A block of mass $$10$$ kg starts sliding on a surface with an initial velocity of $$9.8$$ ms$$^{-1}$$. The coefficient of friction between the surface and block is $$0.5$$. The distance covered by the block before coming to rest is : [use $$g = 9.8$$ ms$$^{-2}$$]

We are given: mass $$m = 10$$ kg, initial velocity $$u = 9.8$$ m/s, coefficient of friction $$\mu = 0.5$$, $$g = 9.8$$ m/s$$^2$$.

Calculate the deceleration due to friction: the frictional force is $$f = \mu mg$$, so the deceleration is:

$$ a = \mu g = 0.5 \times 9.8 = 4.9 \text{ m/s}^2 $$

Use the kinematic equation to find the distance: using $$v^2 = u^2 - 2as$$ where $$v = 0$$ (block comes to rest):

$$ 0 = u^2 - 2as $$

$$ s = \frac{u^2}{2a} = \frac{(9.8)^2}{2 \times 4.9} = \frac{96.04}{9.8} = 9.8 \text{ m} $$

Therefore, the correct answer is Option A.