Answer
The block travels $~~2.6~m~~$ along the plane.
Work Step by Step
We can use conservation of energy to find the distance along the plane that the block travels:
$K_f+U_f = K_0+U_0$
$0+U_f = K_0+0$
$mgh = \frac{1}{2}mv^2$
$g~d~sin~\theta = \frac{1}{2}v^2$
$d = \frac{v^2}{2g~sin~\theta}$
$d = \frac{(5.0~m/s)^2}{(2)(9.8~m/s^2)~(sin~30^{\circ})}$
$d = 2.6~m$
The block travels $~~2.6~m~~$ along the plane.
