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I have some questions regarding the physical meaning of the units.

  1. The unit of Planck's constant $h$ is J$\cdot$s= kg$\cdot$m$^2\cdot$s$^{-1}$ in the SI system. My question is: When multiplying by m$^2$ what is the meaning of kg$\cdot$m$^2$? As $\frac{kg}{m^2}$ is kilogram per square meter. So, can we write J$\cdot$s= kg$\cdot$m$^2\cdot$s$^{-1}$ = kg$\cdot$m$^4\cdot$s$^{-1}\cdot$m$^{-2}$? The same question for m$^4$ or m$^3$.

  2. Joule is a unit of energy, J= kg$\cdot$m$^2\cdot$s$^{-2}$. What does per s$^2$ mean? Per s and per s ... per two s? But s$^2 \neq 2$s.

  3. Watt is a unit of power, W= kg$\cdot$m$^2\cdot$s$^{-3}$. What does per s$^3$ mean?

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These units look strange, because they are made up of many simpler quantities. So let's start simpler and look at velocity and acceleration first. The SI units of velocity are $[v]= m/s$, since it denotes a distance travelled over time. Acceleration is the change of velocity with respect to time so it will have units of velocity, divided by units of time $[a]= [v]/s = m/s^2$. I think this should answer your second question. If an object is accelerating at $1 m/s^2$, it means that every second, the object's velocity changes by $1 m/s$ every second. That's the meaning of $s^{-2}$.

Now, before we move on to work, let's look at Newton's second law of motion

$$F=ma.$$

Since acceleration has units of $[a]= m/s^2$, then force has units of $[F]= [m] \,[a] =kg \times m/s^2 $. This quantity is called a Newton, $N= kg\times m/s^2$. It means that an amount of force equal to $1N$ would accelerate an object of mass $1kg$ by a velocity of $1m/s$ every second.

Now we are ready to look at the definition of the Joule. Its value comes from the equation for work $W$, which is a quantity measured in joules.

$$W=Fd,$$

where $d$ is the distance travelled due to some amount of force $F$. So one can see that the units of work are $[W]= [F][d] = N\times m=kg\times m^2/s^2=J$. So you can see that the interpretation in terms of force is easy - A joule is the amount of work done by applying a force of $1N$ over a distance of $1m$. However, thinking of it in more elementary terms, things become more opaque and harder to interpret, but it's not impossible. One can also think of the joule as the energy equivalent to travelling a distance (in metres $m$) by an object of mass (in kilograms $kg$) with an acceleration $a$ (in units $m/s^2$). And this is the interpretation of $m^2$ in the definition of $J$. It has to do with an object travelling a certain distance with a certain acceleration, not an area.

Finally, the definition of power is the rate of change of work $$P= \frac{dW}{dt}.$$

From here we can again deduce that the unit of power will have units of work (energy) divided by units of times $[P]=[W]/[t]=J/s = kg\times m^2/s^3$. One can again start to take the equation apart and think of the change in acceleration due to a force, which pushes an object to some distance over some period of time and it all becomes very complicated.

This is why we define these compound units in the first place. They save you the trouble of thinking about these ideas in an overly complicated reductionist way and they allow you to think instead about concepts such as energy, force and power, instead of objects being accelerated over a distance over a period of time...

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  • $\begingroup$ Great explanation. Thank you for your time. $\endgroup$
    – Wabram
    Jul 2, 2020 at 10:12
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Basic interpretation of Plank constant is that it is "uncertainty quanta", which is best seen from uncertainty principle : $$ \Delta E \Delta t \geq \frac{\hbar}{2} $$

If to be expressed numerically it would be :

$$ \Delta E \Delta t \geq 5.273 \times 10^{-35} ~[Js]$$

Which means that, if you have particle detection time uncertainty of 1 second, this will result in particle energy minimum uncertainty on the order of $10^{-35}~\text{Joules}$. And in reverse,- if you detected particle energy with energy uncertainty of 1 Joule, then time window when particle has had that detected energy will have minimum time duration uncertainty of $10^{-35}~\text{Seconds}$.

EDIT

There may be other interpretations of Planck constant. One is a minimum action, which is defined as integration of Lagrangian over time : $${\mathcal {S}}=\int _{t_{1}}^{t_{2}}L\,dt$$ Action has same units $[Js]$, so you can think about Planck constant as "action quanta". Also same units is for definition of angular momentum, which is defined as : $$ L=I\omega $$ In this case Planck constant corresponds to "spin quanta".

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    $\begingroup$ It's a new "dimension" for me. Kindly thank you for your time. $\endgroup$
    – Wabram
    Jul 2, 2020 at 10:28
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You can consider the kg$\cdot$m$^{-2}$$\cdot$s$^{-1}$ as a flux, how much mass passes per m$^{2}$ per second. For s$^{-3}$ you can go on and say it's the flux of the flux ... but you don't really have to. Sometimes it isn't worth it to try find a physical meaning for these units except that they help you in dimention analysis to check your equations or anticipate some laws.

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