An automobile moving with a speed of 36kmh^ - 1 reaches an upward...

Given, initial speed, $u = 36kmh^{- 1}$

$= \dfrac{36 \times 1000}{60 \times 60} = 10ms^{- 1}$

$\theta = 30 ° , \mu = 0 . 1, s = ?$

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

\left(mgsin\theta + F\right)s = \dfrac{1}{2}mu^{2}$</p><p>\left(mgsin\theta + \muR\right) \times s = \dfrac{1}{2}mu^{2}$

\left(mgsin\theta + \mu mgcos\theta\right) \times s = \dfrac{1}{2}mu^{2}$</p><p>$s = \dfrac{\dfrac{1}{2}mu^{2}}{mg\left(sin\theta + \mu cos\theta\right)} = \dfrac{u2}{2g\left(sin\theta + \mu cos\theta\right)}$</p><p>$= \dfrac{10 \times 10}{2 \times 10 \times \left(\sin 30 ° + 0 . 1\cos 30 °\right)} = 8 . 53m$