Normal reation on the block from the wall will be `(F_("net") =0`, perpendicular to the wall)
`R =F =20N`
Therefore, limiting friction `f_(1) =muR =((1)/(4)) (20) =5N`
Weight of the block is `w = mg =(1)(10) = 10N`
A horizontal force of `10N` is applied to the block. Both weight and this force along the wall the wall. The resultant of these two force will be `10sqrt2N` in the direction shown in figure. Since ,this resultant is greater than the limiting friction. The block willmove in the direction of `F_("net")`with acceleration.
`a =(F_("net")-f_(L)/(m) =(10 sqrt2-5)/(1) =9.14 m//s^(2))`