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Work-Energy Derivation question: F(x(t)) = F(t) ?

  1. Nov 28, 2015 #1
    Does this derivation:

    work-energy_zpsflypwpst.png

    ...imply:

    work-energy2_zpszo01eiy4.png


    My best guess is that x(t) ≠ t
    So I would also guess that F(x(t)) ≠ F(t)

    But then how can this derivation be explained?

    How can F(x(t)) = m(a(t))? What does that actually mean?
    How come it's not: F(x(t)) = m(a(x(t))) ? Why/How does the x just cancel out?


    Thank you
     
  2. jcsd
  3. Nov 28, 2015 #2
    Would this question be better suited for the math section? If so, please feel free to transfer it. It seems to be bordering on a math related question within a physics derivation and I don't think I'll fully understand the derivation without understanding how the math is being work at this part.

    Again, thanks.
     
  4. Nov 28, 2015 #3
    x(t) implies x is a function of t, ie for a particular t there is a particular x.
    F(x) implies that F is a function of x
    F(x(t)) implies ... You should be able to figure that out.
    Same for a(t).

    Hmmmm
     
  5. Nov 29, 2015 #4
    The following might describe more precisely where I'm experiencing some uncertainty,

    Suppose I start with some object of 5kg moving in some way described by x(t) = t^4, then:

    7f70758a-b644-4a88-b672-4c6fce7a6af0_zpslqatzu2p.jpg

    Why do I arrive at 60t^2 for f(t)
    and arrive at 60t^6 for f(x(t))?

    It seems like f(t) = 60(t)^2
    while f(t(x)) = f(t^4) = 60(t^4)^2 = 60t^6
     
    Last edited: Nov 29, 2015
  6. Nov 29, 2015 #5

    Svein

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    You need to fix up your formula. If x is a distance and t is time you cannot have x=t4 (the dimensions do not match). You could get away with x = Ct4, but then the dimension of C would be m/s4.

    As to the rest of your questions: Of course you cannot have F(t) = F(x(t)). The dimensions do not match! If you have a well-behaved function, you can possibly invert the x(t) to t = g(x), and then you will have F(t) = F(g(x)).
     
  7. Nov 29, 2015 #6
    Yes, x is a distance and t is time, so x≠t and x≠ [itex]t^{4}[/itex] , however, I am convinced I included parenthesis everywhere to indicate that x is a function of t. That is, x(t) = t and x(t) = [itex]t^{4}[/itex]. As a function, the dimensions don't have to match, right? That is, displacement is a function of time, then x(t) = t can be written. I do not see where I may have written x = t or x = [itex]t^{4}[/itex]


    Okay, and if I cannot have F(t) = F(x(t)), then why does the formula in the original post say: F(x(t)) = m(a(t)) = m v'(t) ? <---this is what is written in my book.

    So I'm ending up at a contradiction. On one hand, F(x(t)) ≠ F(t), but then my book states F(x(t)) = m(a(t)). Can this be?

    Thank you
     
    Last edited: Nov 29, 2015
  8. Nov 29, 2015 #7

    Nathanael

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    Homework Helper

    When you're speaking of an equation which is physically meaningful, the dimensions must match. With that said, there is no need to worry about dimensions when contriving a particular example like this, because you can assume a particular unit system which leaves the dimensional-constant with unit value.
    You could say F(x(t))=F'(t) where F' is some other function (not the same as F unless x=t).
    That's what I see it as saying: some function F of x of t is equal to (m times) some other function a(t).

    It's really not an interesting statement (and isn't relevant to the Work-KE theorem).

    It also doesn't imply a(x(t))=a(t) (which is the source of your contradiction).


    P.S. (t4)2=t8 (not t6)
     
  9. Nov 29, 2015 #8

    Nugatory

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    Staff: Mentor

    In your example, you went wrong in the step where you wrote $$F(x(t)) = m(a(x(t))) = 60[x(t)^2]^2$$ The first equality is correct, the second is not. The problem is that ##a(t)## and ##a(x)## are different functions: the first is the formula you use to calculate the acceleration at a given time and the second is the formula you use to calculate the acceleration when the object is at a given position and is not equal to ##12t^2##. Whenever you see ##a(t)## you are allowed to substitute ##12t^2##, but you cannot make the same substitution when you see ##a(x)##.

    The same is true of ##x##, ##v##, and ##F##; for example ##x(t)=t^4## but ##x(x)=x##.

    If you work through your example again but use different names (for example ##F_x(x)## and ##F_t(t)## for the force) for the functions of time and of position it will all come out consistently.
     
  10. Nov 29, 2015 #9

    DrGreg

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    Many physics authors and teachers use a convention where, for example, ##F(x)## denotes a force as a function of distance, and ##F(t)## denotes the same force as a function of time. Strictly speaking, from a pure mathematician's point of view, this is nonsense and you should really use different names for the two functions, e.g. ##F_1(x)## to denote force as a function of distance, and ##F_2(t)## to denote force as a function of time. With this notation we have$$
    F_2(t) = F_1(x(t))
    $$
     
    Last edited: Nov 29, 2015
  11. Nov 30, 2015 #10
    Thank you,

    If I were to look at each function graphically, I have something like this:

    f8114d58-f86d-4867-a951-7e260ee09e39_zpsvyhofmjx.jpg

    However, I'm having some difficulty in figuring out how to express the force as a function of distance.

    From starting with a displacement as a function of time: x(t) = [itex] t^{4} [/itex] and arriving at a force with respect to time: f(t) = [itex]60t^{2}[/itex]

    How can I take it one step further and find a function of force with respect to displacement? F(x) or F(x(t)) ?

    My first guess is something like this:

    1d9bd928-e053-40e4-a07e-f3d5b0c46cae_zpsiq8gqcmg.jpg
     
  12. Dec 1, 2015 #11

    Svein

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    Well, first we introduce a dimension-correcting factor B (with dimension m/s4), so that [itex] x(t)=Bt^{4}[/itex]. For x≥0, this is one-to-one, so [itex]t=(\frac{x}{B})^{-4} [/itex].
     
  13. Dec 3, 2015 #12
    Thank you Svein. I've continued to look into this and have made good progress. I must go on a tangent and ask a new question regarding momentum before coming back to the topic of kinetic energy. To be continued...
     
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