Answer
The block travels $~~1.5~m~~$ along the plane.
Work Step by Step
We can use work and energy to find the distance $d$ along the plane that the block travels:
$K_f+U_f = K_0+U_0+W$
$0+U_f = K_0+0+W$
$mgh = \frac{1}{2}mv^2-mg~cos~\theta~\mu_k~d$
$g~d~sin~\theta +g~cos~\theta~\mu_k~d= \frac{1}{2}v^2$
$d~(g~sin~\theta +g~cos~\theta~\mu_k)= \frac{1}{2}v^2$
$d = \frac{v^2}{(2)(g~sin~\theta +g~cos~\theta~\mu_k)}$
$d = \frac{(5.0~m/s)^2}{(2)[(9.8~m/s^2)~(sin~30^{\circ})+(9.8~m/s^2)~(0.40)~(cos~30^{\circ})]}$
$d = 1.5~m$
The block travels $~~1.5~m~~$ along the plane.
