Solve Rain Drop VTerminal Homework: Find Drag Force & Time

[cleverfield]

Homework Statement

The terminal velocity of a 3×10−5kg raindrop is about 5m/s . Assuming a drag force , and Drag Force = -bv

1. Assuming a drag force determine the value of the constant .

2. Assuming a drag force determine the time required for such a drop, starting from rest, to reach 63% of terminal velocity.

The Attempt at a Solution

I solved Q1 with an answer of b=6*10^-5 kg/s

for #2. I used Vfinal = Vinit + at

Equation 1 = V final = 0.63 of Vterm and therefore is 3.15 m/s
V init = 0
a = ?
Time = ?

for a I used the constant b and did FBD and N2

got Fdrag - mg = ma

Work:

-(6E-5kg/s)(3.15m/s) - (3E-5kg)(9.8m/s2) = (3E-5kg)a

but that gives a = -16.1 m/s2 which doesn't make sense to me. And i dropped that number in equation 1 above.

Thanks

 

[Hannisch]

Fdrag - mg = ma

Think about the direction of the forces. The raindrop is falling down, thus accelerating downwards. The drag force is slowing the drop down and is working against the direction of motion. Isn't it gravity, on the other hand, that pulls the drop downwards and therefore causes the acceleration?

 

[AEM]

Homework Statement

The terminal velocity of a 3×10−5kg raindrop is about 5m/s . Assuming a drag force , and Drag Force = -bv

1. Assuming a drag force determine the value of the constant .

2. Assuming a drag force determine the time required for such a drop, starting from rest, to reach 63% of terminal velocity.

The Attempt at a Solution

I solved Q1 with an answer of b=6*10^-5 kg/s

for #2. I used Vfinal = Vinit + at

Equation 1 = V final = 0.63 of Vterm and therefore is 3.15 m/s
V init = 0
a = ?
Time = ?

for a I used the constant b and did FBD and N2

got Fdrag - mg = ma

Work:

-(6E-5kg/s)(3.15m/s) - (3E-5kg)(9.8m/s2) = (3E-5kg)a

but that gives a = -16.1 m/s2 which doesn't make sense to me. And i dropped that number in equation 1 above.

Thanks

The equation of motion for your situation is $$ma = -mg + F_{drag}$$

This means that you cannot use $$V_{final} = V_{init} + at$$ ! That is the solution for motion without drag. You need to integrate the equation of motion with a drag force and use that, or the appropriate relationship derived from it, to find your answer.