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?

Answer 1

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.