I am trying to check if my understanding of conservation of energy is correct. Imagine a pulley problem like so with $m_2$ heavier than $m_1$ and the pulley is ideal (in the original problem I borrowed this from $v$, $m_1$, and $m_2$ are given:
and our goal is to find $h$.
Splitting the energy up we have at rest, all energy must balance so:
$E_i = KE + PE = m_2gh$
Taking the ground as the zero potential plane. Since $m_1$ is at rest at the $0$ potential plane there's no energy there. Since nothing is moving, there is also no kinetic energy.
Now, looking at the energy situation in the final period when we release the system and $m_2$ hits the ground:
$E_f = KE + PE = \frac{m_2v^2}{2} + \frac{m_1v^2}{2} + m_1gh$.
Of course, setting these equal (since energy is conserved) and solving for $h$ gives the answer.
I want to reason out why the kinetic energy is the way it is in the final configuration. Initially I had thought that since the system is once again at rest, the only kinetic energy that would matter would be the kinetic energy of $m_2$. But it seems that I must also consider the kinetic energy used to lift $m_1$ to $h$ as well.
Is this because $m_2$ strikes the ground with the sum of the kinetic energies of $m_1$ and $m_2$ due to conservation, or am I interpreting this result incorrectly?