A cylindrical tank with radius 2 m and height 5 m is filled with water. Find the work required to pump all the water to the top of the tank.
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Step-by-Step Solution
Step 1: Define the problem and variables
We need to calculate the work done to pump water out of the cylindrical tank. The work done (W) is the integral of the force required to lift a small slice of water multiplied by the distance it is lifted. Here, ρ is the density of water, g is the acceleration due to gravity, A(y) is the cross-sectional area of the tank at height y, and (H-y) is the distance each slice of water at height y needs to be lifted to the top of the tank.
Step 2: Identify given values and constants
The radius of the cylindrical tank is r=2 m, and its height is H=5 m. The density of water is ρ=1000 kg/m3, and the acceleration due to gravity is g=9.8 m/s2. The water is filled to the top, so we will integrate from y=0 to y=H.
Step 3: Calculate the cross-sectional area
For a cylindrical tank, the cross-sectional area A(y) is constant at any height y. It is given by the formula for the area of a circle, πr2. Substituting the given radius r=2 m, we get A(y)=4π m2.
Step 4: Set up the integral for work
Now we substitute all the known values into the work integral. The limits of integration are from y=0 (bottom of the tank) to y=5 (top of the tank). The distance to lift each slice is (H−y)=(5−y).
Step 5: Evaluate the integral
First, we pull the constants out of the integral: 1000×9.8×4π=39200π. Then, we integrate (5−y) with respect to y, which gives 5y−2y2.
Step 6: Calculate the final work
Now we evaluate the definite integral by plugging in the upper and lower limits. This gives us 39200π(25−225)=39200π(225). Multiplying these values, we get the total work done.