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I came across a multiple choice A level past paper question (CIE June 04 paper 1 Q15) asking to choose which of the following defined power:

15 What is the expression used to define power?
(A) $\frac{\text{energy input}}{\text{energy output}}$
(B) $\text{energy × time taken}$
(C) $\text{force × velocity}$
(D) $\frac{\text{work done}}{\text{time taken}}$

The answer given was (D), but I am perplexed as to why (C) was not considered correct.

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    $\begingroup$ Maybe they did not accept $F \times v$ because they wanted to emphasize $F \cdot v$ (only the component of force in the direction of velocity counts). Doesn't seem like a good question. (Maybe they even expected you to interpret $\times$ as the cross product, which doesn't seem fair since multiplication would make more sense and be closer to right). $\endgroup$
    – causative
    Aug 9, 2021 at 11:52
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    $\begingroup$ An electric oven or a microwave both use electrical power. What is the "force" and "velocity" that you could multiply together to calculate the power of those devices? $\endgroup$
    – alephzero
    Aug 9, 2021 at 12:56
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    $\begingroup$ I agree with causative ... this is a poorly written question. $\endgroup$
    – David White
    Aug 9, 2021 at 17:10
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    $\begingroup$ Long ago when I did my A-levels, they would emphasise the vector/scalar distinction between the terms "velocity" and "speed". So I'm inclined to agree with @causative, except that I think it's a deliberately pedantic question, which is a reasonable thing for them to have set within the frame of reference of A level physics as it's taught. Had they instead worded C as force times speed, I suspect it could have been offered as the right answer to a question (with no competing option for work divided by time). $\endgroup$
    – tea-and-cake
    Aug 9, 2021 at 22:12
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    $\begingroup$ @Nat -- A levels, or GCE Advanced Levels, are qualifications issued in England and Wales (and as an option in Scotland) for students typically aged 16-18, in between GCSE (high-school qualifications) and undergraduate degrees. Most A-level physics students will also take a single A-level in maths, but there's the option to do "double maths", ie maths and Further Maths as two separate A-levels... $\endgroup$
    – tea-and-cake
    Aug 10, 2021 at 9:19

8 Answers 8

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Both seem to be correct, as another answer also points out. Meaning, both formulas can be used to calculate power.

But the fundamental definition of power is the latter, namely energy per time.

In this general definition, power can be calculated for mechanical systems, thermodynamic systems, electric system etc. The former version includes force and might thus be harder to use in, say, thermodynamic problems.

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  • $\begingroup$ Force and velocity are vectors. At A level, it doesn't look like vector multiplication is defined, so $F \times v$ cannot be used to correctly calculate power. If it could, it'd be $F \cdot v$ and you'd get different answers even in one dimension, for example, if you apply the brake and the accelerator. $\endgroup$
    – user121330
    Aug 10, 2021 at 20:49
  • $\begingroup$ It is funny that in Russia we were taught to think of force as being "more fundamental". The definition of force we were given is "мера взаимодействия между телами". I studied in Physical-Mathematical School (ФМШ) #146 with profound study of Physics, Mathematics and Informatics (Perm, Perm Krai, Russia). Different kinds of energy and work were defined in context and eventually we ended up talking about SI and integration. The point of forces being fundamental was that even temperature was perceived as "how fast particles were moving". Warning: the author had solid 60 for his Physics. $\endgroup$
    – Dmitrii Demenev
    Aug 13, 2021 at 2:34
  • $\begingroup$ @DmitriiDemenev Force might be more fundamental than energy, sure. I wouldn't necessarily disagree. (I guess this is also depending on opinion and not set in stone.) But what is more fundamental is not necessarily more useful. $\endgroup$
    – Steeven
    Aug 13, 2021 at 5:11
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It's because the question asks for the 'definition'

Definition of power

so D is the answer, even though C is a valid formula.

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    $\begingroup$ C is a valid formula in mechanics. But it is not a generally valid formula. It doesn't apply for electric power. $\endgroup$
    – md2perpe
    Aug 10, 2021 at 21:05
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D is the correct definition. C is valid only for mechanics. For instance, you cannot find the power of electrical appliances using $P=Fv$. Though the definition in option D is always correct. More precisely, D is the definition of Power while C is a derivation.

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I am also perplexed. Both are correct, but $P=\vec F \cdot \vec v$ is better, in my opinion, because it gives you the instantaneous power. In contrast $P=\Delta W/\Delta t$ gives you only the average power.

