Question

A tank is full of oil weighing 30 lb/ft3. The tank is an inverted right rectangular pyramid (with the base at the top) with a width of 1 feet, a depth of 1 feet, and a height of 2 feet. Find the work required to pump the water to a height of 3 feet above the top of the tank.

Answer 1

Answer:

Step-by-step explanation:

density of oil, d = 30 lb/ft³

Volume of tank, V = l x w x h = 1 x 2 x 1 = 2 ft³

mass of oil, m = Volume x density of oil = 1 x 30 = 30 lb

Work = m x g x h = 30 x 32 x 3 = 2880 units

Answer 2

By using the classic equation for the work, we will find that the work needed is: 5,760 lb*ft^2/s^2

How to find the work?

The work needed to lift an object of mass M to a height h is just:

W = M*g*h

Where g is the gravitational acceleration; g = 32 ft/s^2

First, we need to get the mass, it will be equal to the density of the oil times the volume of the tank, the density of the oil is:

D = 30 lb/ft^3

The volume of the tank is:

V = 1ft*1ft*2ft = 2ft^3

The mass of oil is:

M = 2ft^3*30lb/ft^3 = 60lb

Then the work needed to pump the water to a height of 3ft is:

W = 60lb*(32 ft/s^2)*3ft = 5,760 lb*ft^2/s^2

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