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## Homework Statement

A crate lies on a plane tilted at an angle θ = 22.0° to the horizontal, with μ

_{kin}= 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

ΣF

_{horizontal}= Ffrict+Fgrav(sin22) = ma

ΣF

_{vertical}= Fnorm + Fgrav(cos22) = ma

V

_{final}

^{2}= V

_{Initial}

^{2}+ 2a(displacement)

## The Attempt at a Solution

ΣF

_{horizontal}= (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

ΣF

_{vertical}= 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...

V

_{final}

^{2}= (0m/s)

^{2}+ (2)(1.079m/s

^{2})(9.3m)

V

_{final}

^{2}=(20.07m/s)

(V

_{final}

^{2})

^{1/2}=(20.07m/s)

^{1/2}

V

_{final}=4.480m/s

Thanks in Advance!