Say we have a waterfall and we know that 10kg of water falls over the ledge per second. We are also given that the acceleration due to gravity is 10m/s^2, and the distance to the ground is 5m. We are asked to calculate the power of the waterfall.

The way that I was taught is this: We know the acceleration and the distance, so we can calculate the time the waterfall takes to hit the ground by using $$d = \frac{1}{2}at^2$$ In this case, t = 1. From this, we know how much kg of water falls over in t seconds. This this case, 1s*10kg/s = 10 kg. Work is equal to Fd = 10*10*5 = 500. This work is done in one second, so power = 500 Watts.

I happen to disagree with this. Water is continuously flowing. This means that at t = 0, barely any water is over the ledge. At t = 1, the first drops of water that went over the ledge finally hit the ground. Furthermore, at t = 1, the final drops comprising the 10kg of water have barely gone over the edge. A diagram at t = 1 would look roughly like this:

We will assume that the drops have a uniformly divided mass (ie. 5 drops means each drop is 2kg)The thing to note here is that d varies for each drop. For the rightmost drop, d = 5. But for the others, d < 5. In comparison, if we were diagraming the 500W case, then we would have one big drop at the very bottom and therefore, d always equal to 5. For me, intuitively, I would think that that would mean a discrepancy in work done, and consequently power. Specifically, the power of the waterfall is < 500 Watts and probably by a good margin too. I also the best answer is something that involves limits, because the more water droplets we have, the more accurate our answer is. Again, this is just intuition.

Is my intuition correct? If not, why?

• Are you extracting the energy from the drops that haven't made it to the bottom yet? Commented Nov 18, 2017 at 6:39
• @DJohnM To be quite honest, I'm not sure. In this example I was thinking of Power as Work/Time. Commented Nov 18, 2017 at 21:09

Well, if there is nothing being said about how the energy should be extracted or that there is friction, there is no need to use the kinetic energy of the water that has already fallen down. So just use $E_{pot}=m*g*h$ and then use the info you have about the mass flow to calculate Power. $P=E*t$