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    $\begingroup$ Agreed. This seems like an instance of poor teaching. It is more elegant in classical mechanics to define the power as ${\bf F} \cdot {\bf v}$ for a single particle. Another user pointed out that the work definition may be extendable to thermodynamics, which may explain the pedantry. Though still, the definition doesn't make sense as power is an instantaneous quantity. $\endgroup$
    – Evan
    Aug 9, 2021 at 15:57
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    $\begingroup$ @Evan Power isn't just for forces pulling things through a distance, it's used for heat, light, electrical energy etc...the 'definition of power' is option D. $\endgroup$
    – John Hunter
    Aug 9, 2021 at 17:52
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    $\begingroup$ I agree. But given the context of Option C, where forces and velocities are mentioned, I assume the class is about Classical Mechanics. In that case, option D does not reconcile with Classical Mechanics, since power is an instantaneous quantity. Perhaps if the question asked about average power it would be appropriate. $\endgroup$
    – Evan
    Aug 9, 2021 at 17:53
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    $\begingroup$ The fact that finite-difference ratio gives only the average power is not a problem whatsoever, given that for any tolerance $\varepsilon>0$ there exists $\Delta t(\varepsilon)$ such that the average power is equal to instantaneous power with error bound of $\varepsilon$ (this is just the definition of limit, paraphrased). $\endgroup$
    – Ruslan
    Aug 10, 2021 at 21:54
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    $\begingroup$ If $\lim\limits_{\Delta t\to0} \frac{\Delta W}{\Delta t}$ doesn't exist, your product won't either, since they are the same quantity. $\endgroup$
    – Ruslan
    Aug 11, 2021 at 11:01
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The strong definition of power is input times output, or, flow times effort. This is valid in many physical domains, electrical, mechanical, etc. The reason that flow (velocity, current, or any other rate of changes) is selected as the output is because it gives us real values (pure dissipation). Position and acceleration contributes to complex values (phase shift only and no dissipation, meaning no work is actually done).

It is definitely a bad question.

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  • $\begingroup$ I'm not sure what the units of input, output, flow or effort are. If input and output have the same units, you've got a very strange system for it to give power as a result. $\endgroup$
    – user121330
    Aug 12, 2021 at 22:50
  • $\begingroup$ The units are of course different. In a mechanical system: the input is force and the output is velocity. In an electrical system: the input is voltage and the output is current. And so on. However, in a mathematical model, we do not care the units. We care about the model. That's why it is the mathematical definition of power. Please see: bondgraph model which is actually a port-controlled Hamiltonian system. For example from here: groups.csail.mit.edu/drl/journal_club/papers/… Take a look at page 2. $\endgroup$
    – manurunga
    Aug 13, 2021 at 23:48
  • $\begingroup$ Interesting way to look at it! Not sure the value of this perspective at this level. Also, any complex valued system can be mapped back to a real-valued system - position and acceleration typically being real-valued vectors. $\endgroup$
    – user121330
    Aug 14, 2021 at 23:26
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Within the frame of reference of A-level physics, I think they want you to reject "force × velocity" because they're expecting you to associate the term "velocity" only with a vector quantity, not with speed. This is a pedantic point of terminology that they certainly emphasised in my day.

They've used the division sign "÷" in answer D, so in that context I reckon there's no doubt that, by the multiplication symbol "×" in answer C, they mean some sort of elementary multiplication. Either the product of two scalars, or of one vector and one scalar. They want you to notice that you can't multiply (in that sense) two vectors like force and velocity, and so reject that answer. But even if you interpret it to mean the cross product instead, that makes the answer even wronger.

Within the domain of A-level physics as taught, I think "force × speed" would have been an acceptable answer (with answer D correspondingly replaced by something wrong).

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  • $\begingroup$ Power, being a scalar, can never be $F \times |v|$ because that would be a vector as well. $\endgroup$
    – user121330
    Aug 10, 2021 at 20:36
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Power is "defined" as work done per unit time , so option (d) is correct.

Now coming to the third option. Look the instaneous power can be given as

$$P=\frac{dw}{dt}=\frac{d(\vec F\cdot \vec s)}{dt}= \vec F\cdot \vec {v}+ \vec {s}\cdot \frac{\vec {dF}}{dt}$$

So power = force × velocity is incomplete . May be that's why they made the third option wrong.

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  • $\begingroup$ I believe you meant $F\cdot v + s \cdot \dot{F}$? In any case this doesn't quite make sense because $\vec{s}$ is not a well-defined instantaneous quantity $\endgroup$
    – Kai
    Aug 9, 2021 at 21:04
  • $\begingroup$ @Buraian thanks for pointing out that silly mistake.. $\endgroup$
    – Ankit
    Aug 10, 2021 at 5:30
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Velocity and force are both vectors (in this case, a number each for forward, left and up). The standard ways to multiply vectors are $F \cdot v$, which yields a scalar, the cross product, $F \times v$, which yields a vector and an outer product, $F \otimes v$, which yields a matrix.

This site on A levels covers vectors and there's no mention of how to multiply. Since they didn't say how to do it, don't. FYI, dividing vectors is meaningless and illegal forever. This site makes a baffling and unforgivable abuse of notation describing power in terms of velocity and force, but it's clearly a derivation.

If you're going to mention the $F \cdot v$ version of power, the germ of vectors must be included: it's the difference between the accelerator and the brake (ignoring the fact that the friction between the wheels and the road move the car).

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