A train of weight `10^(7)N` is running on a travel track with uniform speed of `36 km h^(-1)`. The frictional force is 0.5 kg f per quintal. If `g = 10 ms^(-2)`, power of engine is

Options

A. `500 kW`
B. `50 kW`
C. `5 kW`
D. `0.5 kW`

Answer

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

### Step 1: Convert the weight of the train from Newtons to kilogram-force (kgf)
The weight of the train is given as \(10^7 \, \text{N}\). We know that \(1 \, \text{kgf} = 10 \, \text{N}\). Therefore, we can convert the weight to kgf:

\[ \text{Weight in kgf} = \frac{10^7 \, \text{N}}{10 \, \text{N/kgf}} = 10^6 \, \text{kgf} \]

### Step 2: Calculate the frictional force
The frictional force is given as \(0.5 \, \text{kgf}\) per quintal. Since \(1 \, \text{quintal} = 100 \, \text{kg}\), we can convert the weight of the train to quintals:

\[ \text{Weight in quintals} = \frac{10^6 \, \text{kgf}}{100 \, \text{kg/quintal}} = 10^4 \, \text{quintals} \]

Now, we can calculate the total frictional force:

\[ \text{Frictional force} = 0.5 \, \text{kgf/quintal} \times 10^4 \, \text{quintals} = 0.5 \times 10^4 \, \text{kgf} = 5000 \, \text{kgf} \]

### Step 3: Convert the frictional force to Newtons
Now, we convert the frictional force from kgf to Newtons:

\[ \text{Frictional force in N} = 5000 \, \text{kgf} \times 10 \, \text{N/kgf} = 50000 \, \text{N} \]

### Step 4: Convert the speed from km/h to m/s
The speed of the train is given as \(36 \, \text{km/h}\). We can convert this to meters per second (m/s) using the conversion factor:

\[ \text{Speed in m/s} = 36 \, \text{km/h} \times \frac{5}{18} = 10 \, \text{m/s} \]

### Step 5: Calculate the power of the engine
Power is calculated using the formula:

\[ \text{Power} = \text{Force} \times \text{Velocity} \]

Substituting the values we have:

\[ \text{Power} = 50000 \, \text{N} \times 10 \, \text{m/s} = 500000 \, \text{W} = 500 \, \text{kW} \]

### Final Answer
The power of the engine is \(500 \, \text{kW}\).

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