A block of mass `m` is lying on a wedge having inclination angle `alpha = tan^(-1)((1)/(5))` wedge is moving with a constant acceleration `a = 2 ms^(-2)` The minimum value of coefficient of friction `mu` so that `m` remain stationary w.r.t. wedge is

Options

A. `2//9`
B. `5//12`
C. `1//5`
D. `2//5`

Answer

b

Explanation

FBD of m in frame of wedge
`N = mg cos alpha - ma sin alpha`
Now `f = mu N = ma cos alpha + mg sin alpha`
`mu = (a cos alpha + g sin alpha)/(g cos alpha - a sin alpha)`
`= (a + g tan alpha)/(g - tan alpha) = (5)/(12)`