A block of mass 10kg starts sliding on a surface with an initial...

Given initial velocity $u = 9 . 8m{s}^{- 1}$.

Acceleration $a = \mug = 0 . 5 \times 9 . 8 = 4 . 9m{s}^{- 2}$

Using, ${v}^{2} = {u}^{2} + 2as$, here, final velocity $v = 0$.

So, $0 = {\left(9 . 8\right)}^{2} + 2\left(4 . 9\right)s ⇒ s = 9 . 8m$