A skier traveling 9.4 m/s reaches the foot of a steady upward 16 ∘ incline and glides 12 m up along this slope before coming to rest. What was the average coefficient of friction?

The  average coefficient of friction is 0.104

Explanation:

Given;

velocity of the skier, v = 9.4 m/s

angle of inclination, θ = 16°

height of the slope, h = 12 m

Calculate the kinetic energy of the skier at the given velocity;

KE =¹/₂mv²

KE = ¹/₂m(9.4)²

KE = 44.18m

Calculate the potential energy of the skier up the slope;

PE = mghsinθ

PE = m x 9.8 x 12 x sin16

PE = 32.411 m

Calculate the friction force on the inclined plane;

Fk = μk x mgCosθ

Fk =  μk x m x 9.8 x Cos16

Fk =  μk x m x 9.421

Calculate the work done by friction;

Work done = Fk x d

Work done = μk x m x 9.421 x d

Work done = μk x m x 9.421 x 12

Work done = 113.052mμk

Apply the principle of conservation of energy;

KE = PE + work done by friction

44.18m = 32.411 m + 113.052mμk (factor out m and divide through by it)

44.18 = 32.411  + 113.052μk

44.18 - 32.411  = 113.052μk

11.769 = 113.052μk

μk = 11.769 / 113.052

μk = 0.104

Therefore, the  average coefficient of friction is 0.104

probably not if net force is zero