Question

An engine of power $58.8kW$ pulls a train of mass $2\times 10^{5}kg$ with a velocity of $36km{h}^{-1}$. 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,
$P=58.8kW=58.8\times 10^{3}W$
Mass of the train, $m=2\times 10^{5}kg$
Velocity of train, $v=36km{h}^{-1}$
$=\frac{36\times 10^3}{60\times 60}=10m{s}^{-1}$
Power, $P=F\cdot V$
$\therefore F=\frac{P}{V}=\frac{58.8\times 10^3}{10}$
$=5880N$
Let the coefficient of friction is $\mu$.
$\therefore F=f_s=\mu mg$
$\Rightarrow \mu =\frac{F}{mg}=\frac{5880}{2\times 10^5\times 10}$
$=0.003$