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mathminor88

Member

Registered: 2007-09-16

Posts: 12

Calculus/Physics Work

An aquarium 2 m long, 1 m wide, and 1 m deep is full of water. Find the work needed to pump half of the water out of the aquarium. (Use the fact that the density of water is 1000 kg/m^3)

This was what I thought, but the answer is: 2450 J

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Calculus/Physics Work

Almost, you just misread the question a bit.

The pool is 1m deep, and the question asks how much energy will drain half of it. So your limits of integration are 0 and 1/2.

With those new limits put into what you have, the right answer should come out.
Edit: No it doesn't. I wrote this in a hurry.

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mathminor88

Member

Registered: 2007-09-16

Posts: 12

Re: Calculus/Physics Work

Actually if you used 1/2 and 0 as the limits then you would end up with 1225 J

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Calculus/Physics Work

using limits of 0 and 1/2 is correct, what is incorrect is the function you are integrating the correct integral is

times 2, because the pool is 2m×1m laterally, not 1m×1m so for every metre you lift out depth wise, you have 2m³ of water, not 1m³

if you evaluate that integral you get the correct answer if 2450J

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mathminor88

Member

Registered: 2007-09-16

Posts: 12

Re: Calculus/Physics Work

Thank you. I have a similar problem, but don't have access to the answer and want to make sure I'm doing it properly:

A circular swimming pool has a diameter of 24 ft, the sides are 5 ft high, and the depth of the water is 4ft. How much work is required to pump all of the water out over the side? (use the fact that water weights 62.5 lb/ft^3)

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Calculus/Physics Work

cross sectional area of pool is 72pi for each area, you have to lift it (x+1) feet, acceleration due to gravity is ~32 for imperial units.

so:

in general for this sort of question you have:

where A is cross sectional area, g is acceleration due to gravity, and p is density
obviously in a more complicated question where cross sectional area is a function of the depth aswell, you'd have to move the A into the integral aswell

Last edited by luca-deltodesco (2008-03-13 07:17:49)

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mathminor88

Member

Registered: 2007-09-16

Posts: 12

Re: Calculus/Physics Work

Where did you get 72 * pi? If you need find the area of each cross section (of a circle) wouldn't it be pi * r^2 (144 * pi)?

Also, you don't need to factor in the 32 (gravity) when dealing with lbs. Pounds is weight and not mass like kg.

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Calculus/Physics Work

man, i was sure that the equation was ½pi×r^2, getting confused with other equations And i didn't know that about the pound, i've never used imperial units in any realm of physics, I assumed it was a unit of mass like kilograms since we 'weigh' ourselves with it sorry about that.

looking into it, imperial units are even more confusing, pound IS a unit of mass. and the pound-force is the unit of force. the problem being that the proper symbol lbf or lb[sub]f[/sub] is very rarely used, nor its name and its often just called the pound aswell

Last edited by luca-deltodesco (2008-03-13 08:14:43)

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leiner19

Guest

Re: Calculus/Physics Work

I'm pretty sure its this
the integral from x = 0 to x = 4   62.5(pi)(12^2)(5-y) dy

leiner19

Guest

Re: Calculus/Physics Work

the integral from y = 0 to y = 4   62.5(pi)(12^2)(5-y) dy
Sorry i didn't mean x