Answer:
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
