A circular swimming pool has a diameter of 14 m The circular

A circular swimming pool has a diameter of 14 m. The circular side of the pool is 3 m high, and the depth of the water is 2.5 m. (The acceleration due to gravity is 9.8 m/s 2 and the density of water is 1000 kg/m 3 How much work (in Joules) is required to: pump all of the water over the side? pump all of the water out of an outlet 2 m over the side?

Solution

The circular swimming pool has diameter 14 m. and the depth of the water is 2.5 m.

So, the volume of the water (V) = (2*Pi*7)*2.5 m³ =70*Pi m³. The density of water is 1000 kg/m³.

Therefore the mass of the water (M) = V*1000 kg = 35000*Pi kg.

The height of the pool is 3 m and the acceleration due to gravity is 9.8 m/s².

So, the total force (F) = M*9.8*3 Newton. Since pump all of the water out of an outlet 2 m over the side, so

the total work done = F*2 Joules=M*9.8*3*2 Joules=35000*Pi*58.8 Joules= 35*Pi*58.8 killo Joules =6465.40 killo Joules.