Engery- Falling Coffee Filter Problem

In summary, the coffee filter falls faster the higher up it is dropped, and takes longer to hit the ground the higher it is dropped.
  • #1
quickclick330
83
0
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?
Fair = 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 = 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 ___s

Where do i need to start to begin this problem?? Thanks for the help! :-)
 
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  • #2
What is the net force on the filter at terminal velocity?
 
  • #3
zero
 
  • #4
because there isn't any more change in momentum
 
  • #5
quickclick330 said:
zero

exactly. So you have the upward air resistance force and downward gravitational force acting on the object... so what is the air resistance force equal to at terminal velocity?
 
  • #6
ahhh...i got it. mg = Fair. Now I'm having issues with part c? would you just divide 47 by 3?
 
  • #7
What is the equation of air resistance force in terms of velocity?

What is the terminal velocity in the case of the single filter... using the case of the single filter, what is the height?

What is the terminal velocity in the case of 3 filters... what is the time using the height calculated previously?
 
  • #8
In the book they give us an equation for approximate air resistance?
 
  • #9
quickclick330 said:
In the book they give us an equation for approximate air resistance?

what's the equation?
 
  • #10
can someone elaborate on how to get c? the previous posts don't help at all
 
  • #11
rubberduck said:
can someone elaborate on how to get c? the previous posts don't help at all

What learningphysics asked you for is reasonable. What relationship have they given you to work with, giving some proportionality between the force of air resistance and the speed of the object through the air. Is it F = k·v? F = k·(v^2)? The constant k is the "coefficient of drag", which you will be able to eliminate when you compare the terminal velocities for parts (b) and (c). You know what the resistance force equals at terminal velocity in both cases, so a ratio of the forces will give you a ratio involving the two different terminal velocities. (You don't need to know the height the filters fall through. Just call it H, since it will also cancel out in your comparison ratio.)
 

1. What is the "Energy- Falling Coffee Filter Problem"?

The Energy- Falling Coffee Filter Problem is a physics problem that involves calculating the amount of potential and kinetic energy of a falling coffee filter.

2. How is potential energy involved in this problem?

Potential energy is involved in this problem because the coffee filter has potential energy due to its position above the ground. As it falls, the potential energy is converted into kinetic energy.

3. How is kinetic energy involved in this problem?

Kinetic energy is involved in this problem because the coffee filter gains kinetic energy as it falls towards the ground due to its motion.

4. What factors affect the amount of potential and kinetic energy in this problem?

The factors that affect the amount of potential and kinetic energy in this problem are the mass of the coffee filter, the height at which it is dropped, and the force of gravity.

5. How can this problem be solved?

This problem can be solved by using the equations for potential and kinetic energy, and plugging in the appropriate values for the mass, height, and gravitational force. The solution will give the total energy of the falling coffee filter.

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