Maximum velocity of the block!

(Apply conservation of Energy.)

(Note that a body attains it's maximum speed when all the external forces acting on it gets balanced.)

$\Rightarrow kx = \mu mg$ where $x$ is the compression in the spring at the time of maximum velocity of the block.

$\text {Once the spring force is equal to the friction force, friction will absorb} \\\text {K.E.and so the velocity will then go on decreasing. }\\So ~ V_{max} ~\text {will take place when spring and friction forces are equal.}\\\text{At this point, that is at elongation of } ~X_{vmax}, ~~k*X_{vmax}=\mu*m*g. \\\text {And the block will stil lhave P.E. of } ~k*X_{vmax}^2=\dfrac{(\mu*m*g)^2}{k}...(1) \\\text {So the P.E. lost in friction PLUS converted to K.E.} \\=~Initial~ P.E. ~-~ P.E.~ left~ at ~V_{max} ~given ~ in ~(1)\\=\dfrac 1 2 *k* \{ (3\mu mg)^2 - (\mu mg)^2 \} =\color{#D61F06}{\dfrac { 4* (\mu mg)^2}{k}}.\\\text {P.E . absorbed by friction PLUS that converted to K.E.} \\=\mu mg*\dfrac{ \{3\mu mg \}}{k} + \dfrac 12 m*V_{max}^2\\ =\dfrac {\color{#D61F06}{ 3}* (\mu mg)^2}{k}+ \dfrac 12 m*V_{max}^2=\color{#D61F06}{\dfrac { 4* (\mu mg)^2}{k}} .\\V_{max} =~~~~~~\Large \color{#3D99F6}{2 \mu g \sqrt{\dfrac{m}{k}}}$
This is how I would like to explain $\color{#BA33D6}{Nishant ~Rai's }$ idea.

You can directly use fact of SHM , that
$\displaystyle{{ V }_{ max }=A\omega \\ kA=3\mu mg-\mu mg=2\mu mg\\ \omega =\sqrt { \cfrac { k }{ m } } \\ { V }_{ max }=\cfrac { 2\mu mg }{ k } \times \sqrt { \cfrac { k }{ m } } =2\mu g\sqrt { \cfrac { m }{ k } } }$