Question

A point particle of mass $m$, moves along the uniformly rough $PQR$ as shown in the figure. The coefficient of friction, between the particle and the rough track equals . The particle is released, 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 and the distance $x(=QR)$, are respectively close to.

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

Answer 1

The correct option is B $0.29$ and $3.5m$

Let's say the velocity of the particle at point Q is $v$.

Energy at P $=E_P=mgh$

Energy at Q $=E_Q=\frac{1}{2}m{v}^2$

Energy at R $=E_R=0$, assuming datum at $R$.

Loss in energy is due to negative work done by dissipative force, i.e. friction. Thus:

$E_P-E_Q=\mu mg\cos {30}^{\circ }\times PQ\to \text{(1)}$

$E_Q-E_R=\mu mg\times x\to \text{(2)}$

It is given that the losses of energies are equal in parts $PQ$ and $QR$. Thus:

$\mu mg\cos {30}^{\circ }\times PQ=\mu mg\times x$

$\Rightarrow x=\cos {30}^{\circ }PQ=\frac{\cos {30}^{\circ }}{\sin {30}^{\circ }}h$

$\therefore x=h\times \sqrt{3}\approx 3.5m$

From $\text{(1)}$ and $\text{(2)}$, we have:

$E_P-E_R=\mu mg(PQ\cos {30}^{\circ }+x)=2\sqrt{3}\mu mgh$

But $E_P-E_R=mgh$.

$\therefore \mu =\frac{1}{2\sqrt{3}}\approx 0.29$

$\therefore \mu \approx 0.29$ and $x\approx 3.5m\to$ option $\text{(B)}$.

$\to QED.$