Question

In the figure shown below, $m_A=2\text{kg}$ and $m_B=3\text{kg}$. A force $\text{F}=40\text{N}$ is applied on $2\text{kg}$ block as shown. Find the acceleration of COM of the system. (Neglect friction everywhere and assume the string to be inextensible)

• A. $1.2{\text{m/s}}^2$
• B. $1{\text{m/s}}^2$
• C. $2.4{\text{m/s}}^2$
• D. $2{\text{m/s}}^2$

Answer 1

The correct option is A $1.2{\text{m/s}}^2$

FBD's of Blocks


From F.B.D of $2\text{kg}$ block
$m_{A}g+F-T=m_{A}a$
$\Rightarrow 2g+40-T=2a...\left(i\right)$

For $3\text{kg}$ block
$T-m_{B}g=m_{B}a$
$\Rightarrow T=3g+3a....\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$
$2g+40-(3g+3a)=2a$

$\Rightarrow 5a=30\Rightarrow a=6{\text{m/s}}^2$

$a_{com}=\frac{m_A{a}_A+m_B{a}_B}{m_A+m_B}=\frac{2\times (-6\hat{j})+3\times 6\hat{j}}{5}$
$a_{com}=\frac{6}{5}\hat{j}{\text{m/s}}^2=1.2\hat{j}{\text{m/s}}^2$