You drop a single coffee filter of mass 1.5 grams from a very tall building, and

Question

You drop a single coffee filter of mass 1.5 grams from a very tall building, and it takes 50 seconds to reach the ground. In a small fraction of that time the coffee filter reached terminal speed. What was the upward force of the air resistance while the coffee filter was falling at terminal speed? Next you drop a stack of 4 of these coffee filters. What was the upward force of the air resistance while this stack of coffee filter was falling at terminal speed? 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?

Explanation / Answer

(a) At terminal speed,the speed of coffee filter becomes constant.So,net force must be equal to zero.

So the upward force of air resistance = downward force due to weight

  Fair   = mass* gravity = 1.5*10-3*9.8 =0.0147 N

(b) Here also,at terminal speed,the speed of the stack of four coffee filters becomes constant.So,net force must be equal to zero.

So the upward force of air resistance = downward force due to weight of stack of 4 coffee filters

Fair = mass* gravity = 4*1.5*10-3*9.8 =0.0588 N

(c) for air friction: Fdrag = -1/2 cpAv^2
where c is drag coefficient ,p= density ,A = cross sectional area ,v= velocity
Since c, p , A are the same for 4 filters as for one, we can drop these variables.
Fdrag= -1/2 v^2

So if one coffee filter is 0.0147 N= -1/2 v^2
then 4 coffee filters is 0.0588 N = -1/2 v^2

Solving for velocity for 4 filters is twice the speed of 1 coffee filter, so stack will hit the ground in 25 seconds.

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