A force applied by an engine of a train of mass `2.05xx10^6`kg changes its velocity from 5 m/s to 25 m/s in 5 minutes. The power of the engine is

Options

A. 1.025 MW
B. 2.05 MW
C. 5 MW
D. 6 MW

Explanation

To find the power of the engine, we can follow these steps:

### Step 1: Identify the given values
- Mass of the train, \( m = 2.05 \times 10^6 \, \text{kg} \)
- Initial velocity, \( u = 5 \, \text{m/s} \)
- Final velocity, \( v = 25 \, \text{m/s} \)
- Time taken, \( t = 5 \, \text{minutes} = 5 \times 60 = 300 \, \text{s} \)

### Step 2: Calculate the change in kinetic energy
The work done on the train can be calculated using the change in kinetic energy formula:
\[ \text{Work done} = \Delta KE = \frac{1}{2} m (v^2 - u^2) \]
Substituting the values:
\[ \Delta KE = \frac{1}{2} \times 2.05 \times 10^6 \times (25^2 - 5^2) \]

### Step 3: Calculate \( v^2 \) and \( u^2 \)
Calculating the squares:
\[ v^2 = 25^2 = 625 \]
\[ u^2 = 5^2 = 25 \]
Now, calculate \( v^2 - u^2 \):
\[ v^2 - u^2 = 625 - 25 = 600 \]

### Step 4: Substitute back into the work done formula
Now substituting back into the work done equation:
\[ \Delta KE = \frac{1}{2} \times 2.05 \times 10^6 \times 600 \]

### Step 5: Calculate the work done
Calculating the work done:
\[ \Delta KE = 1.025 \times 10^6 \times 600 = 6.15 \times 10^8 \, \text{J} \]

### Step 6: Calculate the power
Power is defined as work done per unit time:
\[ P = \frac{\text{Work done}}{t} = \frac{6.15 \times 10^8}{300} \]

### Step 7: Final calculation for power
Calculating the power:
\[ P = 2.05 \times 10^6 \, \text{W} \]

### Step 8: Convert to Megawatts
Since \( 1 \, \text{MW} = 10^6 \, \text{W} \):
\[ P = 2.05 \, \text{MW} \]

### Final Answer
The power of the engine is \( 2.05 \, \text{MW} \).

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