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cutemouse

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A particle of unit mass falls from rest in a medium with resistive force , where k is a constant. Prove that the distance x fallen when the velocity is v, is given by:



Thanks!
 

cutemouse

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Hmm okay...

Just one thing aside though...

If the resistive FORCE,

But as F=ma, how can a = kv^2 ? :S
 

gurmies

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F = ma

F is composed of g and kv^2

ma = m(g-kv^2)

As is to my knowledge...I haven't done mechanics yet though.
 

Revacious

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im guessing you get

a = (mg-kv^2)/m

what ive found is that you look at the question, and determine whether or not to use that or use (g-kv^2)

i think its really stupid and ambiguous as well (perhaps its just my ignorance), but yeah, use the simpler one for most questions, and if it doesnt work, do the other one.
 

cutemouse

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I've read a few threads here that it doesn't matter if you use mkv or kv as the mass is constant... But how would I do this using the actual resistive FORCE as kv?
 

cutemouse

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If the resistive FORCE (which is mass * accel) = kv^2

then how can the ACCELERATION suddenly become mkv?
 
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medhat tawfik

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If it says resistive force = kv^2, then ma = kv^2 and a = kv^2 / m.

Sometimes they say a force per unit mass, so kv^2 / 1 Kg = mkv^2 / m Kg.

And sometimes they say retardation, which is acceleration, so a = kv^2, F = mkv^2.

Differences are subtle but important.
 

untouchablecuz

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what i always assumed was that:

if F=kv^2, k constant, then

F = Kmv^2, where k = Km

hence a = Kv^2
 

cutemouse

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I clarified it with a really good teacher today.

He basically said that in the HSC it is clear when the 'm' terms cancel out or not.

If it does cancel out then you get a nice ln function, if not then some inverse tan function I believe in most cases.

what i always assumed was that:

if F=kv^2, k constant, then

F = Kmv^2, where k = Km

hence a = Kv^2
I mentioned this a while back, it seems to be how Coroneos does it. He said that it's simply wrong.
 

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