The two blocks in an Atwood machine have masses 2.0 kg and 3.0 kg. Find the work done by gravity during the fourth second after the system is released from rest.

$Given,m_1=3kg,$

$m_2=2kg,t=during4thsec$

From the freebody diagram,

$T-3g+3a=0$...(i)

$T-2g-2a=0$ ...(ii)

Equating (i) and (ii), we get,

$3g-3a=2g+2a$

$\Rightarrow a=\frac{g}{5}m/se{c}^2$

Distance travelled in 4th sec is given by,

$S4th=\frac{a}{2}(2n-1)$

$=\frac{\left(\frac{g}{5}\right)}{2}(2\times 4-1)$

$=\frac{7g}{10}=\frac{7\times 9.8}{10}m$

Net mass'm' $=m_1-m_2=3-2=akg$

So, decrease in P.E.=mgh

$1\times 9.8\times \left(\frac{7}{10}\right)\times 9.8=67.2J$