An engine of one metric ton is going up an inclined plane, 1 in 2 at the rate of 36 kmph. If the co-efficient of friction is $1/\sqrt{3}$ , the power of engine is

Incline plane 1 in 2. $\Rightarrow \mathrm{sin\theta }=\frac{1}{2}$

The force required to move up the incline with constant speed must be equal to the weight component and frictional force.

$\mathrm{F}=\mathrm{mg}\left(\mathrm{sin\theta }+\mathrm{\mu cos\theta }\right)={10}^3\mathrm{x}9.8\mathrm{x}\left(\frac{1}{2}+\frac{1}{\sqrt{3}}.\frac{\sqrt{3}}{2}\right)=9800\mathrm{N}$

$\mathrm{P}=\mathrm{F}.\mathrm{v}=9800.10=98\mathrm{kW}$