Posted by **zamdirhy** on Thursday, April 19, 2012 at 6:46am.

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 -
**Elena**, Thursday, April 19, 2012 at 10:51am
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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