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Newton's Problems

  1. Mar 30, 2010 #1
    1. The problem statement, all variables and given/known data
    A 1.25kg object is moving in the x direction at 17.4m/s. Just 3.41s later, it is moving at 26.8m/s at 34.0 degrees to the x-axis. What are the magnitude and direction of the force applied during this time?


    2. Relevant equations
    F=ma

    3. The attempt at a solution
    Firstly i drew a diagram and it was in vector components so i calculated the resultant.

    1st component
    H: 17.4m/s
    V: 0m/s

    2nd component
    H: 26.8cos(34)
    V: 26.8sin(34)

    Resultant component:
    Horizontal = 17.4 + 26.8cos(34) = 39.62 m/s
    Vertical = 26.8sin(34)

    so [tex]\sqrt{39.62^2+22.22^2}[/tex]
    direction Resultant = [tex]tan^{-1}(\frac{22.22}{39.62})[/tex]

    direction = 30 degrees

    find the magnitude [tex]F = m * \frac{v}{t}[/tex]

    [tex]F = 1.25 * \frac{45.43}{3.41}[/tex]
    F=16.65N

    This is incorrect can someone tell me where i have gone wrong.

    P.S
     
  2. jcsd
  3. Mar 30, 2010 #2

    rl.bhat

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    Homework Helper

    Horizontal = 17.4 + 26.8cos(34) = 39.62 m/s
    You have to find the change in horizontal component of velocity (vf - vi)
     
  4. Mar 30, 2010 #3
    is this the same in the vertical component?

    edit never mind its not the same
     
    Last edited: Mar 30, 2010
  5. Mar 30, 2010 #4

    rl.bhat

    User Avatar
    Homework Helper

    Yes. But initial vertical component of the velocity is zero.
     
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