1. Interpret the force law

   The retarding force in the rough region is given as $F_r=-kx,$ where $k=10\,\text{N/m}$.

   The block enters this region at $x=0.1\,\text{m}$ and leaves at $x=1.9\,\text{m}$.

2. Use work-energy theorem

   Work done by the retarding force from $x=0.1$ to $x=1.9$ is $W=\int_{0.1}^{1.9} F\,dx=\int_{0.1}^{1.9} (-kx)\,dx.$

   Substituting $k=10$: $W=-10\int_{0.1}^{1.9} x\,dx$ $=-10\left[\frac{x^2}{2}\right]_{0.1}^{1.9}$ $=-5\left(1.9^2-0.1^2\right).$

   Now, $1.9^2=3.61, \quad 0.1^2=0.01$ so $W=-5(3.61-0.01)=-5(3.60)=-18\,\text{J}.$

3. Initial kinetic energy

   $K_i=\frac{1}{2}mv_i^2=\frac{1}{2}(1)(10^2)=50\,\text{J}.$

4. Final kinetic energy

   By work-energy theorem, $K_f=K_i+W=50-18=32\,\text{J}.$

5. Find final speed

   $K_f=\frac{1}{2}mv_f^2$ $32=\frac{1}{2}(1)v_f^2$ $v_f^2=64$ $v_f=8\,\text{m/s}.$

6. Check options

   • A: $4\,\text{m/s}$ ❌
   • B: $10\,\text{m/s}$ ❌
   • C: $8\,\text{m/s}$ ✅
   • D: $6\,\text{m/s}$ ❌

Therefore, the correct answer is Option C.