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 static friction is

Options

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

Answer

To find the coefficient of static friction (μ) for the train being pulled by the engine, we can follow these steps:

### Step 1: Convert the velocity from km/h to m/s
Given:
- Velocity (V) = 36 km/h

To convert km/h to m/s, we use the conversion factor:
\[ 1 \text{ km/h} = \frac{1}{3.6} \text{ m/s} \]

So,
\[ V = 36 \text{ km/h} \times \frac{1}{3.6} = 10 \text{ m/s} \]

### Step 2: Write down the power equation
The power (P) produced by the engine is used to overcome the friction force (F_friction) acting on the train. The power can be expressed as:
\[ P = F_friction \times V \]

### Step 3: Express the friction force in terms of the coefficient of friction
The friction force can be expressed as:
\[ F_friction = \mu \times m \times g \]
where:
- \( m = 2 \times 10^5 \text{ kg} \) (mass of the train)
- \( g = 9.8 \text{ m/s}^2 \) (acceleration due to gravity)

### Step 4: Substitute the friction force into the power equation
Substituting the expression for friction force into the power equation gives:
\[ P = (\mu \times m \times g) \times V \]

### Step 5: Rearrange the equation to solve for the coefficient of friction (μ)
Rearranging the equation for μ:
\[ \mu = \frac{P}{m \times g \times V} \]

### Step 6: Substitute the known values
Now, substituting the known values:
- \( P = 58.8 \text{ kW} = 58.8 \times 10^3 \text{ W} \)
- \( m = 2 \times 10^5 \text{ kg} \)
- \( g = 9.8 \text{ m/s}^2 \)
- \( V = 10 \text{ m/s} \)

So,
\[ \mu = \frac{58.8 \times 10^3}{(2 \times 10^5) \times 9.8 \times 10} \]

### Step 7: Calculate the value of μ
Calculating the denominator:
\[ (2 \times 10^5) \times 9.8 \times 10 = 19.6 \times 10^6 \]

Now substituting back:
\[ \mu = \frac{58.8 \times 10^3}{19.6 \times 10^6} \]

Calculating:
\[ \mu = \frac{58.8}{19.6} \times \frac{10^3}{10^6} = \frac{58.8}{19.6} \times 10^{-3} \]

Calculating \( \frac{58.8}{19.6} \):
\[ \frac{58.8}{19.6} \approx 3 \]

Thus,
\[ \mu \approx 3 \times 10^{-3} = 0.003 \]

### Final Answer
The coefficient of static friction (μ) is approximately **0.003**.

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