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 sin
θ ,
µ cos θ = 2 sin
θ ,
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 ) .
