(Answer check) for a work problem

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The discussion revolves around a work problem related to calculating the force required to lift water from a tank. Participants note discrepancies in their answers, suggesting that the vessel may have been incorrectly modeled as a cone instead of a cylinder. The conversation highlights that the required sucking force increases as the water level decreases. One participant emphasizes a varying force approach, which accounts for lifting the entire column of water. The need for clarification and verification of calculations is evident throughout the exchange.
helloword365
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Homework Statement
A cylinder with radius r = 2 meters and height 4 meters is filled with water up to the top surface of the container. Calculate the minimum amount of energy necessary to remove all the water from the cylinder by pumping it out through the top. (Assume g = 9.8 m/s2 & ρ = 1000 kg/m3)
Relevant Equations
work = force * distance
I talked to other ppl about this problem, and we've all gotten pretty wide-ranging answers, so I was wondering if someone could try and do this so I could see whether my answer is right/wrong. (Answer key does exist but its not that great for this problem).

My work (if needed):
Screenshot 2025-04-12 203902.png
 
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You seem to have treated the vessel as a cone instead of a cylinder.
 
helloword365 said:
My work (if needed):
What is your calculated number?
Note that the needed sucking force increases as the level of the tank gets lower.
 
Lnewqban said:
Note that the needed sucking force increases as the level of the tank gets lower.
That is just a different way of analysing it.
@helloword365 considers the work required to lift a parcel of water from a depth of (4m-h) to the surface. The varying force approach considers the force required to lift the whole column of water in the pipe by dy.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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