A block of mass 2kg moving on a horizontal surface with speed of 4...

Acceleration of the block $a = \dfrac{f}{m} = - \dfrac{12x}{2} = - 6x$ (Here the block is retarding)

Now, using, $a = v\dfrac{dv}{dx}$
$⇒ vdv = adx$
On integrating, we get

${∫}_{4}^{v}v d v = - 6{∫}_{0 . 5}^{1 . 5}x d x$

$⇒ \dfrac{{v}^{2} - {4}^{2}}{2} = - 6\left(\dfrac{{\left(1 . 5\right)}^{2} - {\left(0 . 5\right)}^{2}}{2}\right)$

$⇒ {v}^{2} = 16 - 12 ⇒ v = 2m{s}^{- 1}$