A rectangular tank that is 4 feet long, 2 feet wide and 15 feet deep is filled with a heavy liquid that weighs 60 pounds per cubic foot. In each part below, assume that the tank is initially full. Your answers must include the correct units.

(a) How much work is done pumping all of the liquid out over
the top of the tank?
(b) How much work is done pumping all of the liquid out of a
spout 3 feet above the top of the tank?
(c) How much work is done pumping two-thirds of the liquid
out over the top of the tank?
(d) How much work is done pumping two-thirds of the liquid
out of a spout 3 feet above the top of the tank

a) W₁ = 54000 Lb-ft

b) W₂ = 77760 Lb-ft

c)  W₃ = 24000 Lb-ft

W₄ = 40560 Lb-ft

Step by step

W= ∫₁² ydF         1  and  2 are the levels of liquid

Where dF is the differential of weight of a thin layer

y is the height of the differential layer and

ρ*V  = F

Then

dF = ρ* A*dy*g

ρ*g  =  60 lb/ft³

A= Area of the base then

Area of the base is:

A(b) = 4*2  =  8 ft²

Now we have the liquid weighs  60 lb/ft³

Then the work is:

a)

   W₁ = ∫₀¹⁵ 8*60*y*dy     ⇒   W₁ =480* ∫₀¹⁵ y*dy

W₁ =480* y² /2 |₀¹⁵      ⇒  480/2 [ (15)² - 0 ]

W₁ = 240*225

W₁ = 54000 Lb-ft

b) The same expression, but in this case we have to pump 3 feet higher, then:

W₂ = ∫₀¹⁸ 480*y*dy     ⇒ 480*∫₀¹⁸ydy    ⇒ 480* y²/2 |₀¹⁸

W₂ =  480/2 * (18)²

W₂ =  240*324

W₂ = 77760 Lb-ft

c) To pump two-thirds f the liquid we have

2/3* 15  =  10

W₃ = 480*∫₀¹⁰ y*dy   ⇒  W₃ = 480* y²/2 |₀¹⁰

W₃ = 240*(10)²

W₃ = 24000 Lb-ft

d)

W₄ =480*∫₀¹³ y*dy

W₄ =480* y²/2 |₀¹³

W₄ = 240*(13)²

W₄ = 240*169

W₄ = 40560 Lb-ft

c. newton's third law of motion