First slide
Connected bodies
Question

A constant force F = m2g/2 is applied on the block of mass m1 as shown. The string and the pulley are light and the surface of the table is smooth. Find the acceleration of m1.

Moderate
Solution



T - F = {m_1}a.........(1)
{m_2}g - T = {m_2}a.........(2)
(1) + (2)
m2g – F = (m1 + m2)a
{m_2}g - {m_2}\frac{g}{2} = ({m_1} + {m_2})a
\Rightarrow \frac{{{m_2}g}}{2} = ({m_1} + {m_2})a
\therefore \,\,a = \frac{{{m_2}g}}{{2({m_1} + {m_2})}}

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