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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 .

Answer :

Olajidey

Answer:

= 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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