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I've been trying to solve a problem for some time. I have been given conflicting information by both literature, colleagues and people on this very forum.

It's a very simple question: What is the definition of power?

In particular, I would like to know how constant power would accelerate a constant mass over a fixed distance, assuming that no resistive forces are at work, i.e. no friction, no air resistance, etc.

I know that power is the rate at which work is done with respect to time, i.e. $P = \operatorname{d}\!W/\operatorname{d}\!t$.

One formula that I have seen is that $P = Fv$, i.e. power is the product of force and velocity. According to Wikipedia, this requires $v$ to be constant. Someone on this site, and someone I work with have both said that I need $v$ to be constant for $P=Fv$ to hold.

However, I found an article from 1930, which found equations of motion for constant power. Just like the standard equations of motion $v=u+at$, $v^2=u^2+2as$, $s=ut+\frac{1}{2}at^2$, and $s=\frac{1}{2}(u+v)t$ all assume a constant force, i.e. a constant acceleration, L. W. Taylor (1930) found six analogous formulae for motion under constant power. He used $P= Fv$ as his definition for power.

Can I work with $P=Fv$, assuming that $F$ and $v$ vary, provided their product is constant?

Fly by Night
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2 Answers2

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The equation $P = Fv$ always holds, even when neither the force, the velocity nor their product are constant. If you write:

$$ P(t) = F(t)v(t) $$

Then this is true at all times $t$ and gives the instantaneous power at the time $t$. Obviously if the velocity and/or force vary with time then the power will also vary with time.

John Rennie
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    Damn, Josh's answer is better, as usual :-) – John Rennie Sep 24 '13 at 17:41
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    John don't be ridiculous. Our answers are exactly the same not to mention I pilfered an old post of mine which is pretty shady :) – joshphysics Sep 24 '13 at 17:49
  • How can one justify $P = Fv$, when $P = \dot{W}$ and $W = Fs$. The product rule gives $\dot{W} = \dot{F}s+F\dot{s} = m\dot{a} + Fv$ Don't we need $\dot{a}=0$, i.e. $v$ constant? – Fly by Night Sep 24 '13 at 18:17
  • @FlybyNight: this is discussed in the question Josh linked, http://physics.stackexchange.com/questions/77935/why-isnt-the-product-rule-used-in-the-definition-of-mechanical-work. – John Rennie Sep 25 '13 at 05:44
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There is a precise definition of power in the following question: Why isn't the product rule used in the definition of mechanical work?. Here is a reproduction of that text:

For a particle traveling along a parameterized curve $\vec x(t)$ and under the influence of a force $\vec F(\vec x,t)$ which is explicitly dependent on both position in space and time, the work performed on the particle by this force from a time $t_0$ to a time $t$ is defined as follows: \begin{align} W_{t_0}(t) = \int_{t_0}^{t} \vec F(\vec x(t'),t')\cdot \dot{\vec x}(t') \,dt' \end{align} Note that the expression on the right is often written $\int \vec F\cdot \vec dx$, but this is really schematic, the the mathematically precise definition is what I have written above in terms of a parameterized path with an integral over some range of parameter values. The definition of instantaneous power is then \begin{align} P(t) = \dot W_{t_0}(t) \end{align} Taking the derivative of both sides with respect to $t$, and using the fundamental theorem of calculus, we obtain the desired expression for the power \begin{align} P(t) = \vec F(\vec x(t),t)\cdot \dot{\vec x}(t) \end{align}

Notice that the expression $\vec F\cdot \vec v$ is the instantaneous power for arbitrarily changing force and velocity. There is not even a restriction on the constancy of their product.

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joshphysics
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