Two bodies of masses m_1 = 2 kg and m_2 = 4 kg hang on a…

Two bodies of masses $m_1 = 2\ \mathrm{kg}$ and $m_2 = 4\ \mathrm{kg}$ hang on a rope over a massless, frictionless pulley. Initially the heavier one is $h = 2\ \mathrm{m}$ higher than the lighter one; they start from rest. Find the kinetic energy of the system when the two are level.

Explanation of Method

Kinetic energy in an Atwood machine.

Step 1: Energy conservation

They are level after each has moved $h/2 = 1\ \mathrm{m}$: $m_2$ down, $m_1$ up. The kinetic energy equals the net loss of potential energy: \[E_k = m_2 g\,\frac{h}{2} - m_1 g\,\frac{h}{2} = (m_2 - m_1)\,g\,\frac{h}{2} = 2 \cdot 9.81 \cdot 1.\]

\[E_k \approx 19.6\ \mathrm{J}\]

$2$ $4$ $h = 2\ \mathrm{m}$ $\text{meet}$ $E_k = (m_2 - m_1)g\,\tfrac{h}{2}$

The bodies meet halfway: the 4 kg body goes down 1 m and the 2 kg body goes up 1 m. The net loss of potential energy, $(m_2 - m_1)g \cdot 1\ \mathrm{m}$, becomes kinetic energy.

Result

\[\boxed{\,E_k \approx 19.6\ \mathrm{J}\,}\]

Remark

Check. The original result is correct. The common speed then is $v = \sqrt{2E_k/6} \approx 2.56\ \mathrm{m/s}$.