Answer:
μ = Sin θ * d₁ / (d₂ - Cos θ*d₁)
d₂ = (d₁*Sin θ) / μ
Step-by-step explanation:
a) We apply The work-energy theorem
W = ΔE
W = - Ff*d
Ff = μ*N = μ*m*g
<em>Distance 1:</em>
- Ff*d₁ = Ef - Ei
⇒ - (μ*m*g*Cos θ)*d₁ = (Kf+Uf) - (Ki+Ui) = (Kf+0) - (0+Ui) = Kf - Ui
Kf = 0.5*m*vf² = 0.5*m*v²
Ui = m*g*h = m*g*d₁*Sin θ
then
- (μ*m*g*Cos θ)*d₁ = 0.5*m*v² - m*g*d₁*Sin θ
⇒ - μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ <em>(I)</em>
<em>Distance 2:</em>
<em />
- Ff*d₂ = Ef - Ei
⇒ - (μ*m*g)*d₂ = (0+0) - (Ki+0) = - Ki
Ki = 0.5*m*vi² = 0.5*m*v²
then
- (μ*m*g)*d₂ = - 0.5*m*v²
⇒ μ*g*d₂ = 0.5*v² <em>(II)</em>
<em />
<em>If we apply (I) + (II)</em>
- μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ
μ*g*d₂ = 0.5*v²
⇒ μ*g (d₂ - Cos θ*d₁) = v² - g*d₁*Sin θ <em> (III)</em>
Applying the equation (for the distance 1) we get v:
vf² = vi² + 2*a*d = 0² + 2*(g*Sin θ)*d₁ ⇒ vf² = 2*g*Sin θ*d₁ = v²
then (from the equation <em>III</em>) we get
μ*g (d₂ - Cos θ*d₁) = 2*g*Sin θ*d₁ - g*d₁*Sin θ
⇒ μ (d₂ - Cos θ*d₁) = Sin θ * d₁
⇒ μ = Sin θ * d₁ / (d₂ - Cos θ*d₁)
b)
If μ is a known value
d₂ = ?
We apply The work-energy theorem again
W = ΔK ⇒ - Ff*d₂ = Kf - Ki
Ff = μ*m*g
Kf = 0
Ki = 0.5*m*v² = 0.5*m*2*g*Sin θ*d₁ = m*g*Sin θ*d₁
Finally
- μ*m*g*d₂ = 0 - m*g*Sin θ*d₁ ⇒ d₂ = Sin θ*d₁ / μ
