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Lets talk about the water pump problem where you must pump water up and out of a tank and figure out how much work was needed to accomplish this.

So assume we have a cubic container with length 1 on all sides just to make the math easy. Why is it that we assume that the force is equal to the weight of the water?

Gravity is acting downward on the water so the weight of the water should be irrelevant. Unless we are talking about how much work the earth is producing on the water which would be negative. If we are talking about how much work the pump is doing we need to know how much pressure the pump is producing on the water in the tank. Unless it is assumed that the pump has area equal to the top of the tank and is producing a force upward twice the weight of the water we shouldn't be able to answer the question.

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  • $\begingroup$ Think about it this way. For the water to move at a constant velocity, what must be true of the net force on the water? Gravity is providing a force, so how does one reach the necessary net force? $\endgroup$ – Rushabh Mehta Sep 10 '18 at 14:16
  • $\begingroup$ If water is moving at a contant velocity then the net force is zero. But that would mean no work is being done. Are you saying that the pump has an equal and opposite force to gravity (because the water would simple float)? For the pump to do as much work as the problem assumes the pump would have to be at the top of the tank pulling the water up at twice it's weight. Or at the bottom of the tank pushing up at twice the waters weight.@RushabhMehta $\endgroup$ – user160110 Sep 10 '18 at 14:25
  • $\begingroup$ @How does net force being zero mean that no work is being done. The pump is certainly exerting a force... $\endgroup$ – Rushabh Mehta Sep 10 '18 at 14:26
  • $\begingroup$ No, the pump initially at slightly more than gravity's force (but you can ignore this) and then pushes at exactly gravity's force. The point is there is no net force, but that doesn't mean the pump isn't doing work. It means that the total energy in the system is being converted directly from physical work to potential energy. $\endgroup$ – Rushabh Mehta Sep 10 '18 at 14:28
  • $\begingroup$ @RushabhMehta Why would we assume that the pump has a force exactly equal to that of gravity?(Clearly $F_P>=W$) Why not state that part in the problem? Rather than infer that from the small parts where they say "(and don't forget that the force of gravity is $9.8\frac{m}{s^2}$)" being that the answer is completely different assuming $F_P>W$? $\endgroup$ – user160110 Sep 10 '18 at 14:46
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If you consider a small volume of water $dV$, gravity is pulling down on it with a force $\rho g dV$ where $\rho$ is density and $g$ is the acceleration of gravity. If you want to lift the water you need to overcome that force so you must push upward with a force of (at least) $\rho g dV$. You then integrate over the total volume, computing the amount you have to lift each small volume. A small volume at the surface will not need to be lifted at all, so the work is zero. A small volume at the bottom will need to be lifted $1$. This is exactly the difference in potential energy between having the water in the tank and having all the water at the height of the top of the tank.

Note that you do not need a force that corresponds to the total mass of the water. You can use a small force as long as you pump slowly and take a long time. The force times the mass flow rate is the power. Power integrated over time is energy. As you do the problem you need to keep clear which you are talking about.

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  • $\begingroup$ While this is correct, I do believe OP's error lies in his misunderstanding of the meaning of force, rather than the mathematics. $\endgroup$ – Rushabh Mehta Sep 10 '18 at 14:32

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