A block of mass 5 kg is kept against an accelerating wedge with a wedge angle of `45^(@)` to the horizontal. The coefficient of friction between the block and the wedge is `mu = 0.4`. What is the minimum absolute value of the acceleration of the wedge to keep the block steady? [Assume, `g = 10 m/s^(2)`]
 

Options

A. `(60)/(7)m//s^(2)`
B. `(30)/(7)m//s^(2)`
C. `(30)/(sqrt(7))m//s^(2)`
D. `(60)/(sqrt(7))m//s^(2)`

Answer

B

Explanation

Let a = acceleration of wedge, then force on block due to acceleration of wedge = ma.

Then, from FBD we have,
For equilibrium of block,

`"mg "sin a="ma "cosalpha+mu("ma "cosalpha+"ma "sin alpha)`
`impliesg=a+mu(g+a)implies(0.6)/(1.4)g=9`
`implies" "a=(30)/(7)"ms"^(-2)`