I don't understand part b

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 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 = ZZ MV ? EKENRE- the need w fk fk = MK n need component Fx =. 4 ( 9.8 xin 30 0) perpendicular to the surface fx = 1. 9 19 of the ramp I V 2 = ( 9.8 ) ( 8 sin 30 0 ) - 1.96 IF = ma V 2 = 79.48 - 8 the mar X 9.8 ( os 30 0 ) - 1.96 = V = 8. 63 / s b. What is the coefficient of kinetic friction for the horizontal surface? (15 pts.) $6 = MKh EF= ma Vi V ! " = v . 2 + 29 (x f - X : ) keep going ... - 7 0:8.63 + 29 ( 4 ) 29 = 18. 62 9 :- 9. 3 7 / 5 2