Question

The base of a rectangular tank measures 10 ft by 20 ft. The tank is 16 ft tall, and its top is 10 ft below ground level. The tank is full of water weighing 62.4 lb/ft3. How much work does it take to empty the tank by pumping the water to ground level? Give your answer to the nearest ft ∙ lb.

Answer 1

Answer:

Step-by-step explanation:

Given a rectangular tank with dimension (10ft by 20ft by 16ft)

Then the volume of the tank is

Volume =length × breadth ×height

Volume=10×20×16=3200ft³

V=3200ft³

Then,

∆F= weight density × volume

∆F= 62.4×3200

∆F= 199,680 lb

Then,

Let the 0-point on the x-axis be at the bottom of the tank, so the level of the water ranges from x = 0 to x = 16ft. (It would just as well to let 0 be ground level and let x range from x = −26ft to x − 0.) Then a slice of water at level x is raised (26− x)ft

Then ∆x=(26-x)ft

Work is given as

W= -∫F∆xdx. From 0 to 16

W= -∫199,680(26-x)dx From 0 to 16

W=-199,680∫26-x dx From 0 to 16

W=-199,680 (26x-x²/2). From 0 to 16

W=-199,680(26×16-0.5×16² -0-0)

W=-199,680(288)

W=-57,507,840J

The work done to empty the tank is 57,507,840J

Answer 2

Answer:

2 X 10∧6 ft.lb

Step-by-step explanation:

Volume of rectangular tank =  Base area X Height = (10 X 20 X 16)ft³ = 3200ft³

Mass of tank filled with water = 3200ft³ X 62.4 lb/ft³ = 199680lb or 90706.37kg

Hence, work done due to gravity,Δg = m * g * h, where, m = mass of tank, g = gravity = 9.8m/s² and h = height = 10ft or 3.05 meter(m)

∴ Δg = 90706.37 * 9.8 * 3.05 =  2.711 X10∧6 J or 2 X 10∧6 ft.lb