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 s

Question

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?

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Dâu 5 years 2021-08-23T11:21:27+00:00 1 Answers 27 views 0

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  1. minhkhang
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    2021-08-23T11:23:11+00:00
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    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.

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