Question

A mass of block $8\text{kg}$ is attached by a rope with other block of mass $4\text{kg}$ as shown in the figure. Find the acceleration of COM of the system.
(Neglect friction everywhere and neglect the mass of the rope and pulley)

• A. $\frac{g}{9}$
• B. $\frac{g}{3}$
• C. $\frac{g}{27}$
• D. $\frac{g}{5}$

Answer 1

The correct option is A $\frac{g}{9}$

Consider blocks of masses $8\text{kg}$ & $4\text{kg}$ as a system.
Let acceleration of blocks $=a$ as shown in the fig.


For $8\text{kg}$ block


$mg-T=ma$
$8g-T=8a......\left(1\right)$

For $4\text{kg}$ block,


$T-mg=ma$
$T-4g=4a.....\left(2\right)$

By adding equation $\left(1\right)$ & $\left(2\right)$,
$4g=12a$
$\Rightarrow a=\frac{g}{3}$
As we know,
$a_{COM}=\frac{m_1\times a_1+m_2\times a_2}{m_1+m_2}$
Here
$m_1=8\text{kg}$
$a_1=\frac{g}{3}$
$m_2=4\text{kg}$
$a_2=\left(\frac{-g}{3}\right)$
$\therefore a_{COM}=\frac{8\times \left(\frac{g}{3}\right)+4\times \left(\frac{-g}{3}\right)}{8+4}$

$\Rightarrow a_{COM}=\frac{g}{9}$