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The coefficient of friction between the block of mass m1 = 3.00 kg and the surface in the figure below is ìk = 0.365. The system starts from rest. What is the speed of the ball of mass m2 = 5.00 kg when it has fallen a distance h = 1.90 m?

  • physics -

    The equations of the motion in vector form are
    m1•⌐a=m1•⌐g+⌐N +⌐F(fr) +⌐T,
    m2•⌐a = m2•⌐g + ⌐T,
    Projections on x- and y- axes:
    m1•a= T – F(fr),
    m1•g = N,
    m2••a = m2g – T.
    F(fr) = k•N.
    Solving for acceleration a, we obtain
    a = g• (m2 –km1)/(m1+m2) = 4.78 m/s^2.
    From kinematics
    a = v^2/2•h,
    then
    v =sqrt(2•a•h) = sqrt(2•4.78•1.9) = 4.26 v/s.

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