Question

An engine of mass $m$ is moving up a slope of inclination $\theta$ at a speed $v$. The coefficient of friction between engine and the rail is $\mu$. If the engine has an efficiency $\eta$ then the energy spent by engine in time $t$ is

• A. $\eta mg(\sin \theta +\mu \cos \theta )vt$
• B. $\frac{mg\left(\mu \cos \theta \right)vt}{\eta }$
• C. $\frac{mg(\sin \theta +\mu \cos \theta )vt}{\eta }$
• D. $\frac{mg}{2}\left(\frac{\sin \theta }{\eta }\right)vt$.

Answer 1

Answer: (C)

$\frac{mg(\sin \theta +\mu \cos \theta )vt}{\eta }$

For exerted by engine $F=mg\sin \theta +\mu mg\cos \theta =mg(\sin \theta +\mu \cos \theta )$
Power exerted by engine $=$ $Fv$
Work done by engine $=$ $Fvt$
Efficiency of engine $\eta =\frac{output}{input}$
$=\frac{Fvt}{input}$
$\therefore$ Input of engine
$=\frac{Fvt}{\eta }=\frac{mg(\sin \theta +\mu \cos \theta )vt}{\eta }$