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We have this pulley system where $m_1 = m_2$ = 1 kg. Considering the rope AB is cut, what would be the work done by the resultant force acting on $m_1$ during 1 second after cutting?

The naive approach would be to just consider $m_1$ to be in free falling, so only weight will act on it, and you will get $mgh$, where $h$ is the distance traveled in 1 sec: $h = 1/2\ gt^2 = 5$ m, if $g = 10 m/s^2$.

This would lead to $W = 50$ J, which is not the answer.

Another approach would be to just consider somehow the tension. But I don't seem to get the right answer (which is 72 J). The pulley is assumed to be ideal.

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Because of the rope wrapped around the pulley, if $m_2$ falls a distance $x$ after the rope is cut then $m_1$ falls a distance $2x$ (because the pulley falls a distance $x$ and the length of rope between the pulley and $m_1$ is also decreased by an amount $x$). So if the acceleration of $m_2$ is $a$ then the acceleration of $m_1$ is $2a$.

The difference in accelerations must be accounted for by a tension $T$ in the rope. The force acting on $m_1$ is $m_1g+T$, and the force acting on $m_2$ is $m_2g - 2T$. So we have

$2m_1a = m_1g + T \\ m_2a = m_2g - 2T$

Eliminating $T$ from these equations we get

$4m_1a + m_2a = 2m_1g + m_2g \\ \displaystyle a = \frac {2m_1 + m_2}{4m_1 +m_2} g$

Since $m_1=m_2=1$, we have $a = 0.6g$, and so the acceleration of $m_1$ is $2a = 1.2g$. From this you can find the work done by the resultant force on $m_1$ in the first second.

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