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October 1st, 2012, 03:35 AM   #1
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a) A tank (as shown in the diagram) is full of water. Find the work required to pump the water out of the spout. The density of water is 1000kg/m3.
b) suppose that for the tank of part a, the pump breaks down after 4.7*10^5 J of work has been done. what is the depth of the water remaining in the tank?

i have done part a but i have no idea how to do part b.
Can anyone help with detail?
Thanks a lot.
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 October 1st, 2012, 10:47 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: work You want to solve: $78480\int_0\,^h 5x-x^2\,dx=4.7\times 10^5$ for $h$ (taking the root $0\le h\le3$).
 October 1st, 2012, 02:48 PM #3 Member   Joined: Oct 2012 Posts: 30 Thanks: 0 Re: work can i ask where is the 5x-x^2 come from?
 October 1st, 2012, 03:05 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: work I figured that would be familiar to you since you solved the first part of the problem...how did you solve the first part?
 October 1st, 2012, 07:51 PM #5 Member   Joined: Oct 2012 Posts: 30 Thanks: 0 Re: work first use the Mass=density * volume F=mg displacement= (2+x) work = integrate from 0 to 3 [78400(3-x)(2+x)] dx that's my equation
 October 1st, 2012, 08:13 PM #6 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 635 Thanks: 96 Math Focus: Electrical Engineering Applications Re: work Leaving units out which is not a good idea, but $work=mass \cdot g \cdot height_{rise}=mgh_r$ $mass=m=area \cdot \Delta h \cdot density=AD\Delta h$ $A=0$ at $h=0$ and $A=24$ at $h=3$ so $A=8h$ where $h$ is the height from the bottom of the trough. $m=AD\Delta h=8000h\Delta h$ $work=mgh_r=8000h\Delta h \cdot 9.81 \cdot (2+3-h)=78480h(5-h)\Delta h$ $w=78480(5h-h^2)\Delta h$ integrate:
 October 1st, 2012, 08:15 PM #7 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 635 Thanks: 96 Math Focus: Electrical Engineering Applications Re: work Darn it, hit submit instead of preview. But you should get the idea. One other thing, shouldn't the right side of the equation for part b) be the total work minus 470000?
 October 1st, 2012, 08:18 PM #8 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: work @summychan: Okay, your equation is just as valid, we just set it up differently. I let $x$ be the depth of the water, while you let $x$ be the distance the depth has decreased. Your approach is better for part b) as your integration moves from the top down, while mine moves from the bottom up, so you actually want to solve: $78400\int_0\,^{3-h} (3-x)(2+x)\,dx=4.7\times10^5$ for $h$, where $0\le h\le3$.
 October 1st, 2012, 08:28 PM #9 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 635 Thanks: 96 Math Focus: Electrical Engineering Applications Re: work Just to check answers, using g = 9.8 m/s^2, I get a) = 1058400 J b) = 2.02856 m Are these the same results that you get?
 October 1st, 2012, 08:33 PM #10 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: work Yes, I get the same results.

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# suppose that the tank shown below is full of water. find the work required to pump the water

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