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 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.)
Fall time is approximately

Explanation / Answer

Terminal velocity equals the force of gravity on the coffee filter.

a)for terminal velocity the downward and upward force must balance hence,

Force = .0015 kg x 9.8 m/ s2

so this force is equal to the Air resistance Force =0 .0147N

b)
5times the weight will give a force og 5* 0.0015 * 9.8 N
=5 x 0.0147 = .0.0735N

c) for air friction:
f drag = -1/2 cpAv2
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, you can drop those variables and get
f drag = -1/2 v2

So if one coffee filter is0.0147 N= -1/2 v^2

so v = 0.171m/s

so, distance = 0.171m/s* 47 = 8.05 m
then 5 coffee filters is 0.0735N = -1/2 v2

so, the stack will have a speed of 0.383m/s

so, time = 8.05/ 0.383 m/s = 21.04 secs

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