A 25kg box in released on a 27° incline and accelerates down the incline at 0.3 m/s2. Find the friction force impending its motion? What is

Question

A 25kg box in released on a 27° incline and accelerates down the incline at 0.3 m/s2. Find the friction force impending its motion? What is the coefficient of kinetic friction?
A block is given an initial speed of 3m/s up a 25° incline. Coefficient of friction

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Helga 4 years 2021-09-05T03:42:37+00:00 1 Answers 14 views 0

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  1. Amity
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    2021-09-05T03:44:01+00:00
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    Answer:

    a)  μ = 0.475
    , b)   μ = 0.433

    Explanation:

    a) For this exercise of Newton’s second law, we create a reference system with the x-axis parallel to the plane and the y-axis perpendicular to it

    X axis

         Wₓ – fr = m a

    the friction force has the expression

         fr = μ N

    y Axis

         N – W_{y} = 0

    let’s use trigonometry for the components the weight

         sin 27 = Wₓ / W

         Wₓ = W sin 27

         cos 27 = W_{y} / W

         W_{y} = W cos 27

         N = W cos 27

         W sin 27 – μ W cos 27 = m a

         mg sin 27 – μ mg cos 27 = m a

          μ = (g sin 27 – a) / (g cos 27)

          very = tan 27 – a / g sec 27

          μ = 0.510 – 0.0344

          μ = 0.475

    b) now the block starts with an initial speed of 3m / s. In Newton’s second law velocity does not appear, so this term does not affect the result, the change in slope does affect the result

             μ = tan 25 – 0.3 / 9.8 sec 25

             μ = 0.466 -0.03378

             μ = 0.433

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