Let's divide the aquarium into tiny layers of water. Each layer is located at a different depth and will require a different amount of work to be done to be moved to the top of the aquarium. Therefore integration is required to calculate the total work to move half of the water out.
Recall this equation for the work required to move an object to a certain height:
W = mgh
W = work done, m = mass, g = gravitational acceleration, h = height
Focus on one of the tiny layers of water. It has a tiny mass dm which requires a tiny amount of work dW to be moved a distance equal to its depth h:
dW = dm(g)(h)
The tiny mass dm can be represented as the water's density p multiplied by a tiny bit of volume dV:
dm = p(dV)
Make a substitution:
dW = p(dV)(g)(h)
Each tiny layer of water has a volume dV equal to the base of the aquarium multiplied by a tiny height dh:
dV = 4(3)(dh) = 12dh
Make another substitution:
dW = 12pgh(dh)
Now let's integrate both sides:
W = ∫12pgh(dh)
Set the integration bounds to h = [0m, 1m] since the aquarium is 2m deep and we are only removing half of its water.
W = 6pgh² evaluated between h = 0m and h = 1m
W = 6pg
Given values:
p = 1000kg/m³, g = 9.8m/s²
Plug in and solve for W:
W = 6(1000)(9.8)
W = 58800J
The work needed to pump half of the water out of the aquarium is 58,800 J.
The given parameters;
The work done in pumping out half of the water form the aquarium is calculated as;
[tex]W = \int\limits^1_0 {(\rho gA)} h\, dh[/tex]
where;
[tex]W = \rho gA [\frac{h^2}{2} ]^1_0\\\\W = (1000 \times 9.8 \times 4\times 3)(\frac{1^2}{2} - 0)\\\\W = 58,800 \ J[/tex]
Thus, the work needed to pump half of the water out of the aquarium is 58,800 J.
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