An engine of power `58.8 k W` pulls a train of mass `2xx10^(5)kg` with a velocity of `36 kmh^(-1)`. The coefficient of kinetic friction is

Options

A. 0.3
B. 0.03
C. 0.003
D. 0.0003

Answer

To solve the problem, we need to find the coefficient of kinetic friction (\( \mu_k \)) acting on a train being pulled by an engine. Here are the steps to arrive at the solution:

### Step 1: Convert the velocity from km/h to m/s
The velocity of the train is given as \( 36 \, \text{km/h} \). We need to convert this to meters per second (m/s).

\[ \text{Velocity} = 36 \, \text{km/h} \times \frac{1000 \, \text{m}}{1 \, \text{km}} \times \frac{1 \, \text{h}}{3600 \, \text{s}} = 10 \, \text{m/s} \]

### Step 2: Write down the power equation
The power generated by the engine is given as \( 58.8 \, \text{kW} \), which can be converted to watts:

\[ \text{Power} = 58.8 \, \text{kW} = 58.8 \times 10^3 \, \text{W} \]

### Step 3: Relate power to frictional force
Since the train is moving at a constant velocity, the power generated by the engine is used to overcome the frictional force. The power due to friction can be expressed as:

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

### Step 4: Express the friction force
The friction force (\( F_k \)) can be expressed in terms of the coefficient of kinetic friction (\( \mu_k \)) and the normal force. The normal force for the train is equal to its weight:

\[ F_k = \mu_k \cdot (m \cdot g) \]

Where:
- \( m = 2 \times 10^5 \, \text{kg} \) (mass of the train)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

### Step 5: Substitute the values into the power equation
Now substituting the expression for friction force into the power equation:

\[ 58.8 \times 10^3 = \mu_k \cdot (2 \times 10^5 \cdot 9.8) \cdot 10 \]

### Step 6: Solve for \( \mu_k \)
Rearranging the equation to isolate \( \mu_k \):

\[ \mu_k = \frac{58.8 \times 10^3}{(2 \times 10^5 \cdot 9.8) \cdot 10} \]

Calculating the denominator:

\[ 2 \times 10^5 \cdot 9.8 \cdot 10 = 19.6 \times 10^6 \]

Now substituting back into the equation for \( \mu_k \):

\[ \mu_k = \frac{58.8 \times 10^3}{19.6 \times 10^6} = \frac{58.8}{19.6} \times 10^{-3} \]

Calculating \( \frac{58.8}{19.6} \):

\[ \frac{58.8}{19.6} = 3 \]

Thus,

\[ \mu_k = 3 \times 10^{-3} = 0.003 \]

### Final Answer
The coefficient of kinetic friction (\( \mu_k \)) is \( 0.003 \).

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