Question

As shown in the figure the applied force F is equal to half of the weight of m2. The acceleration of m2 is
A) m2g2(m1+m2)downwards
B) m2g(m1+m2)upwards
C) m2g2(m1+m2)upwards
D) 3m2g2(m1+m2)upwards

Answer 1

$F-T = m_{1}a$

again,

$T -m_{2}g = m_{2}a$

Add the equations:

$F - m_{2}g =( m_{1} + m_{2})a$

$= > a = \dfrac{ F - m_{2}g }{( m_{1} + m_{2})}$

$= > a = \dfrac{(m_{2}g/2) - m_{2}g }{( m_{1} + m_{2})}$

$= > |a| = \dfrac{m_{2}g}{2( m_{1} + m_{2})}$

Acceleration is downwards.

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