Question

System shown in figure is in equilibrium. Find the magnitude of net change in the string tension between two masses just after, when one of the springs is cut. Mass of both the blocks is same and equal to $m$ and spring constant of both the springs is $k$

• A. $\frac{mg}{2}$
• B. $\frac{mg}{4}$
• C. $\frac{mg}{3}$
• D. $\frac{3mg}{2}$

Answer 1

The correct option is A $\frac{mg}{2}$

$T_i=mg$
$2kx=2mg$
$\therefore kx=mg$
One $kx$ force (acting in upward direction) is suddenly removed. So net downward force on system will be $kx$ or $mg$. Therefore net downward acceleration of system,
$a=\frac{mg}{2m}=\frac{g}{2}$
$\text{Free body diagram of lower block gives the equation,}$
$mg-T_f=ma=\frac{mg}{2}$
$\therefore T_f=\frac{mg}{2}$
$\text{From these two equations we get,}$
$△T=\frac{mg}{2}$