2 The coefficient of friction between the block of mass $m_1=3.00\U{kg}$
3 and the surface in Figure~P8.22 is $\mu_s=0.400$. The system starts
4 from rest. What is the speed of the bal of mass $m_2=5.00\U{kg}$ when
5 it has fallen a distance $h=1.50\U{m}$?
6 % m1-block-on-table -- pulley -- hanging m2
10 Because they are linked by a taut string, the blocks will move at the
11 same speed. Therefore, $m_1$ drags a distance $h$ across the table,
14 \Delta E_\text{int} = -W_f = -\mu_k mgh\cos(180\dg) = \mu_k mgh
21 K_{1i} + U_{1i} + K_{2i} + U{2i}
22 &= K_{1f} + U_{1f} + K_{2f} + U{2f} + E_\text{int} \\
24 &= \frac{1}{2} m_1 v_f^2 + 0
25 + \frac{1}{2} m_2 v_f^2 - m_2 gh + \mu_k m_1 gh \\
26 (m_2-\mu_k m_1) gh &= \frac{m_1+m_2}{2} v_f^2 \\
27 v_F^2 &= 2gh\frac{m_2-\mu_k m_1}{m_1+m_2} \\
28 v_f &= \pm\sqrt{2gh\frac{m_2-\mu_k m_1}{m_1+m_2}}
29 = \sqrt{2gh\frac{m_2-\mu_k m_1}{m_1+m_2}}
32 where we dropped the $\pm$ because we only want the magnitude of the
33 velocity, not its direction.