A box with a mass of 0.6-kg falls from a very tall building. While free-falling it is subjected to a frictional drag force given by Fdrag =

Question

A box with a mass of 0.6-kg falls from a very tall building. While free-falling it is subjected to a frictional drag force given by Fdrag = bv2, where v is the speed of the object and b=5.5 N•s2/m2 What is the terminal speed that the box will reach?

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Thái Dương 4 years 2021-07-27T11:27:05+00:00 2 Answers 9 views 0

Answers ( )

  1. thongdat2
    0
    2021-07-27T11:28:47+00:00
    Reply

    Answer:

    1.03m/s

    Explanation:

    According to the question

    Fdrag = bv²

    now Fdrag = weight of the box

    that is Fdrag = mg

    where m = mass of the box, 0.6 Kg

    and g = acceleration due to gravity

    thus,

    Fdrag = bv²   becomes

    mg = bv²

    making v the subject of formula

    v = \sqrt{\frac{mg}{b} }

    v = \sqrt{\frac{0.6 * 9.8}{5.5} }

    v = 1.03m/s

  2. Latifah
    0
    2021-07-27T11:28:49+00:00
    Reply

    Answer:

    the terminal speed that the box will reach v = 1.0355 m/s

    Explanation:

    The drag force is the resultant force component that acts in the relative motion direction of the body.

    Here we have Fdrag = bv²

    But Fdrag = mg

    Therefore F = 0.6 × 9.81 =5.886 N

    v² = Fdrag / b = 5.886 N/(5.5 N s²/m²)

    v = 1.0355 m/s

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