A point particle of mass m, moves long the uniformly rough track PQR as shown in 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 value of the coefficient of friction `mu` and the distance x `(=QR)`, are, respectively close to:

Options

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

Answer

C

Explanation

As energy lost over `PQ=` energy lost over `QR`
`:. (mu mg cos theta)PQ=(mu mg)QR`
or `QR=PQcos theta` .....(i)
From figure, `sin theta=sin 30^(@)=(2)/(PQ)=(1)/(2)`
`PQ=4m`
From (i), `QR=4 cos 30^(@)=4 (sqrt(3))/(2)=2sqrt(3)m=3.5m`
Again, decrease in `P.E.=` loss of energy due to friction in `PQ` and `QR`
`mgh=(mu mg cos theta) PQ+mu mg xx QR`
`h=mu cos thetaxxPQ+mu mg xx QR`
`h=mu cos thetaxx PQ +mu xx QR`
`2= mu cos 30^(@)xx4+muxx2sqrt(3)`
`=mu(4xx(sqrt(3))/(2)+2sqrt(3))=muxx4sqrt(3)`
`mu=(2)/(4sqrt(3))=(1)/(2sqrt(3))=0.29`