Answer
$d = 6.7~m$
Work Step by Step
We can use work and energy to find the speed at the top of the ramp:
$K+U_g = U_e+W_{ext}$
$K = U_e+W_{ext}-U_g$
$\frac{1}{2}mv^2 = \frac{1}{2}kx^2-mg~cos~\theta~\mu_k~s-mgh$
$v^2 = \frac{kx^2}{m}-2g~cos~\theta~\mu_k~\frac{h}{sin~\theta}-2gh$
$v = \sqrt{\frac{kx^2}{m}-2g~cot~\theta~\mu_k~h-2gh}$
$v = \sqrt{\frac{(1000~N/m)(0.15~m)^2}{0.200~kg}-(2)(9.8~m/s^2)~(cot~45^{\circ})~(0.20)~(2.0~m)-(2)(9.8~m/s^2)(2.0~m)}$
$v = 8.09~m/s$
We can find the horizontal distance $d$:
$d = \frac{v^2~sin~2\theta}{g}$
$d = \frac{(8.09~m/s)^2~sin~[(2)(45^{\circ})]}{9.8~m/s^2}$
$d = 6.7~m$
