Question

Block of $3kg$ is intially in equilibrium and is hanging by two identical springs $A$ and $B.9F$ spring $A$ is cut from lower end at $t=0$ then, find acceleration of block in $m{s}^{-2}$ at $t=0$?

Answer 1

From the first fig we can see that at equilibrium

$2kx=3g\Rightarrow kx=15N$

Now at $t=0$, just after one spring is cut situation will be as shown in the fig 2 (when spring is cut)

using Newton's equation we get :$3g-kx=3a\Rightarrow a=\frac{15}{3}=5\frac{m}{s^2}$