To solve the problem, we need to calculate the average power of the pump that lifts 20 kg of water per second from a depth of 10 m and projects it with a velocity of 10 m/s. We will use the Work-Energy Theorem to find the work done by the pump and then calculate the power.
### Step-by-Step Solution:
1. **Identify the Given Data:**
- Mass of water lifted per second, \( m = 20 \, \text{kg} \)
- Depth of the well, \( h = 10 \, \text{m} \)
- Velocity of water projected, \( v = 10 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
2. **Calculate the Change in Kinetic Energy (\( \Delta KE \)):**
The change in kinetic energy is given by:
\[
\Delta KE = \frac{1}{2} m v^2 - \frac{1}{2} m u^2
\]
where \( u \) is the initial velocity (which is 0 since the water starts from rest).
\[
\Delta KE = \frac{1}{2} \times 20 \, \text{kg} \times (10 \, \text{m/s})^2 - 0
\]
\[
\Delta KE = \frac{1}{2} \times 20 \times 100 = 1000 \, \text{J}
\]
3. **Calculate the Work Done Against Gravity:**
The work done against gravity is given by:
\[
W_{\text{gravity}} = mgh
\]
\[
W_{\text{gravity}} = 20 \, \text{kg} \times 10 \, \text{m/s}^2 \times 10 \, \text{m} = 2000 \, \text{J}
\]
4. **Calculate the Total Work Done by the Pump:**
According to the Work-Energy Theorem:
\[
W_{\text{pump}} = \Delta KE + W_{\text{gravity}}
\]
Since the work done against gravity is negative (as it is done against the gravitational force):
\[
W_{\text{pump}} = 1000 \, \text{J} + 2000 \, \text{J} = 3000 \, \text{J}
\]
5. **Calculate the Average Power:**
Power is defined as work done per unit time. Since the pump lifts water at a rate of 20 kg per second, we can consider the time \( t = 1 \, \text{s} \).
\[
P = \frac{W_{\text{pump}}}{t} = \frac{3000 \, \text{J}}{1 \, \text{s}} = 3000 \, \text{W} = 3 \, \text{kW}
\]
### Final Answer:
The average power of the pump is **3 kW**.
