Air resistance

[sunshine02]

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" A dog 7 kg falls 16.5 meters with his stomach pointing down. He is 0.2 meters tall, 0.15 meters wide, and 0.4 meters from nose to tail. The density of air is 1.29kg/m^3."

What is the air resistance by the time the dog gets to the bottom?

The book said to find the terminal velocity and plug it into the drag force equation. But by the time the dog gets to the bottom, it would have already reached terminal velocity, so the drag force should equal the force of gravity, which is 70 N. So air resistance should be 70 N too right? However the correct answer is 5 N. I don't really understand why you can't just set air resistance equal to force of gravity?

Thanks!

 

[Patassa]

" A dog 7 kg falls 16.5 meters with his stomach pointing down. He is 0.2 meters tall, 0.15 meters wide, and 0.4 meters from nose to tail. The density of air is 1.29kg/m^3."

What is the air resistance by the time the dog gets to the bottom?

The book said to find the terminal velocity and plug it into the drag force equation. But by the time the dog gets to the bottom, it would have already reached terminal velocity, so the drag force should equal the force of gravity, which is 70 N. So air resistance should be 70 N too right? However the correct answer is 5 N. I don't really understand why you can't just set air resistance equal to force of gravity?

Thanks!

Are you assuming he reached terminal velocity or does the answer state that the dog in fact reached terminal velocity? If he didnt, then you'll need the final velocity and the surface area to calculate drag. However, this problem seems a bit beyond the scope of mcat physics.

 

[sunshine02]

Hmm...It didn't explicitly say he reached terminal velocity, but I just thought that by the time you reach the bottom, you should have already reached terminal velocity, right?

The question had two parts. The first part asked you to calculate terminal velocity IGNORING air resistance: 18m/s. The second part asked what the air resistance is by the time the dog gets to the bottom, and they used 18m/s as the velocity to plug into the drag force/air resistance eqn (F=CdAV^2; C=0.2, d=density, A=cross-sectional area, V=velocity).

Thanks so much for your help!

 

[Patassa]

Hmm...It didn't explicitly say he reached terminal velocity, but I just thought that by the time you reach the bottom, you should have already reached terminal velocity, right?

The question had two parts. The first part asked you to calculate terminal velocity IGNORING air resistance: 18m/s. The second part asked what the air resistance is by the time the dog gets to the bottom, and they used 18m/s as the velocity to plug into the drag force/air resistance eqn (F=CdAV^2; C=0.2, d=density, A=cross-sectional area, V=velocity).

Thanks so much for your help!

I think your definition is causing the problem here. Terminal velocity only applies when there is air resistance, it's an effect of drag, no air resistance=no terminal velocity. "Terminal" does not mean "final". Final could be terminal, but terminal is when drag = g.