Answer:
To find the work required to empty the tank by pumping all of the water over the top of the tank, you can use the principles of work and gravitational potential energy.
First, you need to find the volume of water in the tank. The tank has the shape of an inverted circular cone, so you can use the formula for the volume of a cone:
\[V = \frac{1}{3}πr^2h\]
Where:
- \(V\) is the volume of the water.
- \(π\) is approximately 3.14159.
- \(r\) is the base radius of the cone, which is 2 m.
- \(h\) is the height of the water, which is 7 m.
Plug in these values to find the volume of water:
\[V = \frac{1}{3} * 3.14159 * (2^2) * 7 = \frac{1}{3} * 3.14159 * 4 * 7 = 29.32 \, \text{cubic meters}\]
Next, you need to find the mass of the water. You can use the formula:
\[m = \text{density} \times \text{volume}\]
Given that the density of water is 1000 kg/m³, you can calculate the mass of water:
\[m = 1000 \, \text{kg/m³} \times 29.32 \, \text{m³} = 29,320 \, \text{kg}\]
Now, you can calculate the gravitational potential energy (GPE) of the water using the formula:
\[GPE = m \times g \times h\]
Where:
- \(GPE\) is the gravitational potential energy.
- \(m\) is the mass of the water, which is 29,320 kg.
- \(g\) is the acceleration due to gravity, which is 9.8 m/s².
- \(h\) is the height to which the water is lifted, which is 15 m (the height of the tank).
Plug in these values to calculate the GPE:
\[GPE = 29,320 \, \text{kg} \times 9.8 \, \text{m/s²} \times 15 \, \text{m} = 4,311,960 \, \text{Joules}\]
Now, convert the energy from Joules to kilojoules:
\[4,311,960 \, \text{Joules} = 4311.96 \, \text{kilojoules}\]
So, the work required to empty the tank by pumping all of the water over the top of the tank is approximately 4312 kilojoules (rounded to the nearest kilojoule).