1. The problem statement, all variables and given/known data A crate lies on a plane tilted at an angle θ = 22.0° to the horizontal, with μkin = 0.12. a.) determine the acceleration of the crate as it slides down the plane b.) if the carton starts from rest 9.30m up the plane from its base, what will be the carton's speed when it reaches the bottom? this is the most confusing so far.... 2. Relevant equations ΣFhorizontal = Ffrict+Fgrav(sin22) = ma ΣFvertical = Fnorm + Fgrav(cos22) = ma Vfinal2= VInitial2 + 2a(displacement) 3. The attempt at a solution ΣFhorizontal = (0.12)(Fnorm)+Fgrav(sin22) = ma = 0.12Fn + 0.375Fg=ma = 0.12(0.279Fg + ma) = ma = 0.111Fg + 0.12ma = ma = 0.111Fg = 0.88ma =Fg = 7.928ma ΣFvertical = Fnorm + Fgrav(cos22) = ma = Fn - 0.927Fg = ma =Fn=0.927Fg+ma = Fn= 0.927(7.928ma) =Fn = 7.349ma since I didnt really know what to do next.. 7.928ma/7.349ma= 1.079 so i assumed that was the acceleration then... Vfinal2 = (0m/s)2 + (2)(1.079m/s2)(9.3m) Vfinal2=(20.07m/s) (Vfinal2)1/2=(20.07m/s)1/2 Vfinal=4.480m/s Thanks in Advance!