Answer
$0.097m$
Work Step by Step
We know that
$W_{nc}=\frac{1}{2}mv_f^2-\frac{1}{2}kd^2$
$\implies -f_k\Delta x=\frac{1}{2}mv_f^2-\frac{1}{2}kd^2$
$-\mu_kmg\Delta x=\frac{1}{2}mv_f^2-\frac{1}{2}kd^2$
This simplifies to:
$d=\sqrt{\frac{2m}{K}(\frac{1}{2}v_f^2+\mu_kg\Delta x)}$
We plug in the known values to obtain:
$d=\sqrt{\frac{2(1.2)}{730}(\frac{1}{2}(2.3)^2+(0.44)(9.81)(0.050))}=0.097m$