Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

up vote 2 down vote accepted

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.

share|improve this answer
    
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
    
Dear joshphysics, it is usually frown upon to directly copy-paste identical answers. (The problem is if everybody start to copy-paste identical answers en mass.) In general in such situations, please consider one of the following options: (i) Delete one of your answers. (ii) Flag for duplicate posts and delete one of your answers. (iii) If you think the two posts are not duplicates, then personalize each answer to address the two different specific questions. –  Qmechanic Sep 24 '13 at 18:36
    
@Qmechanic Yes I totally understand. In this case, which I think falls under (iii), I copy/pasted part of a previous answer and did, in fact, add a bit to personalize. I certainly could have added more, but it's not entirely clear to me how much more would deem my response here appropriate. Also, is copying any component of a previous answer frowned upon in itself? Perhaps I should move these questions to meta... –  joshphysics Sep 24 '13 at 18:51
    
@FlybyNight The question that I linked to is an answer to precisely this product rule concern. The main point is that $W=Fs$ is not the general expression for work, so even before you attempt to use the product rule, you are doing something wrong. –  joshphysics Sep 24 '13 at 18:53
    
@joshphysics: I worry that the copy-n-paste quote style of self-reference would become commonplace. Copy-paste of an equation is OK. Copy-paste of text less so. Alternatively, in such cases, instead of posting an answer, one could leave a comment pointing to the other post, and say that even though the question is different, the answer is essentially the same. –  Qmechanic Sep 24 '13 at 18:57

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.

share|improve this answer
3  
Damn, Josh's answer is better, as usual :-) –  John Rennie Sep 24 '13 at 17:41
1  
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, physics.stackexchange.com/questions/77935/…. –  John Rennie Sep 25 '13 at 5:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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