1. Given data

• Mass of block: $m = 2\,\text{kg}$
• Initial speed before entering rough region: $u = 4\,\text{m s}^{-1}$
• Rough region extends from $x=0.5\,\text{m}$ to $x=1.5\,\text{m}$
• Retarding force in this region: $F(x) = -kx, \quad k = 12\,\text{N m}^{-1}$

We need the speed when the block just leaves the rough region at $x=1.5\,\text{m}$.

2. Use work-energy theorem

The change in kinetic energy is equal to the work done by the retarding force:

$K_f = K_i + W$

Initial kinetic energy:

$K_i = \frac{1}{2}mu^2 = \frac{1}{2}(2)(4^2)=16\,\text{J}$

3. Calculate work done by variable force

Work done from $x=0.5$ to $x=1.5$ is

$W = \int_{0.5}^{1.5} F(x)\,dx = \int_{0.5}^{1.5} (-kx)\,dx$

Substitute $k=12$:

$W = -12\int_{0.5}^{1.5} x\,dx$

$W = -12\left[\frac{x^2}{2}\right]_{0.5}^{1.5}$

$W = -6\left[(1.5)^2-(0.5)^2\right]$

$W = -6(2.25-0.25)$

$W = -6(2.0) = -12\,\text{J}$

4. Find final kinetic energy

$K_f = 16 + (-12) = 4\,\text{J}$

Now,

$K_f = \frac{1}{2}mv^2$

$4 = \frac{1}{2}(2)v^2$

$4 = v^2$

$v = 2\,\text{m s}^{-1}$

5. Match with options

The speed of the block as it just crosses the rough surface is

$\boxed{2.0\,\text{m s}^{-1}}$

So the correct option is C.