Question

A rectangular tank that is 4 feet long, 5 feet wide and 15 feet deep is filled with a heavy liquid that weighs 110 pounds per cubic foot. In each part below, assume that the tank is initially full. Your answers must include the correct units. (You may enter lbf or lb*ft for ft-lb.)

How much work is done pumping all of the liquid out over the top of the tank?

Answer 1

247500 J

Explanation:

We are given that

Length,l=4 feet

Width,b=5 feet

Height,h=15 feet

Weigh,w=110
`pounds/ft^3`

We have to find the work done in pumping all of the liquid out over the top of the tank.

Area of cross-section,A=
`l* b=4* 5=20ft^2`

Force increment,
`dF=20* 110dy=2200dy`

Work increment ,
`dW=2200ydy`

Total work done,W=
`\int_(0)^(15)2200ydy`

`W=2200[(y^2)/(2)]^(15)_(0)`

`W=2200* ((15)^2)/(2)=247500 J`

Answer 2

247500 Joule

Explanation:

length of tank, l = 4 feet

width of the tank, w = 5 feet

depth of the tank, h = 15 feet

density of liquid, d = 110 pounds per cubit foot

Area of crossection, A = length x width

A = 4 x 5 = 20 ft²

Let the height is dh

Force due to height dy

dF = A x dy x density

dF = 20 x 110 x dy = 2200 dy

Work done in increasing the level by y is

dW = 2200 x y x dy

Total work done is

`W = \int dW = \int_(0)^(15)2200 y dy`

`W = 1100[y^(2)]_(0)^(15)`

W = 1100 ( 15 x 15 - 0 )

W = 247500 Joule