To calculate the power of the pump, we need to break down the problem into two parts:

1. Potential Energy (PE): This is the energy required to raise the water to a height of 20 m.
2. Kinetic Energy (KE): This is the energy required to deliver the water at a speed of 8 m/s.

After calculating both energies, we can combine them and adjust for the pump's efficiency to find the actual power.

1. Potential Energy (PE)

The formula for potential energy is: $PE = m \cdot g \cdot h$ where:

• $m = 1000 \, \text{kg}$ (mass of water per minute),
• $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity),
• $h = 20 \, \text{m}$ (height).

$PE = 1000 \times 9.8 \times 20 = 196,000 \, \text{Joules}$

2. Kinetic Energy (KE)

The formula for kinetic energy is: $KE = \frac{1}{2} m v^2$ where:

• $m = 1000 \, \text{kg}$ (mass of water),
• $v = 8 \, \text{m/s}$ (velocity of water).

$KE = \frac{1}{2} \times 1000 \times (8^2) = 32,000 \, \text{Joules}$

3. Total Energy per minute

The total energy required per minute is the sum of the potential energy and kinetic energy: $\text{Total Energy} = PE + KE = 196,000 + 32,000 = 228,000 \, \text{Joules}$

4. Power Output

Power is energy per unit time. Since we're working with energy per minute, we need to convert it to energy per second (Watts). Divide the total energy by 60 seconds (since there are 60 seconds in a minute): $P_{\text{output}} = \frac{228,000}{60} = 3,800 \, \text{Watts}$

5. Power Input (Accounting for Efficiency)

Given that the efficiency of the pump is 40%, the input power must be greater than the output power. Efficiency is defined as: $\text{Efficiency} = \frac{P_{\text{output}}}{P_{\text{input}}}$ Solving for $P_{\text{input}}$: $P_{\text{input}} = \frac{P_{\text{output}}}{\text{Efficiency}} = \frac{3,800}{0.40} = 9,500 \, \text{Watts}$

Thus, the power input to the pump is 9,500 Watts or 9.5 kW.

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Would you like further clarification or more details on any of these steps?

Here are five related questions to explore:

1. How would the power change if the efficiency increased to 60%?
2. What happens to the power if the depth is increased to 30 meters?
3. How would reducing the velocity of water to 5 m/s affect the total power?
4. If the mass of water pumped per minute doubled, what would be the new power required?
5. What is the significance of efficiency in energy-related problems?

Tip: Efficiency always relates the useful energy output to the total energy input—understanding this concept helps in analyzing real-world systems.