A 92-kg skier is sliding down a ski slope that makes an angle of 40 degrees above the horizontal direction. the coefficient of kinetic friction between the skis and the snow is 0.293. neglecting any air resistance, what is the acceleration of the skier?

a) 5.8 m/s2

b) 4.9 m/s2

c) 9.8 m/s2

d) 4.1 m/s2

e) 2 m/s2

d) a = 4.1 m/s²

Explanation:

m = 92 Kg

∅ = 40º

μ = 0.293

a = ?

We can obtain the forces in the direction x' (of motion)

Wx' = W*Sin ∅ = m*g*Sin ∅

Ffriction = μ*N = μ*m*g*Cos ∅

then we use Newton's 2nd Law as follows:

Wx' - Ffriction = m*a

m*g*Sin ∅ - μ*m*g*Cos ∅ = m*a

⇒ a = g * (Sin ∅ - μ*Cos ∅)

⇒ a = (9.81 m/s²) * (Sin 40º - 0.293*Cos 40º)

⇒ a = 4.1038 m/s²