Answer: $0.33 m/s$

According to the work-energy theorem, $W = \Delta KE + \Delta PE$

Here, $W = -f_k x = -\mu_k mgx $

$\Delta KE = \large\frac{1}{2}$$ mv_f^2 - 0 = \large\frac{1}{2}$$ m v_f^2$

$\Delta PE = 0 - \large\frac{1}{2}$$kx^2 = -\large\frac{1}{2}$$kx^2$

$\Rightarrow -\mu_k mgx =\large\frac{1}{2}$$ m v_f^2 -\large\frac{1}{2}$$kx^2$

$\Rightarrow -0.2 \times 2 Kg \times 9.8 m/s^2 \times 0.1m = \large\frac{1}{2} $$ (2 Kg \times v_f^2 - 100 N/m \times 0.1^2 m^2)$

Rearranging and solving for $v_f$, $v_f = \sqrt {0.5 - 0.392} m/s = \sqrt{0.108} = 0.33 m/s$