A block of mass m is kept on an inclined plane of a lift moving down with acceleration of `2 ms^(-1)`.What should be the coefficient of friction to let the block move down with constant velocity relative to lift.

Options

A. `mu=(1)/(sqrt(3))`
B. `mu=0.4`
C. `mu=0.8`
D. `mu=(sqrt(3))/(2)`

Answer

A

Explanation

(a) `m(g-a)sin theta=mu m(g-a)cos theta`
lt brgt `implies" " mu=tan theta=tan 30^(@)=(1)/(sqrt(3))`