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Find Work Done Using Two Different Integrals

  1. Apr 6, 2015 #1
    1. The problem statement, all variables and given/known data
    a rigid body with a mass of 2 kg moves along a line due to a force that produces a position function x(t)= 4t^2, where x is measured in meters and t is measured in seconds. Find the work done during the first 5 seconds in two ways.

    2. Relevant equations

    x(t)= 4t^2
    Work is ->
    W = Integral from a to b of the force F(x) dx


    3. The attempt at a solution
    SInce t is seconds, I tried plugging in 5 for the given eq. of x(t) = 4t^2
    So I got x(5) = 4(5)^2 =
    4(25)= 100 m

    So then I attempted integrating 4t^2 from time 0 to time 5 and got a completely different answer. Help anyone? Thanks!
     
  2. jcsd
  3. Apr 6, 2015 #2

    BvU

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    A completely different answer for a completely different quantity ? There is some difference between ## x(5) ## and $$\int_0^5 x(t) \; dt$$ so why would they be the same ? And wasn't the exercise to calculate the work done ?
     
  4. Apr 6, 2015 #3

    Dick

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    You haven't even found the force yet. Use Newton's law, ##F=ma##.
     
  5. Apr 6, 2015 #4
    As one of the above posters mentioned, you don't integrate x(t) alone. You need to get acceleration function from it first to be able to derive the force. From there you have to integrate the work differential you posted with respect to t (you'll need to rewrite dx in terms of dt).
     
  6. Apr 8, 2015 #5
    So this was also given in the problem, which I forgot to mention.
    a) Note that x^n(t)=8; then use Newton's second law (F= ma = mx^n(t)) to evaluate the work integral.

    (Given)
    F= ma = mx^n(t))
    x^n(t)=8

    F= (2)(8)
    F=16

    W = Integral of F(x)dx from 0 to 5

    I integrated 16x from 0 to 5 and got:
    16(5)-16(0)= 80

    For part b) is asks us to
    Change variables in the work integral and integrate with respect to t. Make sure that your answer agrees with your answer in part a.

    Now I know that I'd have to change the variables, so would just exchange the x with t's in the integral? Help!
     
  7. Apr 8, 2015 #6

    Mark44

    Staff: Mentor

    What do you mean by mx^n(t)?

    This looks to me like ##mx^{n(t)}##, but in the context of this problem this makes no sense to me.
     
  8. Apr 8, 2015 #7
    The integral that you just did does not take into account the fact that x(t) is not a constant function. You have to write the differential dx in terms of dt (time). Once you have the differential in terms of t, then integrate. Does that clarify things?

    Also, if you want to show a second derivative, I would suggest doing it in Leibnizian notation in LaTex, or just use a quotation mark as a double prime (e.g. a(t)=x"(t)).
     
  9. Apr 8, 2015 #8
    Also, could you write the verbatim text of each problem part?
     
  10. Apr 8, 2015 #9
    They asked for 2 different methods. One method is to find the force and integrate it with respect to distance. In this case, the force is constant, because the acceleration is constant, so the force comes out from under the integral. What is the acceleration? What is the force?

    The second method is to make use of the work-energy theorem. Do you know what that is, and how to apply it?

    Chet
     
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