A box is dropped onto a conveyor belt moving at 3.2 m/s. If the coefficient of friction between the box and the belt is 0.28, how long will

Question

A box is dropped onto a conveyor belt moving at 3.2 m/s. If the coefficient of friction between the box and the belt is 0.28, how long will it take (in s) before the box moves without slipping

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Khải Quang 4 years 2021-08-06T01:16:54+00:00 1 Answers 243 views 0

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  1. haidang
    0
    2021-08-06T01:17:54+00:00
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    Answer:

    t = 1.16 s.

    Explanation:

    Given,

    speed of conveyor belt, v = 3.2 m/s

    coefficient of friction,f = 0.28

    Using newton second law

    f = ma

    and we also know that frictional force

    f = μ N

    f = μ m g

    equating both the force equation

    a = μ g

    a = 0.28 x 9.81

    a = 2.75 m/s²

    Using Kinematic equation

    v = u + at

    3.2 = 0 + 2.75 x t

    t = 1.16 s.

    Time taken by the box to move without slipping is 1.16 s.

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