In previous question, if `mu=0.3` the acceleration of the block will be:

Options

A. zero
B. `(g)/(10)uarr`
C. `(g)/(4)darr`
D. `(g)/(5)darr`

Answer

D

Explanation

If `mu=0.3` ltbr. Maximum value of resisting force will be
`F_("lim")=muN=0.3mg`
But have driving force is `mg//2` . Therefoer the friction will be kinetic nature, the block will slid down. Hence acceleration of the block
`a=((mg)/(2)-0.3mg)/(m)=0.2g=(g)/(5)darr` .