Question

The two blocks in an Atwood machine have masses 2 kg and 3 kg. Find the work done by gravity during the fourth second after the system is released from rest.

Answer 1

$$\begin{gathered} \mathrm{Given}, \\ {m}_1=3\mathrm{kg},m_2=2kg,t=\mathrm{during}{4}^{\mathrm{th}}\mathrm{second} \end{gathered}$$



From the free-body diagram,

$$\begin{gathered} T-3g+3a=0...\left(i\right) \\ T-2g-2a=0...\left(ii\right) \end{gathered}$$

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

$$\begin{gathered} 3g-3a=2g+2a \\ \Rightarrow a=\frac{g}{5}\mathrm{m}/s^2 \end{gathered}$$
Distance travelled in the 4^th second,

$$\begin{gathered} \mathrm{s}\left(4^{\mathrm{th}}\right)=\frac{a}{2}\left(2n-1\right) \\ =\frac{\left({\displaystyle \frac{g}{5}}\right)}{2}\left(2\times 4-1\right) \\ =\frac{7g}{10}=\frac{7\times 9.8}{10}\mathrm{m} \end{gathered}$$

Net mass $\text{'}m\text{'}=m_1-m_2=3-2=1\mathrm{kg}$
So, decrease in potential energy,
P.E. = mgh

$P.E.=1\times 9.8\times \left(\frac{7}{10}\right)\times 9.8=67.2\mathrm{J}=67\mathrm{J}$
So, work done by gravity during the fourth second = P.E.= 67 J