Answer:
[tex] W= \int_{0}^8 78000 (9-h) dh[/tex]
[tex] W = 78000 (9h - \frac{h^2}{2}) \Big|_0^8 [/tex]
[tex] W= 78000 (72-32) = 3120000[/tex]
Explanation:
For this case since w ehave a rectangular pool we can find the volume as:
[tex] V = xyh[/tex]
We can approximate the volume with the following formula:
[tex] V \approx 50 ft* 25 ft * \Delta h)[/tex]
The mass is defined [tex] m= \rho V[/tex]
If we replace our approximation for the volume we got:
[tex] m =\rho(\approx 50 ft* 25 ft * \Delta h)[/tex]
The work for this case is defined as:
Work= Density* Force * Distance
We know that the total height is 9 ft but we want to calculate the work until 8ft so then we can express the distance in terms of the heigth h like this [tex] D= 9-h[/tex]
And then the work can be founded like this:
[tex] W = \sum_{i=1}^n \rho (1250 \Delta h) (9-h)[/tex]
And is we convert this into integrals using [tex] \Delta h \to 0[/tex]our integral limits are 0 and 8 and then density for the water is a property [tex] \rho = 62.4 \frac{lb}{ft^3}[/tex] and we have this:
[tex] W= \int_{0}^8 78000 (9-h) dh[/tex]
[tex] W = 78000 (9h - \frac{h^2}{2}) \Big|_0^8 [/tex]
[tex] W= 78000 (72-32) = 3120000[/tex]
