# Work done in a pulley system [closed]

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.

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.