Question

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

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

Answer 1

The correct option is C $0.003$

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, $v=36km{h}^{-1}$
$=\frac{36\times {10}^3}{60\times 0}=10m{s}^{-1}$
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 $\mu$
$\therefore F=f_s=\mu$mg
$\Rightarrow \mu =\frac{F}{mg}=\frac{5880}{2\times {10}^5\times 10}$
$=0.003$.