Question

An engine of power \( 58.8 \) kW pulls a train of mass \( 2 \times 10^5 \) kg with a velocity of \( 36 \text{ km h}^{-1} \). The coefficient of friction is

• A. $0.3$
• B. $0.03$
• C. $0.003$
• D. $0.0003$
• E. $0.04$

Hint:

For constant speed: Power = friction force × velocity.

Answer 1

The Correct Option is B

Concept: Power in uniform motion
When velocity is constant:
• Acceleration = 0
• Net force = 0
• Engine force = friction force ---

Step 1: Convert units
\[ P = 58.8 \text{ kW} = 58800 \text{ W} \] \[ v = 36 \text{ km/h} = 10 \text{ m/s} \] ---

Step 2: Use power relation
\[ P = F \cdot v \] \[ F = \frac{P}{v} = \frac{58800}{10} = 5880 \text{ N} \] ---

Step 3: Friction force
\[ F = \mu N = \mu mg \] \[ 5880 = \mu \times (2 \times 10^5) \times 9.8 \] ---

Step 4: Solve for $\mu$
\[ 5880 = \mu \times 1.96 \times 10^6 \] \[ \mu = \frac{5880}{1.96 \times 10^6} \] \[ \mu = 0.003 \] ---

Step 5: Check carefully
Oops — calculation gives: \[ \mu = 0.003 \] So correct answer is (C), not (B). --- Final Answer: \[ \boxed{0.003} \]