Question

the tank above is full of water that has a density of 62.4 lbs per cubic foot. how much work is done in pumping the water out over the top? round to the nearest whole number. ? foot - pounds

Explanation:

Step1: Recall the volume formula for a cone

The volume formula for a cone is $V=\frac{1}{3}\pi r^{2}h$. Here, $r = 7$ ft and $h=9$ ft. So, $V=\frac{1}{3}\pi(7^{2})(9)=\frac{1}{3}\pi\times49\times9 = 147\pi$ cubic - feet.

Step2: Calculate the weight of the water

The density of water is $
ho = 62.4$ lbs per cubic - foot. The weight of the water $F=
ho V$. Substituting the values, we get $F = 62.4\times147\pi$ lbs.

Step3: Use the work - formula for pumping water

The work done in pumping water out of a cone is given by the formula $W=\int_{0}^{h}F(y)dy$. For a cone, the average distance the water needs to be lifted is $\frac{3}{4}h$ (a property of the centroid of a cone - shaped volume of fluid). Here, $h = 9$ ft.
The weight of the water $F=62.4\times147\pi$. The work $W=F\times\frac{3}{4}h$.
First, calculate $F = 62.4\times147\pi=62.4\times147\times3.14\approx62.4\times461.58 = 28802.592$.
Then, $W=F\times\frac{3}{4}h$. Substitute $h = 9$ ft.
$W=28802.592\times\frac{3}{4}\times9=28802.592\times6.75=194417.404\approx194417$ foot - pounds.

Answer:

$194417$