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Q. A point particle of mass $m$, moves along the uniformly rough track $PQR$ as shown in the figure. The coefficient of friction, between the particle and the rough track equals $\mu$. The particle is released, from rest, from the point $P$ and it comes to rest at a point $R$. The energies, lost by the ball, over the parts, $PQ$ and $QR$, of the track, are equal to each other, and no energy is lost when particle changes direction from $PQ$ to $QR$.
The values of the coefficient of friction $\mu$ and the distance $x(=QR)$, are, respectively close to :Physics Question Image

JEE MainJEE Main 2016Work, Energy and Power

Solution:

From work energy theorem and given condition
$m g h-2 \,\mu m g \,\cos\, \theta \frac{h}{\sin \theta}=0$
$\therefore \mu=\frac{1}{2 \cot 30}=\frac{1}{2 \sqrt{3}}=0.29$
again $\frac{m g h}{2}=\mu \,m g \,. Q R$
$\therefore Q R=\frac{h}{2 \mu}=\frac{2}{2 \times 0.29}=3.5 \,m$