Classical Mechanics - Drag Force

  • Thread starter zeromaxxx
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  • #1
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Homework Statement


Suppose a block sliding on a slippery horizontal surface experiences a drag force F=-cv3/2 where c is a positive constant. At time t=0, the block is at position x=0 with initial positive velocity. Find the velocity and position as a function of time. Derive an expression for the limiting distance the block travels, or show that there is no limit.


Homework Equations


F=-cv3/2


The Attempt at a Solution



For velocity:
I took the integral of the drag force

mdv = -cv3/2 dt

int [dv/v3/2]= -c/m * int[dt]

-2/v1/2 + 2/v01/2 = -c/m * t

Isolating for v:

v= [v01/2/ 1+ (cv01/2t/2m)]2

Can someone kindly check the arithmetic if I got it right.

As for position, I have trouble dealing with the integral of the velocity function. If someone can help me regarding this, it would be greatly appreciated.
 
Last edited:

Answers and Replies

  • #2
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Int [dv/v2/3] is not v1/2 but v1/3. See the error?
 
  • #3
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Int [dv/v2/3] is not v1/2 but v1/3. See the error?
Yes, sorry there's a typo in the original question I put up, the drag force is F = -cv3/2 and not 2/3. Sorry for this, I hope you can go through it again. Thanks!
 
  • #4
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Okay. After you put v = dx/dt, the integral looks like Int [dx] = Int [dt/(a + b*t)2] where a and b are some trivial constants, right? That t integral is easy to do, you know how?

I am sleepy (read sloppy) right now, so I'd rather not fix those constants for you.
 
  • #5
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Hmm, it made more sense the way you put it though I still cannot determine the integral of t.
 
  • #6
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A substitution a + b*t = w (some random variable) brings it in the form Int [dw/w2] with some constants I have suppressed. Does it look familiar now?
 

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