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:

$A$ 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 $N$)

$A$ 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 $t = 0$.

$A$ bead of mass $m = 100 \,g$ is attached to one end of a spring of natural length $L$ and spring constant $k = \frac{(\sqrt{3}+1) mg}{L}$. The other end of the spring is fixed at point $A$ on a smooth vertical ring of radius $R$. The bead is at point $B$ such that the spring makes an angle of $30^{\circ}$ with the horizontal diameter. The normal reaction at $B$ just after it is released to move is (take $g = 9.8 \,ms^{-2}$): (in $\,N$)

If a spring is extended to length $l,$ then according to Hooke's law,

The figure shows two cases. In the first case,a spring (spring constant $K$) is pulled by two equal and opposite forces $F$ at both ends. In the second case,it is pulled by a force $F$ at one end while the other end is fixed. The extensions $(x)$ in the springs will be: