Work Problem: Help Needed.

I need some guidance for this problem:

--x--

Find the work needed to empty the horizontal cylindrical tank if it is full of water. The pump is 2 meters abovce the tank, the tank is 8 meters long (lying on its side), and ther radius of the cylinder is 2 meters.

--x--

Here's what I have so far:

Since work = force * distance, I've gotten the distance as (7-y). The cross section is a rectangle witrh a lenth of 8 meters. To get the width of the rectangle, I used x^2+y^2=25. I solved for y and got: squareroot(25-y^2).

So, the area of the cross-section is 8*squareroot(25-y^2).

It's given by my book that water has a density of 9800 Newtons.

To get the equation of the force, I did: 9800*(7-y)*[8*squareroot(25-y^2)].

Am I supposed to integrate now with boundaries of -5 to 5?

I got stuck in the middle (after factoring out the constant) while trying to integrate it:

78400*[integral from -5 to 5](7-y)*[squareroot(25-y^2)].

I hope everything made sense....

Can someone check my work and help me? Thanks.

The answer is 43.1 million Joules, but I'm not sure how to get it.

Change in potential energy

Quote:

Originally Posted by **CaptainBlack**

It helps if you write the problem correctly. The radius of the tank is 5m.

The minium work needed to empty the tank is equal to the change in

potential energy when the water is lifted by 7m.

The density of water is $\displaystyle \approx 1000\ \mbox{kg/m^3}$, so the mass

of the water in the tank is $\displaystyle m=\pi \times 5^2\times 8 \times 1000 \approx 628000\ \mbox{kg}$

The change in potential energy is:

$\displaystyle

\Delta E=628000 \times 9.91 \times 7 \approx 43.1 \times 10^6\ \mbox{J}

$,

which is the answer.

RonL

I have a doubt about the formula you used.

You have used:

Change in potential energy = m*g*(change in height(h))

The doubt is that is 'h' is same for every element of water?(It is possible only if water is pumped in an identical tank in the same position) :)