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Finding displacement using net work?

  1. Jul 10, 2016 #1
    • This post does not use the homework template because it was originally posted in a non-homework forum.
    1. The problem statement, all variables and given/known data
    The question is: A 2 kg particle is moving in the positive x direction with the speed of 5 m/s. As it passes the origin, a force F = (30 N - 2N/s*t)i is applied to it. Where does the particle come to a stop?

    2. Relevant equations
    F=ma
    W = integral F(x)dx
    K=.5mv^2

    3. The attempt at a solution
    K1 = .5(2)(5)^2 = 25 J
    K2 = .5(2)(0)^2 = 0 J
    W = K2-K1= -25 J

    My main issue is I'm not sure how to rewrite the force equation in terms of x, since I can use it in the formula Work = integral F(x)dx from 0 to x to solve for x.

    Thanks!
     
    Last edited: Jul 10, 2016
  2. jcsd
  3. Jul 10, 2016 #2

    cnh1995

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    I deleted my previous post. I think the suggestion was not correct. You need to first write v as a function of t and find the value of t for which v=0. Then you should write displacement x as a function of time and put the value of t to get the displacement. You'll need differential equations for this. What is the relation between v, F and t?
     
  4. Jul 10, 2016 #3
    F=ma=m(v/t)
    So since F=ma, I divided the force vector by 2kg, giving me an acceleration vector of a=(15-t)i

    Taking the integral of that, I would get a velocity vector of v = (15t - .5t^2 + 5)i. Set v=0, then I get t=30.33 s.

    Does that look right so far?
     
  5. Jul 10, 2016 #4

    cnh1995

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    Yes.
     
  6. Jul 10, 2016 #5
    Ok, so the final result doesn't look right...So taking the integral of v, I get x=[-1.5t^3 + 7.5t^2 + 5t]i . Plugging in 30.33 s, I get x = -384803.29
     
  7. Jul 10, 2016 #6

    cnh1995

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    Check the integration again. The answer comes out to be positive.
     
  8. Jul 10, 2016 #7
    Oops, you're right, it should be (-1/6)t^3 not (-3/2)t^3. And the answer comes out to 2400 m
     
  9. Jul 10, 2016 #8

    cnh1995

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    Right. 2400.82...
     
  10. Jul 10, 2016 #9
    Yup! I only need 2 sig figs, though. Thank you so much for your help!
     
  11. Jul 11, 2016 #10

    cnh1995

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    No problem! And welcome to Physics Forums!
     
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