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 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 the particle changes direction from $PQ$ to $QR$. The values of the coefficient of friction $\mu$ and the distance $x(=QR)$, are respectively

• A. $0.2$ and $6.5\text{m}$
• B. $0.2$ and $3.5\text{m}$
• C. $0.29$ and $3.5\text{m}$
• D. $0.29$ and $6.5\text{m}$

Answer 1

The correct option is C $0.29$ and $3.5\text{m}$

We know, work done by all the non-conservative forces
$W_{nc}=\Delta ME$

Here, the only non-conservative force acting is friction.
Hence, energy lost over path $PQ=\mu mg\cos \theta \times 4$


Energy lost over path $QR=\mu mgx$

Given, $\mu mg\cos {30}^{\circ }\times 4=\mu mgx$ $(\because \theta ={30}^{\circ })$
$x=2\sqrt{3}=3.45\approx 3.5\text{m}$

Initially at point $P$, the ball has only potential energy ($E=mgh$). And finally, ball comes to rest at point $R$ ($E=0$).

From $Q$ to $R$, energy loss is half of the total energy loss.
i.e. $\mu mgx=\frac{1}{2}\times mgh$
$\Rightarrow \mu =\frac{h}{2x}=0.29$

Thus, the values of coefficient of friction $\mu$ and the distance $x(=QR)$ are $0.29$ and $3.5\text{m}$ respectively.