A constant force F = m_2g/2 is applied on the | vedclass

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(A) Let $a$ be the acceleration of the system. Since the string is inextensible,both blocks have the same magnitude of acceleration $a$.For block $m_1

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$\frac{m_2g}{2(m_1 + m_2)}$ towards right

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(A) Let $a$ be the acceleration of the system. Since the string is inextensible,both blocks have the same magnitude of acceleration $a$.For block $m_1$ on the table,the forces acting horizontally are the tension $T$ (towards the right) and the applied force $F = m_2g/2$ (towards the left). Assuming the block $m_1$ moves towards the right,the equation of motion is:$T - F = m_1a$$T - m_2g/2 = m_1a$  --- $(i)$For block $m_2$ hanging vertically,the forces are gravity $m_2g$ (downwards) and tension $T$ (upwards). Since $m_1$ moves right,$m_2$ moves downwards:$m_2g - T = m_2a$  --- $(ii)$Adding equations $(i)$ and $(ii)$:$(T - m_2g/2) + (m_2g - T) = m_1a + m_2a$$m_2g/2 = (m_1 + m_2)a$$a = \frac{m_2g}{2(m_1 + m_2)}$Since the result is positive,the assumption that $m_1$ moves towards the right is correct.

$A$ constant force $F = m_2g/2$ is applied on the block of mass $m_1$ as shown in the figure. The string and the pulley are light and the surface of the table is smooth. The acceleration of $m_1$ is

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