Question

A circular swimming pool has a diameter of 12 m, the sides are 3 m high, and the depth of the water is 1.5 m. How much work (in Joules) is required to pump all of the water over the side? (The acceleration due to gravity is 9.8 m/s^2 and the density of water is 1000 kg/m^3 .)

Answer 1

I'm going to take a rather simplistic view of the situation here. Let's assume that you don't want to move the pump tube at all, so you keep the end of it at the bottom of the pool. No work is done moving water sideways, and when you take some water from the bottom of the pool, the water above and around that position automatically moves to that position without you exerting any work.

The problem then becomes an issue of moving one single mass to a height of 3 metres. What is the single mass? It's the mass of the water. It's in a cylinder of radius 6m, and height 1.5m, so you can get the volume:
\[V_{cylinder} = \pi r^2 h\]The mass of the water is given by:
\[mass=density \times volume\]The work required to move an object of mass \(m\) to a height \(h\) is given by:\[U=mgh\], where m is the mass of the water, g is the acceleration due to gravity, and h is the height of the pool (3m).

Good luck!

Answer 2

total volume of water = pi*(12/2)^2*1.5 cubic metre
mass= volume*density = pi*(12/2)^2*1.5 * 1000 kg
weight of water = m*g = pi*(12/2)^2*1.5 * 1000 * 9.8 kg-m/ s²

force required to lift this volume of water = weight of this volume of water

displacement = 3m

work done = force*displacement = weight*displacement

W=pi*(12/2)^2*1.5 * 1000 * 9.8 *3 kg-m²/ s²

(1 kg-m²/ s² =1 J)