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Dynamics Question #5

  1. May 16, 2008 #1
    1. The problem statement, all variables and given/known data
    A ball of mass [tex]0.60kg[/tex] rolls North at [tex]4.0ms^{-1}[/tex]. It meets a slope which causes a force of [tex]0.18N[/tex] East. This force lasts for [tex]10s[/tex]. Calculate the final velocity of the ball. Neglect any friction effects.

    2. Relevant equations
    [tex]F = m.a[/tex]
    [tex]v = u+at[/tex]

    3. The attempt at a solution
    [tex]m = 0.60kg \ , \ v = 4.0ms^{-1} \ , \ F = 0.18N \ , \ t = 10s[/tex]

    [tex]F = m.a[/tex]
    [tex]a = \frac{F}{m} = \frac{0.18}{0.60} = 0.30ms^{-2}[/tex]

    [tex]v = u + at[/tex]
    [tex]v = 4.0 + 0.3 \times 10[/tex]
    [tex]v = 12 ms^{-1}[/tex]

    Where did I go wrong?
     
  2. jcsd
  3. May 16, 2008 #2

    dx

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    The acceleration is not in the same direction as the initial velocity. Infact, the acceleration is perpendicular to the initial velocity, so the velocity northward doesnt change. What is the eastward component of its velocity at the end?
     
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