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A water tank has the shape of a cylinder with base radius 2 meters and height 6 meters. The tank is positioned vertically, so that its circular base is on the ground, and it has an outlet pipe at the top. It is half full of water. How much work is done to pump all of the water out of the outlet pipe? (Recall that the acceleration due to gravity is g = 9.8 meters per second per second and the mass density of water is ρ = 1000 kilograms per cubic meter.) Your answer should be in Newton-meters. Feel free to leave your answer in terms of g and ρ.

My attempt

Since, the water is only half-filled, I took the integral from $0$ to $3$.

$\int _{ 0 }^{ 3 }{ (1000)(9.8)(4\Pi )(6-y)dy } $

Can anyone verify this answer

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  • $\begingroup$ looks alright to me $\endgroup$ Commented Apr 29, 2018 at 17:23
  • $\begingroup$ Yes, looks correct. $\endgroup$
    – Dan Sp.
    Commented Apr 29, 2018 at 17:27

2 Answers 2

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Since the tank has a uniform cross section, you can also get the answer (or check your integration) by thinking about how far you need to move the center-of-mass.

The tank is full from 0-3 meters, so the center of mass is as 1.5 m and needs to be moved up 4.5 meters.

$$W = mg\Delta h = (1000*4\pi*3)*9.8*4.5$$

It's no mistake that $$3*4.5 = \int_0^3 (6-y)dy$$ which is the only way in which the two methods differ.

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Your calculation is correct. But note that you can empty the tank with almost zero work using a thin hose of length $>12$ meters. Throw this hose over the rim of the tank such that it falls to level $0$ on both sides. Then use suction on the outside until the hose is full of water. Finally just let the water spill on the ground outside until the tank is empty.

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