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 of a train, we will follow these steps:

### Step 1: Calculate the change in kinetic energy (ΔKE)
The formula for kinetic energy (KE) is given by:

\[ KE = \frac{1}{2} mv^2 \]

Where:
- \(m\) is the mass of the train,
- \(v\) is the velocity.

The change in kinetic energy is given by:

\[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \]

Given:
- Mass of the train, \(m = 2.05 \times 10^6 \, \text{kg}\)
- Initial velocity, \(v_i = 5 \, \text{m/s}\)
- Final velocity, \(v_f = 25 \, \text{m/s}\)

Substituting the values:

\[ \Delta KE = \frac{1}{2} (2.05 \times 10^6) (25^2) - \frac{1}{2} (2.05 \times 10^6) (5^2) \]

Calculating \(25^2\) and \(5^2\):

\[ 25^2 = 625, \quad 5^2 = 25 \]

Now substituting back:

\[ \Delta KE = \frac{1}{2} (2.05 \times 10^6) (625) - \frac{1}{2} (2.05 \times 10^6) (25) \]

\[ = \frac{1}{2} (2.05 \times 10^6) (625 - 25) \]

\[ = \frac{1}{2} (2.05 \times 10^6) (600) \]

Calculating the work done:

\[ \Delta KE = 1.025 \times 10^6 \times 600 = 6.15 \times 10^8 \, \text{J} \]

### Step 2: Calculate the time taken in seconds
The time given is 5 minutes. We need to convert this into seconds:

\[ \text{Time} = 5 \, \text{minutes} \times 60 \, \text{seconds/minute} = 300 \, \text{seconds} \]

### Step 3: Calculate the power of the engine
Power (P) is defined as the rate of work done over time:

\[ P = \frac{\text{Work done}}{\text{Time}} = \frac{\Delta KE}{t} \]

Substituting the values we calculated:

\[ P = \frac{6.15 \times 10^8 \, \text{J}}{300 \, \text{s}} \]

Calculating the power:

\[ P = 2.05 \times 10^6 \, \text{W} \]

### Step 4: Convert power 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 MW**.

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