I have a question with the formula power = energy / time. I know it's a simple formula, but when substituting into this formula,

  1. Does the energy have to be in joules or watts (no kJ or something like that)?
  2. Does time usually have to be in seconds?

4 Answers 4


It depends on what you are trying to measure, the constants you're using, and the formula you are using.

From a dimensional standpoint, you need to know what power is. $$\text{Power} = \frac {\text{Energy}}{\text{Time}}$$

You can expand on that and say $$\text {Energy} = \text {Force} \cdot \text {Length}$$ and keep going and substituting until you get $$\text{Power} = \frac {\text {Mass} \cdot \text{Length}^2}{\text{Time} ^3}$$

All you really need is for it to be able to convert properly in whatever system of units you chose.

In SI units for example, the base unit of power is measured in watts where $$\text{Power} (W) = 1 \ \frac {\text{kilogram} \cdot \text{metre} ^2}{\text{second} ^3}$$ There are also various definitions of horsepower such as mechanical horsepower measured in $$\text {Power} ({{HP}_{Mechanical})} = 33,000 \ \frac {\text {foot-pound force}}{\text{minute}}$$

Which can be coverted to watts and vice versa, because a foot pound can be converted to $\frac {\text { pounds (mass)} \cdot \text{feet}^2}{\text{seconds}^3}$ which are all still measures of $\frac {\text {Mass} \cdot \text{Length}^2}{\text{Time} ^3}$

So it doesn't really matter what format your values come it, it depends on what you're trying to measure (watts, kilowatts, horsepower, etc). As long as you know the values in some form, you can convert them into a unit which fits the formula for the result you need.

The best advice I can give with working with units like energy and power, is to convert them to base units like seconds, metres, and kilograms to make sure the relationships still hold if you are unsure. Knowing conversion factors, or where to find them, is also a must.


Watt is the SI unit of power not energy, Joule is the SI unit of energy. Watt is equal to joule per second so yes, if you want to calculate power in SI units, energy should be in joules and time should be in seconds. However, keeping time in seconds if you use kilo-joules for energy, power will be in terms of kilo-watt and likewise.


The equation you mentioned is the definition of Power: $$P = \lim\limits_{\Delta t \rightarrow 0}\frac{\Delta E}{\Delta t} = \frac{dE}{dt}$$ (the case above, the differential form, refers to the instantaneous power but the argument stays the same). Watt ($W$) is the SI derived unit of the power, not force and it is equal to Joule over seconds. $$[W] = \frac{[J]}{[s]}$$ In general every time you want to know how derived unit is made from "more basic" units use the definition and substitute the physical quantities with their units; this process can be done up to the fundamental units and it is called dimensional analysis. For example $W$ is also equal to $kg \cdot m^2 \cdot s^{-3}$.

P.S. Do not get confused by $W$, the unit of power, and $Wh$ (watt hour) a non standard unit for energy and work.

  • $\begingroup$ Hi and welcome to Physics.SE! I edited out some typos, such Newton for Joule, that were spoiling a decent answer. It's worth it being a bit more attentive when posting. $\endgroup$
    – stafusa
    Commented Sep 16, 2017 at 21:50

As long as your units are consistent, it doesn't matter.

If your energy is in joules and your time is in seconds, then the power is in joules per second also known as watts - that is the definition of a watt.

If your energy is in kilowatt-hours and your time is in hours then your power is in kilowatts.

If your energy is in Mars bars and your time is in heartbeats then your power is in Mars bars per heartbeat.

Power is just the rate that energy is used/transferred, in whatever units you prefer.

  • 1
    $\begingroup$ But you didn't specify if you meant an English Mars bar (nougat, caramel, and chocolate) or an American Mars bar (nougat, almonds, and chocolate). $\endgroup$
    – The Photon
    Commented Sep 16, 2017 at 21:53
  • $\begingroup$ @ThePhoton definitely the English one. Almonds, seriously? $\endgroup$
    – IanF1
    Commented Sep 16, 2017 at 22:09

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