In the figure shown acceleration of the block is

Explanation

`N=mg cos theta - 40 sqrt(2)sin 45^(@)`
`=10xx10xx(1)/(2)-40sqrt(2)xx(1)/(sqrt(2))=10 N`
`mu_(s)N=5N, mu_(k)N=4N`
As the downward force is mg sin `theta = 50 sqrt(3)` N and upward force is `40sqrt(2)cos 45^(@)=40N`, the block will move down and force of friction will be `mu_(k)` N and upwards.
Thus, `a=(50sqrt(3)-40-4)/(10)m//s^(2)rArr a=4.3 ms^(-2)`.