To calculate the work required to pump all the water out of the cylindrical tank, we can use the concept of work in physics, specifically for fluids. The formula for work to pump water over the top of a cylindrical tank is:

$W = \int_0^h \rho g A (h - y) \, dy$

Where:

• $W$ is the total work required.
• $\rho$ is the density of water ($\rho = 1000 \, \text{kg/m}^3$).
• $g$ is the acceleration due to gravity ($g = 9.8 \, \text{m/s}^2$).
• $A$ is the cross-sectional area of the tank.
• $h$ is the height of the water.
• $y$ is the height at which the water is located (we integrate from the bottom of the water to the top).
• $h - y$ represents the distance the water at height $y$ needs to be lifted.

Step 1: Calculate the cross-sectional area, $A$

The area of the cross-section of the cylindrical tank is a circle, so:

$A = \pi r^2$

where $r = 4 \, \text{m}$ is the radius.

$A = \pi (4)^2 = 16\pi \, \text{m}^2 \approx 50.265 \, \text{m}^2$

Step 2: Set up the integral

We will integrate from $y = 0$ (the bottom of the water) to $y = 5$ meters (the top of the water). The total work is given by:

$W = \int_0^5 1000 \times 9.8 \times 50.265 \times (5 - y) \, dy$

Step 3: Solve the integral

First, calculate the constants:

$1000 \times 9.8 \times 50.265 \approx 492588 \, \text{J/m}$

Now, integrate $(5 - y)$:

$\int_0^5 (5 - y) \, dy = \left[ 5y - \frac{y^2}{2} \right]_0^5 = \left( 5(5) - \frac{5^2}{2} \right) - (0) = 25 - \frac{25}{2} = \frac{25}{2} = 12.5$

Thus, the work is:

$W = 492588 \times 12.5 = 6157350 \, \text{Joules}$

Step 4: Convert to kilojoules

To convert to kilojoules (kJ), divide by 1000:

$W = \frac{6157350}{1000} = 6157.35 \, \text{kJ}$

Final Answer:

The work required to pump all the water over the top edge of the tank is approximately 6157 kJ.

Let me know if you'd like more details on any part of the process! Here are some related questions:

1. How would the work change if the height of the water was 7 meters instead of 5 meters?
2. How can we derive the formula for the work required to pump water in this case?
3. What assumptions were made in this calculation (such as the density of water and gravity)?
4. How would the work change if the radius of the tank were doubled?
5. How can the method be applied to a tank with a non-cylindrical shape?

Tip: The formula for work in a fluid system often involves integrating the force exerted by the fluid over the distance it needs to be moved.