An aquarium 5 m long, 1 m wide, and 1 m deep is full of water. Find the work needed to pump half of the water out of the aquarium. (Use 9.8 m/s2 for g and the fact that the density of water is1000 kg/m3.)Show how to approximate the required work by a Riemann sum. (Let x be the height in meters below the top of the tank. Enter xi* as xi.)lim n → ∞ n $$1000 i = 1 ΔxExpress the work as an integral.
Question
An aquarium 5 m long, 1 m wide, and 1 m deep is full of water. Find the work needed to pump half of the water out of the aquarium. (Use 9.8 m/s2 for g and the fact that the density of water is1000 kg/m3.)Show how to approximate the required work by a Riemann sum. (Let x be the height in meters below the top of the tank. Enter xi* as xi.)lim n → ∞ n 1000 i = 1 ΔxExpress the work as an integral.
Solution
The work required to pump water out of a tank is given by the formula W = ρghA, where ρ is the density of the water, g is the acceleration due to gravity, h is the height the water is lifted, and A is the area of the cross section of the water being lifted.
In this case, the density of the water is 1000 kg/m^3, g is 9.8 m/s^2, the area A is 5 m * 1 m = 5 m^2, and the height h varies from 0 to 0.5 m (since we're pumping out half of the water).
We can approximate the work done by dividing the tank into n slices of equal thickness Δx = 0.5/n m. The work done to pump out the i-th slice is approximately W_i = ρgAΔxh_i, where h_i = iΔx is the height of the i-th slice.
Summing over all slices, we get the Riemann sum:
W ≈ Σ_{i=1}^{n} ρgAΔxh_i = Σ_{i=1}^{n} 1000 kg/m^3 * 9.8 m/s^2 * 5 m^2 * 0.5/n m * i/n m = Σ_{i=1}^{n} 24500/n^2 * i J.
Taking the limit as n goes to infinity, we get the integral:
W = ∫{x=0}^{0.5} ρgAx dx = ∫{x=0}^{0.5} 1000 kg/m^3 * 9.8 m/s^2 * 5 m^2 * x dx = 12250 J.
So the work needed to pump half of the water out of the aquarium is 12250 Joules.
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