Question

A particle of mass $m$, moves along the rough track $PQR$ as shown in the figure. The coefficient of friction (${\mu }_s={\mu }_k=\mu$), between the particle and the rough track is equal throughout the track. The particle is released from rest, from the point $P$ and it comes to rest again at point $R$. The energy lost by the particle, over the parts $PQ$ and $QR$ of the track are equal and no energy is lost when the particle changes direction from $PQ$ to $QR$. Then, values of the distance $x=QR$ and coefficient of friction $\mu$ are respectively close to:

• A. $x=\sqrt{3}\text{m}$
• B. $\mu =\frac{1}{\sqrt{3}}$
• C. $x=2\sqrt{3}\text{m}$
• D. $\mu =\frac{1}{2\sqrt{3}}$

Answer 1

Answer: (C), (D)

The correct options are
C $x=2\sqrt{3}\text{m}$
D $\mu =\frac{1}{2\sqrt{3}}$


From the figure of the track,
$\sin {30}^{\circ }=\frac{h}{d}=\frac{2}{d}$
$\therefore d=4\text{m}$ (hypotenuse)
$\cos {30}^{\circ }=\frac{\text{base}}{\text{hypotenuse}}=\frac{\sqrt{3}}{2}$
$\therefore$Base $=2\sqrt{3}\text{m}$
FBD of the particle:


$\text{Energy lost over path}PQ=\text{work done by friction}=f_s\times d$
$\Rightarrow \text{Energy lost over path}PQ=\mu N\times d$
$=\mu mg\cos \theta \times d$
$=\mu mg\cos \theta \times 4...\left(i\right)$

Similarly, energy lost over path $QR$ $=f_s\times x$
$\Rightarrow \mu N\times x$
$\text{Energy lost over path}QR=\mu mgx...\left(ii\right)$


Given $\text{Energy lost over path}PQ=\text{energy lost over path}QR$
$\Rightarrow \mu mg\cos \theta \times 4=\mu mgx$
$x=4\cos {30}^{\circ }$
$\therefore x=2\sqrt{3}\text{m}$

At point $R$, particle comes into rest, hence total energy lost by the particle
$=\text{Total energy at}P-\text{Total energy at}R$
$=mgh-0=mgh$

$\because$ along $PQ$ & $QR$, the particle lost the same energy
$\text{Total energy lost}=2\times \left(\text{Energy lost in}QR\right)$
$\Rightarrow mgh=2\mu mgx$
or $h=2\mu x$
or $2=2\times \mu \times 2\sqrt{3}$
$\therefore \mu =\frac{1}{2\sqrt{3}}$