An object whose mass is 1 kg is released from rest at the surface of a fluid. As

Question

An object whose mass is 1 kg is released from rest at the surface of a fluid. As the object falls through the fluid it experiences a drag force proportional to its velocity. After 2 seconds its velocity is 2 m/s. What is its terminal velocity (that is, the limit of its velocity as t rightarrow infinity)? How far under the surface of the fluid is the object when its velocity is 90% of its terminal velocity? (Give the final answers correct to one decimal place, but use four-place accuracy in the intermediate calculations. Take g = 9.8067 m/s^2.)

Explanation / Answer

Weight of the body, W = mg

= 1*9.8067

= 9.8067 N acting downwards

Drag force,

Fd = k*v <--- acting upwards

For terminal velocity, W = Fd

So, mg = k*v

So, v = mg/k = 9.8067/k

Writing equation of motion,

W - k*v = m*a

So, W - k*v = m*dv/dt

So, 9.8067 - k*v = 1*dv/dt

Solving this differential equation, we get :

v = k1*e^(-kt) + 9.8067/k

For t=0, v = 0 m/s

So, 0 = k1 + 9.8067/k ----- (1)

For, t = 2s, v = 12 m/s

So, 12 = k1*e^(-k*2) + 9.8067/k ---- (2)

Solving the two equations above, we get :

k = 0.5393

k1 = -18.18

So, the terminal velocity, Vt = 9.8067/0.5393 = 18.18 m/s <----------answer

From the velocity equation,

v = dx/dt = -18.18*e^(-0.5393*t) + 18.18

So, again integrating,

x = k2 + 18.18*t + 33.71*e^(-0.5393*t)

At, t=0, x = 0

So, 0 = k2 + 33.71

So, k2 = -33.71 m

So, x = -33.71 + 18.18*t + 33.71*e^(-0.5393*t)

For v = 0.9*18.18 = 16.36 m/s

So, -18.18*e^(-0.5393*t) + 18.18 = 16.36

So, t = 4.27 s

So, at this time, x = -33.71 + 18.18*4.27 + 33.71*e^(-0.5393*4.27)

= 47.3 m <-------answer

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