A28 kg rock approaches the foot of a hill with a speed of 15 m/s. this hill slopes upward at a constant angle of 40.0◦ above the horizontal. the coefficients of static and kinetic friction between the hill and the rock are 0.75 and 0.20, respectively. (a) use energy conservation to find the maximum height above the foot of the hill reached by the rock. (b) will the rock remain at rest at its highest point, or will it slide back down the hill? (c) if the rock does slide back down, find its speed when it returns to the bottom of the hill.

a) h = 9.2606 m

b) The rock will slide back down the hill

c) vf = 11.7638 m/s

Explanation:

Given

m = 28 Kg

vi = 15 m/s

∅ = 40º

μs = 0.75

μk = 0.20

a) hmax = ?

Using The principle of conservation of energy we have

Wf = ΔE

where

Wf = - Ff*d = - (μk*N)*d = - μk*(m*g*Cos ∅)*(h / Sin ∅)     (I)

Ff is the frictional force (μk*N) and d (h / Sin ∅) is the displacement

then

ΔE = Ef - Ei = (Kf + Uf) - (Ki + Ui) = (0 + Uf) - (Ki + 0) = Uf - Ki

Ui = 0 since hi = 0   and   Kf = 0   since vf = 0

⇒   ΔE = Uf - Ki = m*g*h - 0.5*m*vi² = m*(g*h - 0.5*vi²)       (II)

we can apply

Wf = ΔE  ⇒  (I) = (II)   ⇒  - μk*(m*g*Cos ∅)*(h / Sin ∅) = m*(g*h - 0.5*vi²)

⇒   h = (0.5*vi²*Sin ∅) / (g*(Sin ∅ + μk*Cos ∅))

⇒   h = (0.5*15²*Sin 40º) / (9.81*(Sin 40º + 0.20*Cos 40º))

⇒   h = 9.2606 m

b) In order to know if the rock will remain at rest at its highest point, or will it slide back down the hill we have to get Ffmax:

⇒   Ffmax = μs*N = μs*(m*g*Cos ∅) = 0.75*28*9.81*Cos 40º = 157.8128 N

and the component of the Weight along a direction down the hill

⇒   Wx' = W*Sin ∅ = m*g*Sin ∅ = 28*9.81*Sin 40º = 176.5609 N

since Ffmax < Wx'  the rock will slide back down the hill

c) Given

m = 28 Kg

vi = 0 m/s

hi = 9.2606 m

hf = 0 m

∅ = 40º

μs = 0.75

μk = 0.20

vf = ?

Using The principle of conservation of energy we have

Wf = ΔE

where

Wf = - Ff*d = - (μk*N)*d = - μk*(m*g*Cos ∅)*(h / Sin ∅)     (III)

then

ΔE = Ef - Ei = (Kf + Uf) - (Ki + Ui) = (Kf + 0) - (0 + Ui) = Kf - Ui

Uf = 0 since hf = 0   and   Ki = 0   since vi = 0

⇒   ΔE = Kf - Ui = 0.5*m*vf² - m*g*h = m*(0.5*vf² - g*h)       (IV)

we can apply

Wf = ΔE  ⇒  (III) = (IV)   ⇒  - μk*(m*g*Cos ∅)*(h / Sin ∅) = m*(0.5*vf² - g*h)

⇒   vf = √(2*g*h*(1 - μk*Ctg ∅))

⇒   vf = √(2*9.81*9.2606*(1 - 0.2*Ctg 40º))

⇒   vf = 11.7638 m/s

the answer would be 32*f