Answer:74880 ft.lb
Step-by-step explanation:
Given
Dimension of pool
base=10 ft
width=20 ft
height=4 ft
Pool is half filled
volume of Pool[tex]=10\times 20\times 2=400 ft^3[/tex]
mass of water[tex]=\rho \times volume[/tex]
mass[tex]=62.4\times 400=24,960 lb[/tex]
This pool is in the form parallelepiped therefore average distance[tex]=\frac{4+2}{2}=3[/tex]
Work [tex]= 249600\times 3=74880 ft.lb[/tex]
