An object of mass 100 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 50 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 90 m/sec? Assume that the acceleration due to gravity is 9.81 m divided by sec squared.

a)  v_f = 9.6138*t / ( 1 + 0.1*t)

b) t = 147 s

Step-by-step explanation:

Given:

- The mass of the object m = 100 kg

- The buoyancy force F_b = W / 50

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

- Water resistance F_w = k*v

- Where, k = 10 Ns/m

Find:

a) find the equation of motion of the object.

b) After how many seconds will the velocity of the object be 90 m/sec?

Solution:

- Construct a FBD of the object, three forces acting on the object. Weight downwards, Buoyancy and drag force upwards. Use Newton's Second law of motion to evaluate the acceleration a.

                                     F_net = m*a

                                W - W/50 - 10*v = m*a

                                 a = 49/50*g - 10/m *v

Plug in the values:

                                 a = 49/50 * 9.81 - 10/100 * v

                                 a = 9.6138 - 0.1*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 = (9.6138 -0.1*v)*t  

                                v_f = 9.6138*t / ( 1 + 0.1*t)

Find the time t when v_f = 90 m/s

                                 90(1 + 0.1*t) = 9.6138*t

                                  90 + 9*t = 9.6138*t

                                  t = 90 / 0.6138 = 147 s

(0.2)+(-8)=0.3+-3+0.3m is the equation i guess? ?

step-by-step explanation: