Answer
Both crates are moving with a speed of $2.91~m/s$ after the crates have moved a distance of 1.4 meters.
Work Step by Step
The kinetic energy of the system of two crates will be equal in magnitude to the net change in the potential energy of the two crates. Note that crate $m_2$ falls a distance of 1.4 meters as crate $m_1$ moves up the incline a distance of 1.4 meters:
$\frac{1}{2}(m_1+m_2)~v^2 = m_2gh_2-m_1gh_1$
$v^2 = \frac{2g~(m_2h_2-m_1h_1)}{m_1+m_2}$
$v = \sqrt{\frac{2g~(m_2h_2-m_1h_1)}{m_1+m_2}}$
$v = \sqrt{\frac{(2)(9.80~m/s^2)~[(16.3~kg)(1.4~m)-(12.4~kg)(1.4~m)~sin~36.9^{\circ}]}{16.3~kg+12.4~kg}}$
$v = 2.91~m/s$
Both crates are moving with a speed of $2.91~m/s$ after the crates have moved a distance of 1.4 meters.
