an open water tank shown below is being emptied by pumping the water up and over the top edge. how much work…

Explanation:

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

Consider a thin horizontal slice of water at a depth $y$ from the top of the cone with thickness $\Delta y$. The radius $r$ of the cone at height $y$ can be related to the radius $R = 4$ ft and height $h=10$ ft of the whole cone using similar - triangles. The ratio of radius to height for the whole cone is $\frac{R}{h}=\frac{4}{10}$. So, $r=\frac{4}{10}y = 0.4y$. The volume of the thin - slice of water $\Delta V=\pi r^{2}\Delta y=\pi(0.4y)^{2}\Delta y = 0.16\pi y^{2}\Delta y$.

Step2: Find the force required to lift the slice

The density of water is $\rho = 62.4$ 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 not given separately and we are using density in lb/ft³, $F=\rho\Delta V=62.4\times0.16\pi y^{2}\Delta y$.

Step3: Find the distance to lift the slice

The distance $d$ that the slice of water needs to be lifted is $y$.

Step4: Find the work done on the slice

The work done $\Delta W$ on the slice is $F\times d$. So, $\Delta W=(62.4\times0.16\pi y^{2}\Delta y)\times y=62.4\times0.16\pi y^{3}\Delta y$.

Step5: Integrate to find the total work

To find the total work done in emptying the full tank, we integrate from $y = 0$ to $y = 10$. [ \begin{align*} W&=\int_{0}^{10}62.4\times0.16\pi y^{3}dy\ &=62.4\times0.16\pi\int_{0}^{10}y^{3}dy\ &=62.4\times0.16\pi\left[\frac{y^{4}}{4}\right]_{0}^{10}\ &=62.4\times0.16\pi\times\frac{10^{4}}{4}\ &=62.4\times0.16\pi\times2500\ &=62.4\times400\pi\ &=24960\pi\ &\approx 78374 \end{align*} ]

Answer:

$78374$