Consider a pool that is 25m by 10m. At the “shallow end” of the pool, the
depth is 2m. The pool has depth 2m for the first 10m of the length of the pool.
For the remaining 15m of the length of the pool, the depth increases linearly to
4m. How much work does it take to pump all of the water out of the pool?
The force of gravity for a kilogram of water is 9.8 Newtons. Therefore, for
a kilogram of water located h meters from the top, the work required to move
this water to the top is 9.8h Joules. Furthermore, the density of water is 1000
kilograms per cubic meter. We need to multiply this value by the cross-sectional
area of the water at height h- if you have a slice with thickness “dh” and then
integrate, then this will give you the total work. The amount of water at height
zero from the top, all the way to 2 meters from the top, is 250m2. Below this
height, the cross-sectional area starts at 150m2 decreases linearly to 0, so we
get that the cross-sectional area must be 250 − 125(h − 2) square meters. So
the total amount of work required to move this water is
Z
0
2
250 ∗ 9800h dh +
Z
4
(150 − 75(h − 2))9800h dh
2
This integral comes out to 8820000 Joules.
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