calculate the work required to pump water into the tank shown below, through a hole at the base of the tank…

Explanation:

Step1: Find the volume of a thin - slice of water

Consider a thin horizontal slice of water at a height $y$ from the base of the tank with thickness $\Delta y$. The radius $r$ of the cross - section of the cone at height $y$ can be found using similar triangles. The ratio of the radius to the height for the whole cone is $\frac{r}{h}=\frac{4}{16}=\frac{1}{4}$. So, $r = \frac{y}{4}$. The volume of the thin slice of water $\Delta V=\pi r^{2}\Delta y=\pi(\frac{y}{4})^{2}\Delta y=\frac{\pi y^{2}}{16}\Delta y$.

Step2: Find the force required to lift the slice

The density of water $\rho = 62.4$ lb/ft³. The mass of the slice of water $m=\rho\Delta V$. The force $F$ required to lift the slice is equal to its weight, so $F = mg=\rho g\Delta V$. Since $g$ is the acceleration due to gravity and we are using the weight - density $\rho = 62.4$ lb/ft³, $F=\rho\Delta V=62.4\times\frac{\pi y^{2}}{16}\Delta y$.

Step3: Find the distance the slice needs to be lifted

The slice of water at height $y$ needs to be lifted a distance $d = y$ feet.

Step4: Find the work done on the slice

The work done $\Delta W$ on the slice is $F\times d$. So, $\Delta W=62.4\times\frac{\pi y^{2}}{16}\times y\Delta y= \frac{62.4\pi y^{3}}{16}\Delta y$.

Step5: Integrate to find the total work

The tank has a height from $y = 0$ to $y = 16$ feet. The total work $W$ is given by the integral $W=\int_{0}^{16}\frac{62.4\pi y^{3}}{16}dy$. First, simplify the integrand: $\frac{62.4\pi y^{3}}{16}=3.9\pi y^{3}$. Then, integrate using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). $W=3.9\pi\int_{0}^{16}y^{3}dy=3.9\pi\left[\frac{y^{4}}{4}\right]_{0}^{16}$. $W = 3.9\pi\times\frac{16^{4}}{4}$. $W=3.9\pi\times\frac{65536}{4}=3.9\pi\times16384$. $W = 63997.6\pi\approx201085$ ft - lb.

Answer:

$201085$