Calculating Terminal Velocity for a Falling Object with Air Resistance

Arcarius
Messages
14
Reaction score
0

Homework Statement


An egg of mass 0.10 kg is dropped from rest at a height h. It is known that the egg dropped experiences a force due to air resistance of the form F = -kv, where v is the velocity of the egg and k is a constant of proportionality equal to 0.01781 N*s/m. What is the terminal velocity of the egg?


Homework Equations



V = sqrt(2mg/pAC)
V_t is terminal velocity,
m is the mass of the falling object,
g is the acceleration due to gravity,
C_d is the drag coefficient,
\rho is the density of the fluid through which the object is falling, and
A is the projected area of the object.


The Attempt at a Solution


The only thing I know about terminal velocity is the formula mentioned above, but it seems to me that I'm missing information (I don't have the projected area of the object, the density of the air, etc.)
The only thing I can think of is that the air resistance covers all of these and can be substituted in, giving me V = sqrt(2mg/-kv). However, I checked this with dimensional analysis and it doesn't yield the correct units. This problem is really confusing me, so I'd appreciate any help on it!
 
on Phys.org
Identify the forces acting on the egg as it falls. How do the forces relate to the acceleration of the egg? At terminal velocity, what is the acceleration?
 
TSny said:
Identify the forces acting on the egg as it falls. How do the forces relate to the acceleration of the egg? At terminal velocity, what is the acceleration?
Thank you for the response!
The forces acting on it are kv going upwards and mg going downwards. I'm taking the downwards direction as positive, meaning that the net force acting on it is mg-kv which is equal to ma. So the acceleration is equal to (mg-kv)/m, and I know that it is 0 when it reaches terminal velocity. So, I got (mg-kv)/m = 0, and solving for v, I got mg/k.
Is this correct?
 
That's correct.
 
  • Like
Likes   Reactions: 1 person
TSny said:
That's correct.

Awesome, thank you for your help! Your questions really helped guide me :)
 

Similar threads

Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 39 ·
2
Replies
39
Views
4K
  • · Replies 13 ·
Replies
13
Views
5K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
8K
Replies
13
Views
25K