Air Resistance of coffee filter

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SUMMARY

The discussion focuses on the air resistance experienced by a coffee filter and a stack of five coffee filters during free fall. The single coffee filter, weighing 1.4 grams, reaches terminal velocity after 47 seconds. The upward force of air resistance at terminal speed is determined to be proportional to the weight of the filters. The relationship between drag force and velocity is clarified, indicating that for light objects like coffee filters, the drag force can be modeled as quadratic (F=kv²) rather than linear (F=kv). This distinction is crucial for accurately calculating the time it takes for the stack of filters to reach the ground.

PREREQUISITES
  • Understanding of terminal velocity concepts
  • Familiarity with the equations of motion and forces
  • Knowledge of drag force equations (F=kv and F=kv²)
  • Basic principles of physics related to free fall
NEXT STEPS
  • Study the effects of air resistance on different shapes and masses
  • Learn about terminal velocity calculations for various objects
  • Explore the differences between linear and quadratic drag force relationships
  • Investigate experimental methods to determine drag coefficients
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and forces, as well as educators looking for practical examples of air resistance and terminal velocity in real-world scenarios.

lebprince
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Homework Statement


You drop a single coffee filter of mass 1.4 grams from a very tall building, and it takes 47 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?

b) Next you drop a stack of 5 of these coffee filters. What was the upward force of the air resistance while this stack of coffee filter was falling at terminal speed?

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.)


The Attempt at a Solution



i was able to get a and b right but part c am having problems with...Thanks
 
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The 'Hint' asks you to consider the relation between speed and the force of air resistance. What have you learned about this relationship between F_drag and v?
 
PhanthomJay said:
The 'Hint' asks you to consider the relation between speed and the force of air resistance. What have you learned about this relationship between F_drag and v?

with the instructor we have i haven't learned anything about the relationship between F drag and V...all my homework i have to dig my way out of it. I assume i should be using the formula F= k.v , so with the increase of V , F will increase. Thats about it
 
lebprince said:
with the instructor we have i haven't learned anything about the relationship between F drag and V...all my homework i have to dig my way out of it. I assume i should be using the formula F= k.v , so with the increase of V , F will increase. Thats about it
Well, that's all right, problem is, sometimes the force of air resistance varies linearly with the speed (F=kv,as you have noted, especially for small light objects or particles at low speeds), and sometimes the relationship is quadratic (F=kv^2, especially for heavier objects with higher speeds). You should have been given the relationship, or perhaps determined it from lab experiments. I believe the relationship between force and speed , for coffee filters, is quadratic.

In either case, let's assume the linear F=kv case for the 5 stacked filters. Since they weigh 5 times more than the single filter, and since you have detremined that the drag force at terminal velocity is 5 times greater than the drag force on the single filter, by how much would the terminal velocity increase? By how much would the time to fall decrease? What if F=kv^2, how would this change the results? (Note that k is a constant for the single or multiple filters cases, since k depends on shape, air density, and surface area exposed to the air, none of which change in either case).
 
Thanks for all ur help there...well since the drag force is 5 times more the 5 filters together i would assume that V would be increased by 5 and the time would decrease by 5 so if i would to divide 47 sec by 5 i would get 9.4s which is giving me a wrong answer.
 
hey buddy i figured it out...i think as you said its a quadratic so instead of dividing by 5 i divided by the sqrt of 5 and got the right answer thanks for all ur help.
 

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