the trough above is full of water that has a density of 62.4 lbs per cubic foot. how much work is done in…

Explanation:

Step1: Consider a thin - slice of water

Let's consider a thin horizontal slice of water of thickness $\Delta y$ at a distance $y$ from the bottom of the trough. The trough has a length $L = 8$ feet, a width $w=3$ feet. The cross - section of the trough is semi - circular with radius $r = 1$ foot. The volume of the thin slice of water $\Delta V$ is approximately $V\approx L\times w\times\Delta y=8\times3\times\Delta y = 24\Delta y$ (since the cross - sectional area of the slice of water is approximately a rectangle with length $L = 8$ and width $w = 3$).

Step2: Calculate the force on the slice

The density of water $\rho=62.4$ lbs/ft³. The force $F$ required to lift the slice of water is equal to its weight. Using the formula $F = mg=\rho Vg$, and since $g$ is already incorporated in the density value for weight - density, the force on the slice $F=\rho\Delta V=62.4\times24\Delta y$ lbs.

Step3: Determine the distance to lift the slice

The distance $d$ that the slice of water needs to be lifted to get out of the trough is $(1 - y)$ feet (the height of the trough is 1 foot).

Step4: Calculate the work done on the slice

The work done $\Delta W$ on the slice is given by $\Delta W=F\times d$. Substituting $F = 62.4\times24\Delta y$ and $d=(1 - y)$, we get $\Delta W=62.4\times24\times(1 - y)\Delta y$.

Step5: Integrate to find the total work

To find the total work $W$ done in pumping out all the water, we integrate from $y = 0$ to $y = 1$. [ \begin{align*} W&=\int_{0}^{1}62.4\times24\times(1 - y)dy\ &=62.4\times24\int_{0}^{1}(1 - y)dy\ &=62.4\times24\left[y-\frac{y^{2}}{2}\right]_{0}^{1}\ &=62.4\times24\left(1-\frac{1}{2}\right)\ &=62.4\times24\times\frac{1}{2}\ &=62.4\times12\ & = 748.8\approx749 \end{align*} ]

Answer:

749