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    Can someone explain to me why the expression for F which it asked you to show (the change in KE version) for a particle moving upwards also applies for a particle moving downwards, while the expression of F given (the exponential form) does not apply for a particle moving downwards? I am under the impression that F(exponential form) directly implies the KE expression for F so how can the latter be a general result?
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    I have to say, it is not clear to me that it has to be true either.

    For normal air-resistance questions, you'd expect the resistance to depend only on v, and the only other force to be g.

    If you assume this, it's quite easy to justify the air resistance being 1/2 b v^2.

    But in principle, I think you could have the air-resistance be k(x)v^3, where k(x) is a function of x chosen to make sure F obeys the given relation in the first part of the question.

    In which case I'm pretty sure the air resistance on the way down is NOT going to be 1/2 b v^2 (though I think it would be very difficult to work out what it is).

    (Original post by ghostwalker)
    ..
    Am I missing something here, do you think?
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    I think we're onto different things here. I'm not sure why the air resistance in the second case has to obey the expression for F (which you had to work out), but not the exponential form of F which was given to you (since the expression you had to work out is a direct consequence of the first).
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    Yes, the 2nd form for F is a direct consequence of the original equation for F.

    As you probably know, the behaviour on the downward journey is different, (because now gravity and air-resistance are acting in opposite directions). If you assume air resistance is simply a function of v, the 2nd form will remain valid (with a change in direction for the gravitational component).

    I don't think you can complete the question without that assumption, and I don't think it has to be true.
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    Why can't F be a different function of v? It's surely possible seeing the change in situation and it seems to me assuming F (in terms of v) is the same for both cases is a shallow assumption.
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    Well, if you assume that the air resistance is a function only of V, then the 2nd form gives you "air resistance = bv^2/2", and from there you can proceed to a sensible answer for the last part.

    (In other words, you're making an assumption, but by no means are you assuming everything).

    But yes, it's an assumption, and if you don't make I don't really think you can get anywhere with the last part.
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    Would I be terribly wrong then to assume that F obeys the exponential form for the second case and write x=maximum height-y? (since x denotes the how high the particle is above the ground).

    This is why applied maths is much less appealing than pure for me, since a lot of times on mechanics you need to make these so called 'sensible assumptions' which often looks like a fudge towards the correct answer.
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    You've found that F = g + bv^2/2 (in downwards direction) on the way up, I would expect that you are supposed to take

    F = g - bv^2/2 on the way down.

    (note that you can't simply use the exponential form, because you need an expression not involving \alpha).

    [Note that I think this is a particularly badly worded question - I wouldn't waste too much time stressing about it].
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    Ah I forgot you're not allowed u for the second part, stupid me >.<

    So they wanted you to assume the expression for F because it conveniently eliminates u...so should the question have allowed u in the answer would assuming the exponential form be ok?
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    I doubt it. You can't ignore the fact that the situation going down is NOT the same as the situation going up (because in one case air resistance and gravity act in the same direction, but in the other case they oppose each other).
 
 
 
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Updated: April 14, 2011
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