Question

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 :

• A. 0.2 and 6.5 m
• B. 0.2 and 3.5 m
• C. 0.29 and 3.5 m
• D. 0.29 and 6.5 m

Answer 1

Answer: (C)

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