## Work problem: pumping water from a tank?

Question

An aquarium 2 m long, 1 m wide, and 1 m deep is half-filled with water. Find the work needed to pump the water out of the aquarium. (Use 9.8 m/s2 for g and the fact that the density of water is

1000 kg/m3.)

I have used F=19,600(delta)x

W=integral (from 0 to .5) 19,600(.5-x)dx

W=2450J

but this is being marked as incorrect

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General Physics
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## Answer ( 1 )

The mass of water in the tank is 1000(2*1*1/2) = 1000 kg.

Water is essentially incompressible so the centroid of the mass is 1/4 m above the bottom of the tank or 3/4 meter below the top of the tank.

The work required to lift the mass of water from its current position to the top of the tank is:

W = mgh = (1000kg)(9.8 m/s²)(3/4 m) = 7350 kg•m²/sec² = 7350 J

By integration, assuming y = 0 is the bottom of the tank and y = 1 meter is the top of the tank, each differential volume of mass is moved from it’s present location, y, to the top of the tank, y = 1. Hence the distance each differential volume is lifted is h = 1-y

W = ∫dW = ∫ρghdV

dV = (1*2)dy

h = 1-y

ρ = 1000 kg/m³

W = 1000(9.8)∫2(1-y)dy

W = 19600(y – y²/2) + C

Since the aquarium is half full, the limits of integration are from the bottom of the tank at y = 0 to the surface of the water at y = 1/2:

W = 19600(1/2 – 0 – 1/8 + 0) = 19600(3/8) = 7350 kg•m²/sec² = 7350 J