A 10 kg box slides down a lane inclined at an angle θ = 30. The plane has a friction of coefficient 0.1. The box starts from the rest and slides down the plane for 2.0 s. What is the distance that the box travels down the plane?

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.