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Sagot :
The work required for the given task of pumping all of the oil out over the top of the tank is 7,500 ft·lb
The known parameters;
The length of the rectangular tank, l = 4 feet
The width of the tank, w = 2 feet
The depth of the tank, h = 5 feet
The weight density of the oil with which the tank is filled, ρ × g = 50 lb/ft³
The unknown parameter
The work required to pump all of the oil out over the top of the tank
Method;
Calculate the force required to lift each slice (layer) of the oil to the top multiplied by the distance, y, the slice moves and summing the result as an integration as follows;
The volume of each slice, [tex]\mathbf{V_i}[/tex] = l × w × dy
The force required to move each slice, [tex]\mathbf{F_i}[/tex] = ρ × g × l × w × dy
The work done, [tex]\mathbf{W_i}[/tex], in moving the slice a distance, y, is given as follows;
[tex]\mathbf{W_i}[/tex] = ρ × g × l × w × y × dy
Therefore, the total work done, W, in pumping all the water located from y = 0, to y = 5, to the top of the tank, is given as follows;
[tex]\mathbf{W = \int\limits^5_0 {(\rho \times g \times l \times w \times y) } \, dy}[/tex]
Therefore;
W = (ρ × g × l × w × y²)/2
Plugging in the values, gives;
W = (50 lb/ft³ × 4 ft. × 3 ft. × (5 ft.)²)/2 = 7,500 ft·lb
The work required to pump all of the oil out over the top of the tank, W = 7,500 ft·lb.
Learn more about the use of integration to calculate the amount of work required for a given task here;
https://brainly.com/question/14318035
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