A block of mass 5kg is placed at rest on a table of rough surface....

Forces acting on block is shown below

From the second equation of motion,

$s = \dfrac{1}{2}a{t}^{2} ⇒ 50 = \dfrac{1}{2}a \times {10}^{2} ⇒ a = 1m{s}^{- 2}$

Applying Newton's second law in vertical direction,

$R - mg = 0 ⇒ R = mg$

(where $R$ is the magnitude of normal reaction on the block due to the surface)

If $\mu$ is the coefficient of kinetic friction between the block and the surface, applying Newton's second law in horizontal direction,

$30 - \muR = ma ⇒ 30 - \mumg = ma ⇒ 30 - 50\mu = 5 ⇒ \mu = \dfrac{1}{2} = 0 . 5$