\begin{problem*}{8.22}
The coefficient of friction between the block of mass $m_1=3.00\U{kg}$
and the surface in Figure~P8.22 is $\mu_s=0.400$.  The system starts
from rest.  What is the speed of the bal of mass $m_2=5.00\U{kg}$ when
it has fallen a distance $h=1.50\U{m}$?
% m1-block-on-table -- pulley -- hanging m2
\end{problem*}

\begin{solution}
Because they are linked by a taut string, the blocks will move at the
same speed.  Therefore, $m_1$ drags a distance $h$ across the table,
loosing
\begin{equation}
  \Delta E_\text{int} = -W_f = -\mu_k mgh\cos(180\dg) = \mu_k mgh
\end{equation}
to friction.

Conserving energy
\begin{align}
  E_i &= E_f \\
  K_{1i} + U_{1i} + K_{2i} + U{2i}
    &= K_{1f} + U_{1f} + K_{2f} + U{2f} + E_\text{int} \\
  0 + 0 + 0 + 0
    &= \frac{1}{2} m_1 v_f^2 + 0
       + \frac{1}{2} m_2 v_f^2 - m_2 gh + \mu_k m_1 gh \\
  (m_2-\mu_k m_1) gh &= \frac{m_1+m_2}{2} v_f^2 \\
  v_F^2 &= 2gh\frac{m_2-\mu_k m_1}{m_1+m_2} \\
  v_f &= \pm\sqrt{2gh\frac{m_2-\mu_k m_1}{m_1+m_2}}
    = \sqrt{2gh\frac{m_2-\mu_k m_1}{m_1+m_2}}
    = \ans{3.74\U{m/s}}
\end{align}
where we dropped the $\pm$ because we only want the magnitude of the
velocity, not its direction.
\end{solution}
