Hi! I don't understand what he is saying I need to do.

mass m is at rest on a rough inclined plane with length 8 m and which makes an angle of 30 with the horizontal. The object is released from rest and it stops on a rough horizontal surface a distance of 4 m from the bottom of the inclined plane. The coefficient of kinetic friction on the inclined plane is 0.4. (30 pts.) 8 m 4 m 30% this is a force , not an energy a. What is the speed of the object at the bottom of the inclined plane? Use energy methods to solve. (15 pts.) KE = 2 mv IKEEPE - need w fk = MKn need component fu = . 4 ( 9.8 xin 30 ) perpendicular to the surface f* = 1. 9 9 of the ramp IV 2 = ( 9.8 ) ( 8 sin 30 ) - 1.96 IF = ma V = 74.48 - 8 X ( 9.8 ( os 309 ) - 1:96 = V = 8.63 m / s