If a spring of stiffness $k$ is cut into two parts $A$ and $B$ of length $l_{A}: l_{B}=2: 3$,then the stiffness of spring $A$ is given by

$A$ spring has length $L$ and force constant $K$. It is cut into two springs of length $L_1$ and $L_2$ such that $L_1 = N L_2$ ($N$ is an integer). The force constant of the spring of length $L_1$ is:

Initially,the spring is at its natural length and both blocks are at rest. Determine the maximum extension in the spring. Given $k = 20 \ N/m$,$m_1 = 0.5 \ kg$,$m_2 = 1 \ kg$,and $F = 1 \ N$.

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:

$A$ massless spring gets elongated by an amount $x_1$ under a tension of $5 \ N$. Its elongation is $x_2$ under a tension of $7 \ N$. For an elongation of $(5x_1 - 2x_2)$,the tension in the spring will be: (in $N$)

The force required to stretch a spring varies with the distance as shown in the figure. If the experiment is performed with the above spring of half length,the line $OA$ will