Classical Mechanics - Drag Force

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The discussion revolves around solving a physics problem involving a block sliding on a surface with a drag force proportional to the velocity raised to the power of 3/2. The user derives the velocity function through integration but expresses uncertainty about the position integral. A correction is noted regarding the drag force equation, clarifying that it is indeed F = -cv^(3/2). Participants suggest using a substitution method to simplify the integral for position, indicating familiarity with the necessary calculus techniques. The conversation highlights the challenges of integrating complex functions in classical mechanics problems.
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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:
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Int [dv/v2/3] is not v1/2 but v1/3. See the error?
 
Sourabh N said:
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!
 
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.
 
Hmm, it made more sense the way you put it though I still cannot determine the integral of t.
 
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?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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