calculate the work required to pump water into the tank shown below, through a hole at the base of the tank the water source is at ground level. hint: density = 62.4 lb/ft^3 4ft 16 ft 4 ft round your answer to the nearest whole number the tank is a cone

[tex]\tt{\green{66,927 ft-lb}}[/tex]

Answered by Nikovax | 2025-07-23

The work required is approximately [tex]\begin{gathered}\begin{gathered} \bold {267,654 \text{ ft-lb}}\end{gathered}\end{gathered}[/tex] (rounded to the nearest whole number).Here's why:Given the problem:Cone tank height, [tex]\begin{gathered} H = 16\end{gathered}[/tex] ftRadius at top, [tex]\begin{gathered}R = 4\end{gathered}[/tex] ft Tank base is [tex]4[/tex] ft above ground.Water density, [tex]\begin{gathered}\begin{gathered}\rho = 62.4\end{gathered}\end{gathered}[/tex]  lb/ft³Water source at ground level ([tex]0[/tex] ft).Pumping hole at tank base ([tex]4[/tex] ft above ground).[tex]\rule{190pt}{1pt}[/tex]Step 1: Define variablesLet [tex]y[/tex] = height from tank base up to water slice (0 to 16 ft)Radius at [tex]y[/tex] by similar triangles:     [tex]\begin{gathered}\begin{gathered}r(y) = \frac{R}{H} y = \frac{4}{16} y = \frac{y}{4}\end{gathered}\end{gathered}[/tex][tex]\rule{190pt}{1pt}[/tex]Step 2: Cross-sectional area of water slice     [tex]\begin{gathered}\begin{gathered}A(y) = \pi r(y)^2 = \pi \left(\frac{y}{4}\right)^2 = \frac{\pi y^2}{16}\end{gathered}\end{gathered}[/tex][tex]\rule{190pt}{1pt}[/tex]Step 3: Volume and weight of slice      [tex]\begin{gathered}\begin{gathered} dV = A(y) dy = \frac{\pi y^2}{16} dy \end{gathered}\end{gathered}[/tex]     [tex]\begin{gathered}\begin{gathered}dW = \text{weight} = \rho dV = 62.4 \times \frac{\pi y^2}{16} dy = \frac{62.4 \pi y^2}{16} dy\end{gathered}\end{gathered}[/tex][tex]\rule{190pt}{1pt}[/tex]Step 4: Distance slice must be pumped Water source at ground ([tex]0[/tex] ft), tank base at [tex]4[/tex] ft; slice at height [tex]\begin{gathered}y\end{gathered}[/tex] above tank base, so total lift distance:     [tex]\begin{gathered}\begin{gathered}d = 4 + y\end{gathered}\end{gathered}[/tex][tex]\rule{190pt}{1pt}[/tex]Step 5: Work for slice     [tex]\begin{gathered}\begin{gathered}dW = weight \times distance = \frac{62.4 \pi}{16} y^2 (4 + y) dy = \frac{62.4 \pi}{16} y^2 (y + 4) dy\end{gathered}\end{gathered}[/tex]Expand:     [tex]\begin{gathered}\begin{gathered}dW = \frac{62.4 \pi}{16} (y^3 + 4y^2) dy\end{gathered}\end{gathered}[/tex][tex]\rule{190pt}{1pt}[/tex]Step 6: Total work by integration    [tex]\begin{gathered}\begin{gathered}W = \int_{0}^{16} \frac{62.4 \pi}{16} (y^3 + 4y^2) dy = \frac{62.4 \pi}{16} \int_{0}^{16} (y^3 + 4y^2) dy\end{gathered}\end{gathered}[/tex]Calculate integrals:     [tex]\begin{gathered}\begin{gathered}\int_{0}^{16} y^3 dy = \left[ \frac{y^4}{4} \right]{0}^{16} = \frac{16^4}{4} = \frac{65536}{4} = 16384. \end{gathered}\end{gathered}[/tex]     [tex]\begin{gathered}\begin{gathered} \int{0}^{16} 4y^2 dy = 4 \left[ \frac{y^3}{3} \right]_{0}^{16} = 4 \times \frac{16^3}{3} = 4 \times \frac{4096}{3} = \frac{16384}{3} \approx 5461.33\end{gathered}\end{gathered}[/tex]Sum:      [tex]\begin{gathered}\begin{gathered}16384 + 5461.33 = 21845.33\end{gathered}\end{gathered}[/tex][tex]\rule{190pt}{1pt}[/tex]Step 7: Final calculation     [tex]\begin{gathered}\begin{gathered}W = \frac{62.4 \pi}{16} \times 21845.33 = 12.25 \times 21845.33 \approx 267,654 \text{ ft-lb}\end{gathered}\end{gathered}[/tex]Final answer:      [tex]\begin{gathered}\begin{gathered}\boxed{267,654 \text{ ft-lb}}\end{gathered}\end{gathered}[/tex]

Answered by Sefton | 2025-07-23