A point particle of mass m, moves along the uniformly rough QPR as shown in the figure. The coefficient of friction, between the particle and rough track equals μ. 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 o the track are equal to each other and no energy is lost when the particle changes direction from PQ to QR. The value of the coefficient of friction μ and the distance x (=QR) are respectively close to




A) 0.2 and 6.5 meterB) 0.2 and 3.5 meterC) 0.29 and 3.5 meterD) 0.29 and 6.5 meter

Given:
• Mass of the particle = mm
• Coefficient of friction = μ\mu
• The particle moves along a uniformly rough track and loses energy due to friction over two parts, PQ and QR, of the track.
• The energy lost by the particle over parts PQ and QR is equal to each other.
• The particle is released from rest at point P and comes to rest at point R.
• No energy is lost when the particle changes direction from PQ to QR.
We are required to find the value of the coefficient of friction μ\mu and the distance x=QRx = QR.
Approach:
1. Energy Lost Due to Friction: The energy lost due to friction over a distance dd is given by:
Efriction=f⋅dE_{\text{friction}} = f \cdot d
Where:
o f=μ⋅Nf = \mu \cdot N, the frictional force, and N=mgN = mg is the normal force (since the track is horizontal).
o So, f=μ⋅mgf = \mu \cdot mg.
o Therefore, the energy lost due to friction over distance dd is:
Efriction=μ⋅mg⋅dE_{\text{friction}} = \mu \cdot mg \cdot d
2. Kinetic Energy: The particle is released from rest, so its initial kinetic energy is 0. As the particle moves, its kinetic energy decreases due to friction.
Let the initial velocity at point P be vPv_P and the final velocity at point R be vR=0v_R = 0. The work-energy principle states that the work done by friction is equal to the change in kinetic energy.
Therefore, for each part of the track:
Energy Lost on PQ=μ⋅mg⋅PQ\text{Energy Lost on PQ} = \mu \cdot mg \cdot PQ
and
Energy Lost on QR=μ⋅mg⋅QR\text{Energy Lost on QR} = \mu \cdot mg \cdot QR
Since the energy lost on PQ and QR is equal, we have:
μ⋅mg⋅PQ=μ⋅mg⋅QR\mu \cdot mg \cdot PQ = \mu \cdot mg \cdot QR
Since the distances PQ and QR are given as PQ=2xPQ = 2x and QR=xQR = x, the equation simplifies as:
2x=x2x = x
Conclusion:
• Thus, the coefficient of friction is 0.2 and distance QR is 6.5m