
The block will remain fixed if horizontal component of tension in the rope is equal to the friction force of block against the ground. i.e.
$T \cos 37^{\circ}=f=\mu\left(m_{B} g-T \sin 37^{\circ}\right)$
$\frac{4 T}{5}=\frac{100 g}{3}-\frac{T}{5}$
$\therefore T=\frac{100 g}{3}$
Now, $m_{A} a=T-m_{A} g$
or, $25 a=\frac{100}{3} g-25 g$
or, $a=\frac{4}{3} g-g=\frac{g}{3}$
(considering $4\ kg$ block doesn't fall on ground)
Consider the following statements
$(i)$ Time when relative motion between them is stopped is $1.4\ second$.
$(ii)$ Time when relative motion between them is stopped in $1.2\ second$
$(iii)$ The common velocity of the two blocks is $8\ m/s$, towards right.
$(iv)$ The displacement of the $4\, kg$ block when relative motion stopped is$10.8\ m$.
Which of the fstatements is/are correct
