1. The problem statement, all variables and given/known data Water is pumped steadily out of a flooded basement at 5.0 m/s through a hose of radius 1.0 cm, passing through a window 3.0 m above the waterline.What is the pump’s power? 2. Relevant equations Volume and mass flow rates are constant [tex]R_v = Av; \quad M_v = \rho A v[/tex] Work-energy Theorem [tex]W = \Delta E = \Delta U + \Delta K[/tex] 3. The attempt at a solution I've seen one solution leading up to the same answer as in the back of the book, 66 Watts. It goes by imagining a small chunk of water of mass Δm and the work done on it by the pump of a time Δt. Several attempts at that method can be found here: https://www.physicsforums.com/threads/power-of-a-pump.551946/ However, this seems odd to me. I would think that the water in the hose from the start should be handled differently from the water in the basement that is not (yet) in the hose. The water at the very top of the hose doesn't need to go anywhere, and the water only a little bit below it only needs to go a small distance and then it vanishes. The water at the bottom of the hose needs to be lifted a distance of 3m. The water not yet in the hose just sits there until space is made for it and then only makes it up the hose a certain distance. My thought about this was to think of, again, a small chunk of water and the work done on it, but to split the two cases where it's in the hose and where it isn't yet. To model the work done on the water I reason that, in time Δt at a height h from the water level, the work done on the water is [tex]\rho \pi r^2 g(3-h)\Delta h[/tex] as this accounts for the linear variation of the amount of height that the water in the hose has to travel. This would then lead to the integral [tex]\int_0^3 \rho \pi (0.01^2) g (3-h)dh[/tex] Then you'd have to do something similar for the water that began not in the hose and add the two calculations of work together. Is this a good argument?