Physics, 10.12.2019 13:31 xuzixin2004

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

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When an object is moving at terminal velocity which is constant, we have by Newton's First Laws of Motion that no net force is acting on the object.

(a) The force of air is approximately 0.14715 N

(b) The force of air is approximately 0.44145 N

Reasons:

(a) The upward force of the air at the terminal speed = The weight of the coffee filter.

The weight of a single coffee filter = 0.015 kg × 9.81 m/s² = 0.14715 N

(b) The weight of three coffee filters = The force of air

Therefore

The weight of three coffee filters = The force of air = 3 × 0.015 kg × 9.81 m/s² = 0.44145 N

The force of the air resistance when there are three filter paper is three

times the force for one filter paper paper, therefore;

Let v represent the speed of one filter paper, we have;

F₃ = 3·F₁

F₃ = k·v₃²

∴ F₃ = 3 × k·v² = k·3·v²

v₃² = 3·v²

v₃ = √(3)·v

$Time = \dfrac{Distance}{Velocity}$

The time it would take the three filter paper is therefore;

$t_3 = \dfrac{d}{\sqrt{3} \cdot v} = \dfrac{1}{\sqrt{3} } \cdot \dfrac{d}{v}$

The time for one Coffee filter, $t = \dfrac{d}{v}$

Where;

t = 49 seconds

Therefore;

The time it would take the three filter paper

$t_3 = \dfrac{1}{\sqrt{3} } \cdot \dfrac{d}{v} = \dfrac{1}{\sqrt{3} } \cdot t$

$t_3 = \dfrac{1}{\sqrt{3} } \times 49 \approx 28.3$

The time it takes a stack of three coffee filters is, t₃ ≈ 28.3 seconds

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Answer from: scottbrandon653

(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

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

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

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You drop a single coffee filter of mass 1.5 grams from a very tall building, and it takes 49 seconds...

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