\begingroup$ There is no error, but the time derivative of a vector is the time derivative of it's components. As such, given that we know the trajectory is linear the derivative of each component will be null and so the whole term becomes 0. $\endgroup$
    – PhysH
    Jan 25 at 20:08

1 Answer 1


$W = \vec F\cdot\vec s$ is only true if the force $\vec F$ is constant through the whole displacement $\vec s$. If it is constant, then $d\vec F/dt$ vanishes and there is no discrepancy. If it is not constant, the formulation of work as $W = \int\vec F\cdot d\vec r$ as in (B) is the correct approach and yields the correct power.

A straight line trajectory does not mean we can dispense with the integral. It doesn't make much sense for the work done between time $t = 0$ and time $t = T$ to depend only on the force at $t=T$, and at no prior time. The integral serves to correctly "average" the force in time/distance.

  • 1
    $\begingroup$ This makes so much sense! Of course F has to be constant throughout the whole displacement for (C) to be true! (Or else which value of F would you choose to plug into W =F.s to find the work done over the displacement!) And that obviously leads to dF/dt = 0. Thank you!!! $\endgroup$ Jan 25 at 20:19

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