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Physics Questions

1. Make two free body diagrams for a falling coffee filter: one at the instant when it is released and the other after it has reached terminal velocity. Your diagrams should include numerical values. In three or four sentences, desribe your free body diagrams.

2. Based on your results, how much would air resistance affect the motion of a steel ball as it falls? Assume the ball has a radius of 1 cm and a mass of 50 grams.

3. If you had a coffee filter with twice the area but the same mass as this one, would you expect the terminal velocity for the larger filter to be smaller, larger, or the same as the filter that you used? How much larger or smaller? What if it was twice the mass and twice the cross sectional area? Explain your answer.

4. If a car experiences 4000 N of air resistance while traveling at 20 m/s, how much air resistance will it experience when traveling at 10 m/s?

5. How would the results of a lab experiment change if the fluid through which the objects were dropped are more viscous? How would the graph be affected?

Comments

Hi Meg;
I updated everything and my answer is pending review.  The computer randomly selects answers for such.
 
When I minored in college physics, we did not study cross-sectional area as it applies to gravitational acceleration.  I researched everything.  The formula is...
F=(mass density)(velocity of object)(drag coefficient)(cross-sectional area)
If the cross-sectional area doubles, so too does the force.
 
Please bear in mind that the radius is presented in cm.  We need mks, meters, not centimeters.
 
I apologize for my mistake.
It's okay. So which answers would that affect? Thank you so much for your help!
Hi Meg;
One more thing.  The velocity is squared.  I cannot edit my comment.
The equation is...
F=(mass density)(velocity of object)2(drag coefficient)(cross-sectional area)
Questions 2 and 3 are affected.  In my answer, I underlined my changes.
Question 2... Calculate the area, not volume, of the sphere in meters.  Bear in mind, this is cross-sectional AREA.  Convert the figure of grams into kilograms, and plug everything into the formula.
Question 3...If cross-sectional area doubles, so too does the force of the object during a gravitational free-fall.
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1 Answer

Hi Meg;
ALL CORRECTIONS ARE UNDERLINED.  WHEN I MINORED IN PHYSICS, WE DID NOT STUDY CROSS-SECTIONAL AREA.  I HAD TO RESEARCH EVERYTHING.
1. Make two free body diagrams for a falling coffee filter: one at the instant when it is released and the other after it has reached terminal velocity. Your diagrams should include numerical values. In three or four sentences, desribe your free body diagrams.
I cannot formulate a diagram here. 
A coffee filter is designed to allow water to pass through it.  Henceforth, it may also allow air to pass through.  Air resistance is a limited issue.  Nonetheless, let's assume this is a used coffee filter which is full of grinds and all the holes are filled.  The coffee filter will be released.
For the diagram...
I would have one vertical arrow from the ground up representing air-resistance.
I would have one arrow from the filter pointing to the ground as it represents the 1st second after beginning the free-fall.
I would have a second arrow from the filter pointing to the ground as it represents the 2nd second after beginning the free-fall.  
I would have a horizontal line representing the moment terminal velocity is released.  I would also exhibit the equation, Fair=Ffilter
For the explanation...
Free-fall acceleration is 9.8 m/s2.  After the 1st second, velocity is 9.8 m/s.  After the 2nd second, velocity is 19.6 m/s.  Eventually, terminal velocity is reached as air-resistance force is equal to the force exerted by the coffee filter.  The force of the filter is calculated by multiplying its mass by its acceleration of 9.8 m/s2.

2. Based on your results, how much would air resistance affect the motion of a steel ball as it falls? Assume the ball has a radius of 1 cm and a mass of 50 grams.
The same philosophy applies.  When the force of air-resistance is equal and opposite to the force of mass multiplied by 9.8 m/s2 in the free-fall, acceleration ceases and terminal velocity is reached.  The mass of 50 grams is applicable.  However, to calculate Newtons, we need to use the mks system, meters-kilograms-seconds system.  Therefore, we must convert 50 grams into 0.05 kg.
F=ma
F=(0.05 kg)(9.80 m/s2)
F=0.49 Newtons
When the force of 0.49 Newtons is equal and opposite to the force exerted by the air-resistance of 0.49 Newtons, acceleration will cease and terminal velocity will be reached.  However, it is possible that air-resistance is not so strong.  Therefore, acceleration will never cease, and there is no terminal velocity.
The radius of the sphere is 0.1 cm.  In meters this is 0.01 meters. The equation to calculate its cross-sectional area is...(pi)(r2)
(3.14)(0.01)(0.01)
0.000304 m2
This applies to this equation...
F=(mass density)(v)(Cd)(cross-sectional area)
F=(1.29 kg/m3)(velocity)2(0.5)(0.000304 m2)
 
 
3. If you had a coffee filter with twice the area but the same mass as this one, would you expect the terminal velocity for the larger filter to be smaller, larger, or the same as the filter that you used? How much larger or smaller? What if it was twice the mass and twice the cross sectional area? Explain your answer.
Size is an issue.  The larger the cross-sectional area, the greater the ability to counter air-resistance.  The equation is...
F=(mass density)(v)2(Cd)(cross-sectional area)
If the cross-sectional area is double size, the force is also double.
Mass is also a non-issue when considering acceleration due to gravity, 9.8 m/s2.  Mass is an issue when considering the issue of terminal velocity only.
The coffee filter with twice the area but same mass would have a higher terminal velocity because it has a superior ability to counter air-resistance.
The coffee filter of twice the mass will take longer to reach terminal velocity because it has a superior ability to counter air-resistance.
4. If a car experiences 4000 N of air resistance while traveling at 20 m/s, how much air resistance will it experience when traveling at 10 m/s?
F=ma
v=at
a=v/t
F=m(v/t)
4000 N=m[(20 m/s)/t]
   x     =m[(10 m/s)/t]
 
The m in the numerator and denominator cancel, because the mass is identical.
The force of the car at the initial speed is double that of the same car at half the speed.
x=2000 Newtons

5. How would the results of a lab experiment change if the fluid through which the objects were dropped are more viscous? How would the graph be affected?
This is no longer a free-fall question.  This is a question of dropping objects through fluid.  The greater the viscosity, the greater the resistance.  The greater the resistance, the lower the acceleration and terminal velocity of the object being dropped.

Comments

For number 2, I forgot to include that Pair=1.29 kg/m3; for a sphere, Cis 0.5. Would that affect the answer you have already given me?
Hi Meg;
I apologize.  Cross-sectional size is an issue when establishing ability to fight air-resistance.  I am updating my answer now.
 

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