Question

A tank has the shape of an inverted circular cone (point at the bottom) with height 10 feet and radius 4 feet. The tank is full of water. We pump out water (to a pipe at the top of the tank) until the water level is 5 feet from the bottom. The work W required to do this is given by W= ? foot-pounds

Answer 1

Answer:

Step-by-step explanation:

Given that tank has the shape of inverted circular cone

Height = 10 feet and radius =  4 feet

After water pumped out height = 5 ft

Hence volume of water pumped out = $\frac{1}{3} \pi R^2 h-\frac{1}{3} \pi r^2 h'$

Here we have r/h is constant always

Hence $\frac{r}{h'} =\frac{R}{h} \\r=Rh'/h = 4(5)/10 = 2$

Substitute to get volume of water pumped out = $\frac{1}{3} \pi (10*16-4*5)=\frac{140\pi}{3}$

Mass of water = density x volume = $62.4(\frac{140\pi}{3})\\= 2912\pi lbs$

Work done = force x displacement

= mass x accen x displacement

Here acceleration = gravity = 32.2 ft/sec^2

Displaement = height reduced = 5 ft

W=2912(32.2)(5)\pi =468832\pi foot -pound