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, form the point P and it comes to rest 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 (= QR), are respectively close to

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

Answer 1

The correct option is C

0.29 and 3.5 m

Using $sin{30}^{\circ }=\frac{h}{PQ}$, we get $PQ=4m$.
As work done by friction is equal over the parts PQ and QR of the track, $\mu mgcos{30}^{\circ }\times 4=\mu mgx$.
Solving we get, $x=2\sqrt{3}m=3.5m$
Now, applying the work-energy theorem, $mgh=\mu mgcos\theta \times 4+\mu mg\times 2\sqrt{3}\Rightarrow \mu =\frac{1}{2\sqrt{3}}=0.29$