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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.
1131379_034f0b10f4584c99bed0e67312970aaa.png

A
0.2 and 3.5 m
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B
0.29 and 3.5 m
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C
0.29 and 6.5 m
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D
0.2 and 6.5 m
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Solution

The correct option is B 0.29 and 3.5 m
Let's say the velocity of the particle at point Q is v.
Energy at P =EP=mgh
Energy at Q =EQ=12mv2
Energy at R =ER=0, assuming datum at R.

Loss in energy is due to negative work done by dissipative force, i.e. friction. Thus:
EPEQ=μmgcos30×PQ(1)
EQER=μmg×x(2)
It is given that the losses of energies are equal in parts PQ and QR. Thus:
μmgcos30×PQ=μmg×x
x=cos30PQ=cos30sin30h
x=h×33.5 m

From (1) and (2), we have:
EPER=μmg(PQcos30+x)=23μmgh
But EPER=mgh.
μ=1230.29

μ0.29 and x3.5 m option (B).
QED.

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