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Motion and Forces

  1. Jan 27, 2008 #1
    A 5.82 kg block located on a horizontal frictionless floor is pulled by a cord that exerts a force F = 12.80 N at an angle θ = 24.0° above the horizontal as shown. What is the speed of the block 5.90 seconds after it starts moving?

    [​IMG]

    F = ma
    Fx = Fcosθ
    Fy = Fsinθ
    v(t) = v0 + at

    Fx = 12.8cos24 = 11.69 N
    Fx = m*ax ax = 2.01 m/s^2
    vx(t) = 0 + 2.01(5.9) = 11.859 m/s

    Fy = 12.8sin24 = 5.21 N
    Fy = m*ay ay = 0.895 m/s^2
    vy(t) = 0 + (.895)(5.9) = 5.2805 m/s

    v(t)^2 = vx(t)^2 + vy(t)^2 = 11.859^2 + 5.2805^2
    v(t) = 12.98 m/s
    I entered 13.0 m/s into LONCAPA and it told me it was incorrect. What am I doing wrong?

    Edit: Also tried a simpler way:

    F=ma
    12.8=5.82a
    a=2.20m/s
    v=v0+at
    v=0+2.20(5.9)
    v=12.98m/s
     
    Last edited: Jan 27, 2008
  2. jcsd
  3. Jan 28, 2008 #2
    The second method is satisfactory.

    YOU ARE MAKING A SIMPLE MISTAKE:

    cos24 degrees=.9135

    >>force=12.80 * .9135 = 11.69N

    now a= 11.69/5.82 = 2.009 m/s^2

    >v= v0+at

    >0+2.009*5.90

    >v= 11.854 m/s
     
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