Answer:
A) 12.08 m/s
B) 19.39 m/s
Explanation:
A) Down the hill, we will apply Newton’s second law of motion in the downward direction to get:
mg(sinθ) – F_k = ma
Where; F_k is frictional force due to kinetic friction given by the formula;
F_k = (μ_k) × F_n
F_n is normal force given by mgcosθ
Thus;
F_k = μ_k(mg cosθ)
We now have;
mg(sinθ) – μ_k(mg cosθ) = ma
Dividing through by m to get;
g(sinθ) – μ_k(g cosθ) = a
a = 9.8(sin 12.03) - 0.6(9.8 × cos 12.03)
a = -3.71 m/s²
We are told that distance d = 24.0 m and v_o = 18 m/s
Using newton's 3rd equation of motion, we have;
v = √(v_o² + 2ad)
v = √(18² + (2 × -3.71 × 24))
v = 12.08 m/s
B) Now, μ_k = 0.10
Thus;
a = 9.8(sin 12.03) - 0.1(9.8 × cos 12.03)
a = 1.08 m/s²
Using newton's 3rd equation of motion, we have;
v = √(v_o + 2ad)
v = √(18² + (2 × 1.08 × 24))
v = 19.39 m/s