The upper half of an inclined plane of inclination `45^(@)` is perfectly smooth while the lower half is rough. A block starting from rest at the top comes back to rest at the bottom. The coefficient of friction for the lower half is

Options

A. `mu =sin theta`
B. `mu =cot theta`
C. `mu =2 cos theta`
D. `mu =2 tan theta`

Answer

D

Explanation

The acceleration of the body on upper half,
`a_(1)=(mg sin theta)/(m) = g sin theta`
Let v be the velocity of the block at the end of smooth half. Using `v^(2) =u^(2) + 2as,` we get
`v^(2) = u^(2) + 2a (l)/(2) =2 (g sin theta) (l)/(2) = g l sin theta`
For lower half, the body comes back to rest at the bottom, it means the body retards on this section.
The ratardation of the body on lower half
`a_(2)=(mumgcostheta-mgsintheta)/(m)=g(mucostheta-sintheta)`
Again using `v^(2) = u^(2) +2as` we get
`0 = u^(2) -2g (mu cos theta - sin theta)(l)/(2)`
`rArr u^(2) =2g (mu cos theta -sin theta) (l)/(2)`
But velocity of the body at the string of this section is equal to final velocity at the end of smooth section.
`:. gl sin theta =2g (mu cos theta -sintheta) (l)/(2)`
`rArr mu cos theta =2sin theta rArr mu =2 tan theta`
.