$A$ spring of length $l$ has a force constant $k$. When a weight $W$ is attached to it,the extension produced is $x$. If the spring is cut into two equal parts and these parts are connected in parallel to support the same weight $W$,what will be the new extension?

$A$ block of $3 \; kg$ is initially in equilibrium and is hanging by two identical springs $A$ and $B$ as shown in the figure. If spring $A$ is cut from the lower point at $t = 0$,find the acceleration of the block in $m/s^2$ at $t = 0$.

Two springs $A$ and $B$ having spring constants $K_{A}$ and $K_{B}$ $(K_{A} = 2 K_{B})$ are stretched by applying forces of equal magnitude. If the energy stored in spring $A$ is $E_{A}$,then the energy stored in spring $B$ will be:

Two massless springs of spring constant $K_1$ and $K_2$ are connected in series,suspended vertically,and a certain mass is attached to the free end. If $e_1$ and $e_2$ are their respective extensions and $f$ is the stretching force,the total extension produced is:

$A$ spring of force constant $k$ is cut into three equal parts. The force constant of each part would be

Two plates each of mass $m$ are connected by a massless spring as shown below. $A$ weight $W$ is placed on the upper plate which compresses the spring further. When $W$ is removed,the entire assembly jumps up. The minimum weight $W$ needed for the assembly to jump up when the weight is removed is just more than ...........$m$.