Answer
$x=0.688m$
Work Step by Step
We can determine the required compression in the spring as follows:
$\frac{1}{2}mv_1^2+\Sigma U_{1\rightarrow 2}=\frac{1}{2}mv_2^2$
$\implies \frac{1}{2}mv_1^2+(-\frac{1}{2}kx^2-\mu_kN(x+2))=\frac{1}{2}mv_2^2$
We plug in the known values to obtain:
$\frac{1}{2}(8)(5)^2+(-\frac{200}{2}x^2-(0.25\times 78.8(x+2)))=0$
This simplifies to:
$x=0.688m$
