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 $x\left(QR\right)$ are, respectively close to

• A. $0.2$ and $3.5m$
• B. $0.25$ and $4m$
• C. $0.28$ and $6.5m$
• D. $0.2$ and $6.5m$

Answer 1

The correct option is B $0.25$ and $4m$

Loss of potential energy from P to Q is

$P.E=mgh=2mg$

Half of this is lost in work done by friction from P to Q.

Then, frictional force between inclined plane and particle is

$f_1=\mu mg\cos \theta$

Work done $N=f_1\times \frac{2}{\sin {30}^{\circ }}=\mu mg\times 4$

and, $W=\frac{P.E}{2}=mg$

$\mu mg\times 4=mg$

$\mu =\frac{1}{4}=0.25$

Now, from Q to R;

$\mu mgx=\frac{P.E}{2}=mg$

$x=4m$

so, $\mu =0.25$ and $x=4m$