A block of mass `5.0kg` slides down from the top of an inclined plane of length `3m`. The first `1m` of the plane is smooth and the next `2m` is rough. The block is released from rest and again comes to rest at the bottom of the plane. If the plane is inclined at `30^@` with the horizontal, find the coefficient of friction on the rough portion.

Options

A. (a) `2/sqrt3`
B. (b) `sqrt3/2`
C. (c) `sqrt3/4`
D. (d) `sqrt3/5`

Answer

B

Explanation

Force method:
From P to Q: `v^2=0^2+2a_1s_1`
where `a_1=gsin30^@`, `s_1=1m`
and from Q to R: `0^2=v^2+2a_2s_2`
where `a_2=g sin 30^@-mu g cos 30^@`, `s_2=2m`
solve to get `mu=sqrt3//2`
Work-energy method:

In terms of energy considerations you can summarize the whole process as loss in gravitational potential energy of the block=work done against friction, or
`(mg)xx(3 sin 30^@)=(mu mg cos theta)xx2`
`3xx1//2muxxsqrt3/2impliesmu=sqrt3/2`