Question

You drop a single coffee filter of mass 1.7 g from a very tall building, and it takes 52 s 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? (b) Next you drop a stack of five of these coffee filters. What was the upward force of the air resistance while this stack of coffee filters was falling at terminal speed? (c) Again assuming that the stack reaches terminal speed very quickly, about how long will the stack of coffee filters take to hit the ground?

Short answer

1. The air resistance force on the single coffee filter is 0.0166 N.
2. The air resistance of five coffee filters is 0.0833 N.
3. The time that the stack of coffee filters to hit the ground is 23.4 s.

Step-by-step solution

Identification of given data

The given data can be listed below,

• The mass of a single coffee filter is, $\mathrm{m}=1.7\mathrm{g}\left(\frac{1\mathrm{kg}}{1000\mathrm{g}}\right)=1.7\times {10}^{-3}\mathrm{kg}$.
• The time taken by coffee filter to reach ground is, t = 52 s

Concept of air resistance force

The friction is a force that is always against the direction of displacement. In the case of a fluid like air, things do not change, even if the resistance depends on the shape in the hypothesis of perfect (frictionless, inviscid) motion.

(a) Determination of the upward force of the air resistance while the coffee filter was falling at terminal speed

When a body attends terminal velocity its resistance force is equal to weight of body and net force is equal to zero.

So, the air resistance force is given by,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{air}}=\mathrm{W} \\ =\mathrm{mg} \end{gathered}$$

Here, m is the mass of the coffee filter and g is the acceleration due to gravity whose value is $9.8m/s^2$.

Substitute values in the above equation.

role="math" $$\begin{gathered} {\mathrm{F}}_{\mathrm{air}}=\left(1.7\times {10}^{-3}\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right) \\ =0.0166\mathrm{N} \end{gathered}$$

Thus, the air resistance force on the single coffee filter 0.0166 N .

(b) Determination of the upward force of the air resistance five stack of coffee filters while it was falling at terminal speed

The mass of five coffee filters is given by,

$$\begin{gathered} m_2=5\left(1.7\times {10}^{-3}kg\right) \\ =8.5\times {10}^{-3}kg \end{gathered}$$

The upward air resistance force is given by,

$F_{air}=m_{2}g$

Here, $m_2$is the mass of five coffee filter and g is the acceleration due to gravity.

Substitute values in the above,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{air}}=\left(8.5\times {10}^{-3}\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right) \\ =0.0833\mathrm{N} \end{gathered}$$

Thus, the air resistance of five coffee filters is$=0.0833\mathrm{N}$ .

(c) Determination of the time that the stack of coffee filters taken to hit the ground

The air resistance force is directly proportional to velocity of the object which is given by,

$F_{air}\alpha v^2$

The air resistance of single coffee filter and five coffee filters is also proportional to the speed of the stick. So, the ratio of these two is given by,

role="math" localid="1657881780900" $$\begin{gathered} \frac{m}{5m}=\frac{v_1^2}{v_2^2} \\ \frac{v_1}{v_2}=\sqrt{\frac{1}{5}} \\ {v}_1=0.45v_2 \end{gathered}$$

Here, is the velocity of single coffee filter and is the velocity of the five coffee filters.

The velocity is given by,

$v=\frac{d}{t}$

Here, d is the distance and t is the time.

From the above equation.

$v\alpha \frac{1}{t}$

So, the ratio of two velocities can be given by,

$\frac{v_1}{v_2}=\frac{t_2}{t_1}$

Substitute all the values in the above,

$$\begin{gathered} \frac{0.45v_2}{v_2}=\frac{t_2}{52s} \\ {t}_2=23.4s \end{gathered}$$

Thus, the time that the stack of coffee filters to hit the ground is $23.4s$.