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# In the arrangement shown the pulley and the strings are ideal.The acceleration of block B is

Eshan
3 years ago
Dear student,

The tensions are correctly shown in the figure.
Now from the contraint relationship, we can find that the acceleration of mass A is half of that of B.
Hence$2a_A=a_B$

From FBD of mass A,

$T-mg=ma_A=m\dfrac{a_B}{2}$

From FBD of mass B,

$mg-\dfrac{T}{2}=ma_B$
$\implies a_B=\dfrac{2}{5}g$