# Calculating WORK...

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• March 13th 2010, 08:30 PM
lazypkjaii
Calculating WORK...
I am having a test the day right after spring break and throughout this time i don't know where to look for help, so i hope someone can help me with this problem...

The diameter of a circle pool is 24ft. the side is 5ft high and the depth of the water is 4ft. How much work is required to pump all of the water out over the side? Note: density of water is 1000kg/m^3, gravity=9.8m/s^2.

I tried to tackle this problem but I am not sure where I have done wrong
I set up the equation like this...
dWork = (dForce)(y) y - the distance from disc to top of the pool
(dWeight)(y)
(dmass)(gravity)(y)
(dVolume)(density)(gravity)(y) dV= (pi)(r^2)(dy)

dWork= (9.8)(1000)(Pi)(144) the integral from 1 to 5 (y)(5-y)(dy)

Hope I can get some guidance from some of the people in the forum
• March 13th 2010, 09:03 PM
Jhevon
Quote:

Originally Posted by lazypkjaii
I am having a test the day right after spring break and throughout this time i don't know where to look for help, so i hope someone can help me with this problem...

The diameter of a circle pool is 24ft. the side is 5ft high and the depth of the water is 4ft. How much work is required to pump all of the water out over the side? Note: density of water is 1000kg/m^3, gravity=9.8m/s^2.

I tried to tackle this problem but I am not sure where I have done wrong
I set up the equation like this...
dWork = (dForce)(y) y - the distance from disc to top of the pool
(dWeight)(y)
(dmass)(gravity)(y)
(dVolume)(density)(gravity)(y) dV= (pi)(r^2)(dy)

dWork= (9.8)(1000)(Pi)(144) the integral from 1 to 5 (y)(5-y)(dy)

Hope I can get some guidance from some of the people in the forum

First we need to match up our units. Note that $1000 \frac {\text{km}}{\text{m}^3} = 62.4 \frac {\text{lb}}{\text{ft}^3}$ (remember, lbs is a unit of force, so gravity is included here).

Now, I hope you have drawn a diagram, let y = 0 be the bottom of the pool, so that y = 5 is at the top. We will compute the work needed to move an infinitesimal slice of water, of thickness $\Delta y$, at a level $y$ above the bottom of the pool, and then integrate to find the total work.

Now, $W = F \cdot D$, where $F$ is force and $D$ is distance.

Note that the distance we want to move this slice will be $D = 5 - y$, it should be easy to see this if you drew your diagram correctly.

Now, the force is $(\text{volume of the slice})(\text{weight of water per unit volume}) = 62.4 \pi (12)^2 \cdot \Delta y$ $= 8985.6 \pi \Delta y$

So that we have $W = 8985.6 \pi (5 - y) \Delta y$

now, the total work is the sum of the works needed to move all the infinitesimal slices, ranging from a level y = 0 to y = 4 of the pool, hence,

$\text{Total Work } = 8985.6 \pi \int_0^4 (5 - y)~dy$

And note that the answer will be in ft-lbs
• March 13th 2010, 09:39 PM
lazypkjaii
I am actually confused at how you converted the units in the beginning...while everything else seems clear now
if the teacher had not given the Density of H2O which is 62.4 lbs/ft^3. Can we still work this problem out?
thanks alot for help me!
• March 13th 2010, 10:12 PM
Jhevon
Quote:

Originally Posted by lazypkjaii
I am actually confused at how you converted the units in the beginning...while everything else seems clear now
if the teacher had not given the Density of H2O which is 62.4 lbs/ft^3. Can we still work this problem out?
thanks alot for help me!

well, we used conversion factors to go from one to the other. presumably you know some. maybe from lbs to kg or ft to m, etc (i used both, or you can look it up :D). which ones do you know? the alternative would probably be to change ft to meters and then you would have force = volume * density * gravity, and your final answer would be in Nm
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