Question

You drop a single coffee filter of mass 1.5 grams from a very tall building, and it takes 49 seconds to reach the ground. In a small fraction of that time the coffee filter reached terminal speed. (a) What was the upward force of the air resistance while the coffee filter was falling at terminal speed? Fair = 0.0147 N (b) Next you drop a stack of 3 of these coffee filters. What was the upward force of the air resistance while this stack of coffee filter was falling at terminal speed? Fair = 0.0441 N (c) Again assuming again that the stack reaches terminal speed very quickly, about how long will the stack of coffee filters take to hit the ground? (Hint: Consider the relation between speed and the force of air resistance.) Fall time is approximately 28.3 s Additional Materials Section 7.10 Practice Another Version Home My Assignments Extension Request Copyright © 2019 Cengage Learning, Inc. All Rights Reserved

Answer 1

(a) 0.0147 N

When the filter reaches the terminal speed, it means that its acceleration is now zero, so the net force acting on it is zero.

There are only two forces acting on the filter:

- The weight of the filter, downward: W = mg, where

$m=1.5 g = 0.0015 kg$ is the mass of the filter

$g=9.8 m/s^2$ is the acceleration due to gravity

- The air resistance, upward, $F_a$

Since the net force is zero, we have

$W-F_a =0$

and solving the equation we find the upward force of air resistance:

$F_a=W=mg=(0.0015kg)(9.8 m/s^2)=0.0147 N$

(b) 0.0441 N

The problem is exactly identical to before, but this time the mass of the stack of filters is 3 times the mass of the single filter, so

$m=3 (0.0015 kg)=0.0045 kg$

And so, the upward force of air resistance is

$F_a = mg=(0.0045 kg)(9.8 m/s^2)=0.0441 N$

(c) 28.3 s

We know that the single coffee filter takes t=49 s to reach the ground, travelling at constant terminal speed of v, so the distance covered (the height of the building) is

$d=vt$

where t = 49 s.

We also know that the air resistance is proportional to the square of the terminal speed:

$F_{air} \propto v^2$

which means

$v\propto \sqrt{F_{air}}$

For the stack of three filters, we found (b) that the air resistance is 3 times the air resistance for the single filter (a). Therefore, the terminal speed of the stack of filter will be $\sqrt{3}$ times larger than the terminal speed of the single filter.

Therefore, the time taken will be:

$t'=\frac{d}{v'}=\frac{d}{\sqrt{3} v}=\frac{t}{\sqrt{3}}$

And since t = 49 s, we have

$t'=\frac{49 s}{\sqrt{3}}=28.3 s$