2
$\begingroup$

The derivative of work is $\bf F\cdot v .$ $$P(t)= \frac{\mathrm dW}{\mathrm dt}= \mathbf{F\cdot v}=-\frac{\mathrm dU}{\mathrm dt}\;.$$

But why not $$(\mathrm d(\mathbf F)/\mathrm dx)\cdot \mathbf x + \mathbf{F\cdot v}\;?$$

$\endgroup$
1
$\begingroup$

For a force to do work, it must act upon an object as that object moves some distance (with a component along or opposed to the direction of that force). Holding an object stationary, but carrying the strength of a given force, adds no more work done. Because of this, we say that

$$W=\int_C\vec{F}\cdot d\vec{r}$$

where C is the curve in space along which the object moves. If power is the rate of change of work, then we would expect a product rule to be applied in the derivative, but no power is expended by a force that does no work, and without a displacement, there is no work and no power.

$\endgroup$
1
$\begingroup$

This is because $P(t)$ as stated is the instantaneous power as a function of time and $W =\mathbf F\Delta \mathbf x$ holds only for constant forces. More generally, recall that a definition of work is the integral:

$$W = \int_C\mathbf F(x)\mathrm d\mathbf x$$

Where $C = C(x,t)$ is some curve in space/time. Expressing in terms of time gives:

$$W = \int_t\mathbf F(t)\frac{\mathrm dx}{\mathrm dt}\mathrm dt = \int_t\mathbf F(t)\mathbf v(t)\mathrm dt$$

From the Fundamental Theorem of Calculus, we have:

$$P \equiv \frac{\mathrm dW}{\mathrm dt}= \mathbf F(t)\mathbf v(t)\;.$$

Apologies for sloppy notation, but that's the idea.

$\endgroup$
  • $\begingroup$ Thanks for answering :) , but i have a follow up question $\endgroup$ – Joe Feb 3 '16 at 6:22
  • $\begingroup$ what if the criteria is given that a body moves under constant power , which means derivative of work with respect to time is constant ,so shouldnt we include in that derivative (according to product rule ) - the derivative (d(F)/dx).x $\endgroup$ – Joe Feb 3 '16 at 6:25
  • $\begingroup$ kb.osu.edu/dspace/bitstream/handle/1811/2458/… this is article im referring to $\endgroup$ – Joe Feb 3 '16 at 6:27
  • $\begingroup$ Hi Joe, the above equation holds whether or not power is constant. The problem is that you only write W = Fd when force is constant (and thus has a vanishing derivative. When force is not constant, you have to resort to the integral definition and in this case, the dF/dx term does not pop up. By the way, thanks for the link. I especially liked the phrase: "... In these days of steam, electric and gasoline transportation..." :) $\endgroup$ – Orko Feb 4 '16 at 6:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.