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

\[\text{ Given } , \]
\[ \text{ m} _1 = 3 \text{ kg, m}_2 = 2 \text{ kg, t = during }  4^{\text{th}} \text{ second}\]

 

From the free-body diagram,

\[\text{ T - 3g + 3a = 0 . . . (i) } \]
\[\text{ T - 2g - 2a = 0 . . . (ii) } \]

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

\[\text{ 3g - 3a = 2g + 2a } \]
\[ \Rightarrow \text{ a }= \frac{\text{g } }{5} \text{m/ s} ^2\]

Distance travelled in the 4^th second,

\[\text{s}( 4^{\text{th}} ) = \frac{a}{2} \left( 2\text{n} - 1 \right)\]
\[ = \frac{\left( \frac{g}{5} \right)}{2}\left( 2 \times 4 - 1 \right)\]
\[ = \frac{7\text{g}}{10} = \frac{7 \times 9 . 8}{10} \text{m}\]

Net mass \['\text{m' = m}_1 - \text{m}_2 = 3 - 2 = 1 \text{kg} \]

So, decrease in potential energy,
P.E. = mgh

\[\text{P . E } . = 1 \times 9 . 8 \times \left( \frac{7}{10} \right) \times 9 . 8 = 67 . 2 \text{J } = 67 \text{J } \]

So, work done by gravity during the fourth second = P.E.= 67 J