We are tasked with finding the work required to pump water out of a conical tank with a spout. To solve this, we'll need to use the following concepts:

• Density of water: $\rho = 1000 \, \text{kg/m}^3$
• Gravitational acceleration: $g = 9.8 \, \text{m/s}^2$
• The formula for work done in moving a volume of liquid against gravity is: $W = \int F \, dy$ where $F$ is the force, and $dy$ represents an infinitesimal change in height.

Step-by-step Plan:

1. Volume Element: We'll first calculate the volume of an infinitesimal slice of water at a height $y$ above the bottom of the cone. Given the geometry of the cone, the radius of the water slice varies with height.

2. Geometric Similarity: Since the tank is a cone, the radius $r$ at any height $y$ is proportional to the height from the tip of the cone. For this cone:

   • At the top $y = 5 \, \text{m}$, the radius is $3 \, \text{m}$.
   • Using the similarity of triangles: $\frac{r}{y} = \frac{3}{5} \quad \text{or} \quad r = \frac{3}{5} y$

3. Infinitesimal Volume $dV$: The infinitesimal volume $dV$ of a slice of water at height $y$ with thickness $dy$ is the volume of a cylinder with radius $r = \frac{3}{5} y$ and height $dy$: $dV = \pi r^2 \, dy = \pi \left( \frac{3}{5} y \right)^2 \, dy = \pi \frac{9}{25} y^2 \, dy$

4. Force to Lift Water: The weight of this infinitesimal volume of water is $\rho g dV$: $dF = 1000 \times 9.8 \times \pi \frac{9}{25} y^2 \, dy$

5. Distance to Lift Water: To pump water out of the spout, the water at height $y$ needs to be lifted a distance of $6 - y$ meters (since the spout is 1 meter above the top of the cone).

6. Work for Infinitesimal Slice: The work to pump this infinitesimal slice is: $dW = dF \times (6 - y) = 1000 \times 9.8 \times \pi \frac{9}{25} y^2 \, (6 - y) \, dy$

7. Total Work: Integrate this expression from $y = 0$ to $y = 2$, as the tank is filled to a height of 2 meters: $W = \int_0^2 1000 \times 9.8 \times \pi \frac{9}{25} y^2 \, (6 - y) \, dy$ Simplifying constants: $W = 35280 \pi \int_0^2 y^2 (6 - y) \, dy$

Step-by-step Integration:

Now, we'll evaluate the integral: $\int_0^2 y^2 (6 - y) \, dy = \int_0^2 6y^2 \, dy - \int_0^2 y^3 \, dy$

Let's calculate each part:

$\int_0^2 6y^2 \, dy = 6 \left[ \frac{y^3}{3} \right]_0^2 = 6 \times \frac{8}{3} = 16$ $\int_0^2 y^3 \, dy = \left[ \frac{y^4}{4} \right]_0^2 = \frac{16}{4} = 4$ Thus, the total integral is: $16 - 4 = 12$

Final Work Calculation:

Now substitute this result into the expression for $W$: $W = 35280 \pi \times 12 = 423360 \pi$ Approximating $\pi \approx 3.1416$, we get: $W \approx 423360 \times 3.1416 = 1,329,305.6 \, \text{J}$

Thus, the total work required to pump all the water out of the tank is approximately $\boxed{1,329,305.6} \, \text{J}$.

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. How would the work change if the height of the spout were increased to 2 meters?
2. What is the effect of changing the cone's base radius on the work required?
3. How would the work differ if the tank were only half-filled with water?
4. What if the water density were lower, like for oil (850 kg/m³)?
5. How does the shape of the tank (e.g., cylindrical vs. conical) affect the work required?

Tip: Always check the dimensions of your parameters to ensure consistency when solving physics-related problems!