Question

Two blocks of $4\text{kg}$ and $6\text{kg}$ are attached by springs. They are hanging in vertical position as shown. Lower spring breaks due to excessive force. Acceleration of $4\text{kg}$ block just after breaking is: $(g=10{\text{m/s}}^2)$

• A. $15{\text{m/s}}^2$
• B. $25{\text{m/s}}^2$
• C. $10{\text{m/s}}^2$
• D. Zero

Answer 1

The correct option is A $15{\text{m/s}}^2$

Before breaking,
in equilibrium,
$\therefore$ Balancing the forces
$k_1{x}_1=4g+k_2{x}_2$
Also
$k_2{x}_2=6g$
After breaking, $k_2{x}_2$ on $4\text{kg}$ becomes zero.
$\therefore F_{\text{net}}$ on $4\text{kg}=k_1{x}_1-mg$
$=(4g+6g)-4g$
$=6g$
$\therefore a_{4\text{kg}}=\frac{6g}{4}=15{\text{m/s}}^2$