Find out the range of force in the above situaion for which 2 kg block does not move on the incline.

Explanation

F.B.D. of 2kg block

Now take different value of F
Force ( F ) F.B.D.
Acceleration Friction Type Direction & magnitude
`a = ( F_("net"))/( m)`

From the above table block doesn't move from `F = 4N ( mg sin theta - mu m cos theat ) ` to `F = 20 N ( mg sin theta + mu m g cos theta ) `. So friction develope a range of force for which block doesn't move