The upper half of an inclined plane with inclination is perfectly smooth while the lower half is rough . A body starting from rest at the top will again come to rest at the bottom . What is the coefficient of friction . . . . . . . . . . ?

1 ) The upper half of an inclined plane with inclination is perfectly smooth while the lower half is rough . A body starting from rest at the top will again come to rest at the bottom .  What is the coefficient of friction . . . . . . . . . . ?

A ) cos θ / sin θ ,                                                B ) cot θ / 3 ,

C ) 2 tan θ ,                                                         D ) None of these ,

Correct Answer : -   C ) 2 tan θ ,

Solution : -

Work done against friction is zero for smooth surface .

So  ,  W = ∆ K.E = 0 . ( initial and final speed is zero ) .

So ,

Work done by friction + Work done by gravity = 0 ,

- µ mg cos θ  ( L / 2 )  +  mg sin θ L  =  0   ,

∴     µ = friction constant  ,  L = length .

So ,

Take  L / 2  as common from equation .

(L / 2) • { - µ mg cos θ  + 2 mg sin θ }  =  0   ,

So ,

- µ mg cos θ  + 2 mg sin θ  =  0   ,

µ mg cos θ  =  2 mg sinθ  ,

µ cos θ  =  2 sinθ  ,

µ  =  2 Sin θ / cos θ  ,

µ  =  2 tan θ  ,  Answer .

OR

 

2 ) We have also formula for friction force . . . . . . . . . : -

f = µ N ,

∴  f = friction force  , 

∴   N = normal force : -   Fn = mg  .

∴  µ = friction constant  or coefficient of friction . 

∴   µ = f / N    ,    ( it has no units ) .