Deriving Coefficient of Kinetic Friction

Question

Using

1) ax = g ( sin θ - μk cos θ)

and

2) l = ( ax t ^ 2 ) / 2

Show that:

μk = tan θ - ( ( 2 l ) / ( g t ^ 2 cos θ ) )

Answer 1

ax=g( sin θ - μk cos θ)

(ax)/(g)=sin θ - μk cos θ

(ax)/(g) - sin θ = - μk cos θ

(ax)/(gcosθ) - (sin θ)/(cosθ) = - μk

(ax)/(gcosθ) - tan θ = - μk

l = ( ax t ^ 2 ) / 2

2I = ax * t^2

ax = (2I)/t^2

(ax)/(gcosθ) - tan θ = - μk

((2I)/t^2)/(gcosθ) - tan θ = - μk

((2I)/t^2)/(gcosθ) = - μk + tan θ

μk + ((2I)/(t^2)(gcosθ) = tan θ

μk = tan θ - (2I)/((t^2)(gcosθ))

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