Question

A circular swimming pool has a diameter of 8 meters. The sides are 3 meters high and the depth of the water is 1.5 meters. How much work is required to pump all of the water over the side? Your answer must include the correct

Answer 1

Answer:

Work done = 2218.97 kJ

Explanation:

Work = Force x Displacement.

Force = Weight of water.

Weight of water = Mass of water x Acceleration due to gravity.

Mass of water = Volume of water x Density of water.

Volume of water = Base area of swimming pool x Depth of water.

Base area of swimming pool

$= \frac{\pi {d}^{2} }{4} = \frac{\pi \times \: {8}^{2} }{4} = 50.27 {m}^{2}$

Depth of water = 1.5 m

Volume of water

$= 50.27 \times1.5 = 75.40 {m}^{2}$

Mass of water = 75.40 x 1000 = 75400 kg

Weight of water = 75400 x 9.81 = 739656 N

Work done = 739656 x 3 = 2218.97 kJ