Question

A constant force $F=\frac{m_{2}g}{2}$ is applied on the block of mass $m_1$ as shown in fig. The string and the pulley are light and the surface of the table is smooth. The acceleration of $m_1$ is

• A. $\frac{m_{2}g}{2(m_1+m_2)}$ towards right
• B. $\frac{m_{2}g}{2(m_1-m_2)}$ towards left
• C. $\frac{m_{2}g}{2(m_2-m_1)}$ towards right
• D. $\frac{m_{2}g}{2(m_2-m_1)}$ towards left

Answer 1

The correct option is A $\frac{m_{2}g}{2(m_1+m_2)}$ towards right

Let a be the acceleration of mass $m_2$ in the downward direction.
Then for $m_1$
$F_{net}=ma$
$T-m_2\left(\frac{g}{2}\right)=m_{1}a.....\left(i\right)$

And for $m_2$
$F_{net}=ma$
and $m_{2}g-T=m_{2}a.....\left(ii\right)$
Adding eqs. (1) and (2), we get
$(m_1+m_2)a=m_{2}g-m_2\left(\frac{g}{2}\right)=\frac{m_{2}g}{2}$
$\therefore a=\frac{m_{2}g}{2(m_1+m_2)}$

Alter,
$a=\frac{\text{net pulling force}}{\text{total mass}}=\frac{m_{2}g-\frac{m_{2}g}{2}}{m_1+m_2}$
$\therefore a=\frac{m_{2}g}{2(m_1+m_2)}$