consider the water tank below. if the tank is half full, find the amount of work required to pump the water…

Explanation:

Step1: Define variables

Let the height of the water - slice be $dh$ and its distance from the top of the tank be $h$. The radius of the cylinder $r = 3$ m, and the height of the water in the half - full tank is $h_0=4$ m. The density of water $\rho=1000$ kg/m³ and the acceleration due to gravity $g = 9.8$ m/s².

Step2: Find the volume of a thin slice of water

The volume of a thin slice of water of height $dh$ in the cylinder is $dV=\pi r^{2}dh=\pi(3)^{2}dh = 9\pi dh$ m³.

Step3: Find the mass of the thin slice of water

Using the formula $m=\rho V$, the mass of the thin slice of water $dm=\rho dV=1000\times9\pi dh$ kg.

Step4: Find the force required to lift the thin slice of water

The force required to lift the thin slice of water is equal to its weight, so $dF = dm\times g=(1000\times9\pi dh)\times9.8$ N.

Step5: Find the work done in lifting the thin slice of water

The work done in lifting the thin slice of water through a distance $h$ is $dW=dF\times h=(1000\times9\pi\times9.8)h dh$ J.

Step6: Set up the integral for the total work

Since the water is half - full and the height of the water column is 4 m, we integrate from $h = 0$ to $h = 4$ to find the total work. $W=\int_{0}^{4}(1000\times9\pi\times9.8)h dh$.

Step7: Evaluate the integral

First, we can take out the constant factors: $1000\times9\pi\times9.8\int_{0}^{4}h dh$. We know that $\int h dh=\frac{1}{2}h^{2}+C$. Evaluating the definite integral: $1000\times9\pi\times9.8\left[\frac{1}{2}h^{2}\right]_{0}^{4}$. $=1000\times9\pi\times9.8\times\frac{1}{2}(4^{2}-0^{2})$. $=1000\times9\pi\times9.8\times8$. $=705600\pi$ J. $\approx 2.22\times 10^{6}$ J.

Answer:

$705600\pi\approx 2.22\times 10^{6}$ J