How to lift water out of a pool

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To calculate the work needed to lift water out of a pool, the correct approach involves using the formula for potential energy, mgh, where m is the mass derived from the water's density and volume. The distance the water needs to be lifted varies based on the center of mass, which can be calculated as 4 m from the top of the pool or adjusted based on the depth of the water. Initial attempts yielded incorrect results, but through integration, the correct work was determined to be approximately 2.4 x 10^7 joules. The weight density of water is 9800 N/m^3, which is essential for accurate calculations. Ultimately, using integration proved to be the effective method for solving this problem.
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Homework Statement


given, diameter and height of pool, depth of water, density of water and gravitational acceleration.
g = 9.8
height is 4m
2r = 20m
depth of water = 3.5m
density =1000kg/m^3
So, I'm supposed to find how much work is needed in joules to lift the water out of the pool.

Homework Equations


could I use density times gravity times volume to get my force?
is my distance for my water to travel 4.5m?


The Attempt at a Solution


Anyways, I've tried to use my volume for the pool and the volume I have of water * density *gravity to come up with my force and then used 4.5m as my distance to come up with my work needed to lift water out of a pool. I ended up with 1.08E7 joules but that's incorrect so any help would be appreciated.

BTW, this is a cal 2 problem/integration problem
 
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if it is a cal2/integration problem why aren't you doing any integration? do you know the formula for work?
 
Use the volume of the water and the density to get mass. It would appear that the pool is a right circular cyliner.

Rasining a mass m some distance h in a gravity field producing a local acceleration g increases the potential energy by mgh.
could I use density times gravity times volume to get my force?
Yes

is my distance for my water to travel 4.5m?
Well that depends, if one is lifting the CM of the pool water 4 m or (4 m + 1.75 m) or 5.75 m, which is 4 m from the top of the pool.
 
I've tried integrating it and I've tried astronuc's advice with no success, any ideas?
 
Last edited:
I'm retarded, integration works. Alright, so with a mixture of the advice, correct answer is 2474002 joules. Thanks again fellas.
 
weight density = 9800 N/m^3

work = integral {0 to 3.5} (9800)(pi*10^2)(4 - y)dy

work = 2.4 * 10^7 Joules
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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