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 particle, 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 3.5 m
• B. 0.29 and 3.5 m
• C. 0.29 and 6.5 m
• D. 0.2 and 6.5 m

Answer 1

The correct option is A 0.2 and 3.5 m

When rolling down the inclined plane,

$N_1=mg\cos \left({30}^o\right)$

$F_{r1}=\mu N_1=\mu mg\cos \left({30}^o\right)$

Energy lost equals the work done by friction force,

$E_{lost}=F_{r1}\times PQ=4\mu mg\cos \left({30}^o\right)$

Using conservation of energy,

$mgh=K{E}_Q+E_{lost}$

$K{E}_Q=mgh-4\mu mg\cos \left({30}^o\right)$

Kinetic energy at $Q$ is completely lost along $QR$. Also, energy lost along $QR$ equals the energy lost along $PQ$.

$\therefore K{E}_Q=E_{lost}$

$mgh-4\mu mg\cos \left({30}^o\right)=4\mu mg\cos \left({30}^o\right)$

${\mu }_{=}\frac{h}{8cos\left({30}^o\right)}$

$\mu =\frac{2}{8\times \sqrt{3}/2}\approx .29$

Along $QR$,

$N_2=mg$

$F_{r2}=\mu mg$

Work done by friction or energy lost along $QR$ is

$E_{lost}=F_{r2}\times QR$

$4\mu mg\cos \left({30}^o\right)=\mu mg\times QR$

$QR=4\times \sqrt{3}/2\approx 3.5m$