Definition of Power


Introduction

We say power is one of the three things people generally want, along with money and happiness. However, the way we think about power is more analgous to thinking about force or respect, rather than actual power. In fact, horses and oxen have much more "power" than humans. There's a reason why car engines are often measured in horsepower and not manpower!

Power is actually the rate at which something does work, or the rate at which it can change the energy of a system. Both definitions are pretty much analogous and can be referred to interchangeably. The SI unit for power is the watt, after the "inventor" of the steam engine, James Watt.

Figure 1: Which one has more power? (Hint: A person produces around $100 ~\textrm{W}$, while the car produces around $100 ~\textrm{kW}$.)


If we don't want to invoke calculus, we have to deal with the average power exerted on a system. This is much like the other average values we've dealt with before and involves the change in the energy in the system over the change in time. It's also equivalent to the amount of work done over the time interval. Most of the time, this is enough.

$$ P = \frac{\Delta E} {\Delta t} = \frac{W}{\Delta t} $$
The more accurate definition of power involves a derivative. Shocking, I know. Instantaneous power is the time derivative of energy or work with respect to time.

$$ P = \frac{dE}{dt} = \frac {dW}{dt} $$
Most of the time, we will be dealing with constant power, so we don't need to worry too much about the effects of a varying power. IAfter all, most engines and motors have a consistent power output rather than a varying one. Moreover, in a lot of problems we only really care about the average power, as figuring out the instantaneous power would require quite a bit of work. There is actually a formula for instantaneous power not requiring calculus that is really useful, however.

$$ P = \vec{F} \cdot \vec{v} $$
The exact reasoning for why this is true requires a bit of calculus, but we can understand it another way. Recall that power could be defined as the rate at which work was done on something. But, work is equivalent to the dot product of force and displacement. The rate of change of displacement is equal to the velocity, which is how we get that the power is equal to the dot product of the force and velocity of an object. This can easily be shown to be true with a bit of basic calculus. Recall how power could be defined as $P = \frac{dW}{dt}$. Work is equal to the dot product of the force and displacement. This allows us to just distribute the derivatve inside and differentiate the displacement with respect to time. This gives us the aforementioned formula since the time derivative of position is velocity. Simple, really.

You will find that this concept is rather simple, which is why sometimes problems will arbitraily add difficulty by asking about power in a roundabout way. For instance, they may ask what the rate of change of the system's energy is. However, if you have a good handle on what power is, this should be no issue.

We have already covered all the concepts behind what power is, so we can move on to a practice problem. This is probably the best way to explain what power actually is, because the actual defintion can be a bit vague and mystical until you actually apply it to a scenario.
A M1 Abrams main battle tank attempts to climb a 30° slope in order to reach a target. The tank has an engine with 1500 horsepower, which is equivalent to around $P = 1100 ~\textrm{kW}$ of power, and has a mass of $M = 66800 ~\textrm{kg}$. Watt is the maximum speed it can climb this slope at (Edward made me put this pun in)?

Figure 2: Several tons of raw freedom gaining elevation.

It might not be obvious where to start with a problem like this, so let me walk you through the process a bit. Begin with considering that at the maximum speed, the object isn't accelerating anymore, so the forces on it must be balanced. On an inclined plane, the component of gravity down the incline seeks to pull the tank back down, so the force that the engine provides (actually, friction is what's technically providing this force, but this wouldn't be possible without an engine) must be upwards and equal.

$$ F = Mg \sin \theta = \dfrac{Mg}{2} $$
Now that we know the force that the engine is producing, we can find the velocity. Recall our formula:

$$ P = \vec{F} \cdot \vec{v} = Fv $$
We can drop the dot product because both the force and the velocity are directed up the incline. This gives us a simple equation that lets us solve for the velocity of the tank.

$$ v = \dfrac{P}{F} = \dfrac{2P}{Mg} = \bbox[3px, border: 0.5px solid white]{3.36 ~\textrm{m/s} } $$
If this number seems a bit low, remember a few things. This is still around 7.5 miles per hour, which is a respectable speed for a thirty-degree incline. That brings me to my second point: the incline is thirty degrees. This is a very steep incline and not something you'll often encounter in the real world. Moreover, tanks are designed with protection in mind and contain a lot of heavy armor, making their overall power-to-weight ratio lower than that of your average civilian car. At the end of a day, folks, it's a literal tank.
There's a classic power problem with a car going up an incline. I think it's a good problem, but I'm going to up it a notch by turning the car into an M1 Abrams main battle tank. We'll say this armored monstrosity has a mass $M$ and its engine has a power $P$. It's going up an incline with incline angle $\theta$. Watt is the max speed it can achieve (Edward forced me to write in this joke)?

Figure 2: Several tons of raw freedom gaining elevation.

Notice how we're specifically asked about velocity. This should ring a bell, since one of the formulas I gave for power earlier had velocity in it. I'll write it again here:

$$ P = \vec{F} \cdot \vec{v}$$

We have the power, but we don't know what the force $F$ should be. We'll be forced to use old force analysis techniques to figure this out. At the maximum speed, the tank no longer accelerates, meaning the forces on it are balanced. We don't have to deal with the vertical direction because it's the same as usual, but in the horizontal direction the force exerted by the engine has to equal the component of gravity down the incline. Thus, we write:

$$ F = Mg \sin \theta $$
Now, we have everything we need to solve this problem. We know the force and velocity are in the same direction, so the previous forumla simplifies to $P = Fv$. Now, we just need to do some substitution.

$$ v = \dfrac{P}{F} = \bbox[3px, border: 0.5px solid white]{\dfrac{P}{Mg \sin \theta} } $$
There we have it! Our answer is complete. That wasn't so bad (even though I did make you do some actual algebra), was it? This is about as hard as power problems realistically can get, since the concpet is very simple compared to some others like force analysis and maybe even energy conservation that we've covered in the past.

This is a short lesson because there is not too much to cover on power. You'll likely see this concept pop up again in the future whenever we have discussions of systems with changing amounts of energy, but overall it's a really simple concept that just requires understanding and applying simple formulas. The next section also only contains one concept, but that's where the similarities end. It's a bit of an abstract concept, but with some effort you'll definitely understand it. If you're ready to dive back into gravitation, move on to the next lesson!