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Net Force

  1. Sep 24, 2009 #1
    1. The problem statement, all variables and given/known data
    The velocity of a 4 kg particle is given by v = (3ti + 6t^2j ) m/s, with time t in seconds. At the instant the net force on the particle has a magnitude of 38 N, what are the direction (relative to the positive direction of the x axis) of (a) the net force and (b) the particle's direction of travel?


    2. Relevant equations



    3. The attempt at a solution
    I took the second derivative of V to get the acceleration which is 3(i) + 12t(j). What should my next step be?
     
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  3. Sep 24, 2009 #2

    rock.freak667

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    Force=ma

    so get that vector and use the fact that the magnitude is 38N to find the time.
     
  4. Sep 24, 2009 #3
    so would that vector be (12(i) + 12t(j))? And I would set that equal to 38?
     
  5. Sep 24, 2009 #4

    rock.freak667

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    no you'd get F=12i+48tj

    and |12i+48tj|=38. Find t
     
  6. Sep 24, 2009 #5
    how did you get 48t??
     
  7. Sep 24, 2009 #6

    rock.freak667

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    a=3i + 12tj

    m=4, finding ma, 12*4=48. Unless I am wrong as I am bad with numbers.
     
  8. Sep 24, 2009 #7
    ok, so T = .54 seconds. what should I do next?
     
  9. Sep 24, 2009 #8

    rock.freak667

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    find the force vector (by subbing t=0.54)

    then find the angle it makes with the x-axis. It's a similar process for part b
     
  10. Sep 24, 2009 #9
    ok, the force vector is F= 1.62i + 1.75j
    And the angle is 47.21 degrees? 47.21 is wrong though, i tried it.
     
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