Question

An engine of power $58.8\, kW$ pulls a train of mass $2 \times 10^{5}\, kg$ with a velocity of $36\, kmh ^{-1}$. The coefficient of friction is

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

Answer 1

Answer: (C)

Power of engine,
$P=58.8\,kW =58.8 \times 10^{3} W$
Mass of the train, $m=2 \times 10^{5} kg$
Velocity of train, $v=36 \,kmh ^{-1}$
$=\frac{36 \times 10^{3}}{60 \times 60}=10\,ms ^{-1} $
Power, $P=F \cdot V$
$\therefore F =\frac{P}{V}=\frac{58.8 \times 10^{3}}{10} $
$=5880\, N$
Let the coefficient of friction is $\mu$.
$\therefore F=f_{s}=\mu \,m \,g $
$\Rightarrow \mu=\frac{F}{m g}=\frac{5880}{2 \times 10^{5} \times 10}$
$=0.003$