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5 January, 14:21

An object of mass 200 kg is released from rest from a boat into the water and allowed to sink. While gravity is pulling the object down, a buoyancy force of StartFraction 1 Over 40 EndFraction times the weight of the object is pushing the object up (weight = mg). If we assume that water resistance exerts a force on the object that is proportional to the velocity of the object, with proportionality constant 10 N-sec/m, find the equation of motion of the object. After how many seconds will the velocity of the object be 50 m/sec? Assume that the acceleration due to gravity is 9.81 m divided by sec squared.

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  1. 5 January, 14:24
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    = 7.08 s

    Explanation:

    Given that,

    - The mass of the object m = 200 kg

    - The buoyancy force F_b = W / 40

    - The weight of the object = W = m*g

    - Water resistance F_w = k*v

    - Where, k = 10 Ns/m

    Use Newton's Second law of motion to evaluate the acceleration a.

    F_net = m*a

    W - W/40 - 10*v = m*a

    a = 39/40*g - 10/m * v

    a = 39/40 * 9.81 - 10/200 * v

    a = 9.56 - 0.05*v

    - Using the first kinematic equation of motion we have:

    v_f = v_i + a*t

    object was dropped from rest, v_i = 0, Hence:

    v_f = (9.56 - 0.05*v) * t

    v_f = 9.56*t / (1 + 0.05*t)

    Find the time t when v_f = 50 m/s

    50 (1 + 0.05*t) = 9.56*t

    50 + 2.5*t = 9.56*t

    t = 50 / 7.06

    = 7.08 s
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