Given, initial speed, $u=36 \mathrm{km} {\mathrm{h}}^{-1}$

$=\frac{36\times 1000}{60\times 60}=10 \mathrm{m} {\mathrm{s}}^{-1}$

$\theta =30°, \mu =0.1, s=?$

Here, work done in moving up the inclined road$= KE$ of the vehicle

$\left(mg \sin \theta +F\right) s=\frac{1}{2}m{u}^2$

$\left(mg \sin \theta +\mu R\right)\times s=\frac{1}{2}m{u}^2$

$\left(mg \sin \theta +\mu mg \cos \theta \right)\times s=\frac{1}{2}m{u}^2$

$s=\frac{\frac{1}{2}m{u}^2}{mg \left(\sin \theta +\mu \cos \theta \right)}=\frac{u^2}{2 \mathrm{g} \left(\sin \theta +\mu \cos \theta \right)}$

$=\frac{10\times 10}{2\times 10\times \left(\sin 30°+0.1 \cos 30°\right)}=8.53 \mathrm{m}$