Answer
$x=18.19ft$
Work Step by Step
According to the work-energy principle
$\frac{1}{2}mv_A^2+U_w-U_f=\frac{1}{2}mv_B^2$
$\frac{1}{2}mv_A^2+W(15)-\mu_k W(\frac{4}{5})(25)=\frac{1}{2}mv_B^2$
We plug in the known values to obtain:
$\frac{1}{2}(\frac{10}{32.2})(5)^2+(10)(15)-(0.2)(10)(\frac{4}{5})(25)=\frac{1}{2}(\frac{10}{32.2})v_{B}^2$
This simplifies to:
$v_B=27.08ft/s$
We know that
$S_{Cy}=S_{By}+V_{By}t+\frac{1}{2}a_y t^2$
$\implies 5=30-27.08(\frac{3}{5})t-\frac{1}{2}(32.2)t^2$
This simplifies to:
$t=0.84s$
Now $x=0+(27.08)(\frac{4}{5})(0.84)$
$\implies x=18.19ft$
