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Homework Statement
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...
Homework Equations
ΣFhorizontal = Ffrict+Fgrav(sin22) = ma
ΣFvertical = Fnorm + Fgrav(cos22) = ma
Vfinal2= VInitial2 + 2a(displacement)
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!