The normal force on the slope is 60gcos37 Newtons and the frictional force resulting from this is 0.15×60gcos37 Newtons. The force down the slope in the absence of friction is 60gsin37 (g=acceleration of gravity=9.81m/s²). The frictional force counteracts the gravitational force so the net force is 60gsin37-0.15×60gcos37 N. F=ma where m=mass and a=acceleration, so 60a=60gsin37-0.15×60gcos37.
We can divide through by 60: a=g(sin37-0.15cos37)=9.81(0.4820)=4.73m/s² approx.