Chapter 8 - Potential Energy and Conservation of Energy - Problems and Conceptual Exercises - Page 251: 98

$0.097m$

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: $d=\sqrt{\frac{2(1.2)}{730}(\frac{1}{2}(2.3)^2+(0.44)(9.81)(0.050))}=0.097m$