Answer:
The distance the box traveled down the plane is 19.28 m
Explanation:
The angle of repose, α, is given by the relation;
tan⁻¹(μ) = α
tan⁻¹(0.1) = 5.7°
Therefore, we have;
M·g·sin(θ) - μ·N = M·a
Where:
M = Mass of the box = 10 kg
g = Acceleration due to gravity = 9.81 m/s²
θ = Angle of inclination of the plane = 30°°
μ = Coefficient of friction = 0.1
a = Acceleration of the box along the incline plane
N = Normal force due to the weight of the box = M·g·cos(θ)
10 × 9.81 × sin30 - 0.1 × 9.81 × cos(30) = 10 × a
48.2 = 10 ×a
a = 48.2/10 = 4.82 m/s²
The distance, s, traveled by the box is given by the relation;
s = u·t + 1/2×a·t²
Where:
u = Initial velocity = 0 m/s
t = Time of motion = 2.0 s
∴ s = 0×2 + 1/2 × 4.8 × 2² = 19.28 m
The box traveled 19.28 m down the plane.
