First slide

Connected bodies

Question

A constant force F = m₂ g / 2 is applied on the block of mass m, as shown in fig. The string and the pulley
are light and the surface of the table is smooth The acceleration of m₁ is

Easy

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

$\frac{m_{2} g}{2 (m_{2} - m_{1})} towards left$

Solution

Let a be the acceleration of mass m₂ in the downward direction. Then T -m₂(g/2) = m₁a …(1)
and m₂ g -T = m₂ a …(2)
Adding eqs. (1) and (2), we get
$(m_{1} + m_{2}) a = m_{2} g - m_{2} (g / 2) = m_{2} g / 2$
$∴ a = \frac{m_{2} g}{2 (m_{1} + m_{2})}$ .

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