$F_{net} = mg sin\Theta – \mu mg cos\Theta$ $\Rightarrow mg sin\Theta – \mu xg cos\Theta$ $a = \frac{F_{net}}{m} = g sin\Theta – \mu_{0} xg cos\Theta$

Therefore, $v * \frac{dv}{dx} = g sin\Theta – \mu_{0} xg cos\Theta$

OR

$\int_{0}^{0}vdv = \int_{x_{m}}^{0} (g sin \Theta – \mu_{0} xg cos\Theta)dx$

Solving this equation we get,

$x_{m} = \frac{2}{\mu_{0}} tan\Theta$

Explore more such questions and answers at BYJU’S.