a
From $FBD,$ both the forces are acting downward along the inclined plane. Hence, there is deceleration of the block while it moves up the inclined plane and velocity decreases and becomes zero at some distance S from the bottom of the inclined plane.
$m a=m g \sin 37^{\circ}+\mu N=m g \sin 37^{\circ}+\mu m g \cos 37^{\circ}$
or $a=10 \sin 37^{\circ}+(0.5) 10 \cos 37^{\circ}=10 \mathrm{m} / \mathrm{s}^{2}$
using formula, $v^{2}=u^{2}-2 a S,$ we get
$0=5^{2}-2(10) S \Rightarrow S=\frac{25}{20}=1.25 \mathrm{m}$
