Question

The two blocks in an Atwood machine have masses $2.0kg$ and $3.0kg$. Find the work done by gravity during the fourth second after the system is released from rest. $(g=10{m/s}^2)$

Answer 1

$a=\frac{m_2-m_1}{m_1+m_2}g=\frac{g}{5}=2m/s^2$

Distance covered by each block in $n^{th}$ second is

$s=u+\frac{a}{2}(2n-1)$ and here $u=0$

$⟹s=\frac{2}{2}(2\times 4-1)=7m$

Total work done=work done on $m_1$ moving up $+$ work done on $m_2$ moving down.

Work=$F\times s=mgs$ as force is gravity.

$W=-m_{1}gs+m_{2}gs=(m_2-m_1)gs$

$⟹W=1\times 10\times 7$

$W=70J$