Fill in the blank: The force constant of a spring is $0.5 \, Nm^{-1}$. The force necessary to increase the length of the spring by $10 \, cm$ 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 $t = 0$.

$A$ body of mass $m$ is suspended from an ideal spring of force constant $k$. The expected change in the position of the body due to an additional force $F$ acting vertically downwards is

Natural length of a spring is $60\, cm$,and its spring constant is $4000\, N/m$. $A$ mass of $20\, kg$ is hung from it. The extension produced in the spring is,.......$cm$ (Take $g = 9.8\, m/s^2$)

Two springs,$P$ and $Q$,of force constants $k_P$ and $k_Q$ (where $k_Q = k_P / 2$) are stretched by applying forces of equal magnitude. If the energy stored in $Q$ is $E$,then the energy stored in $P$ is:

Two masses of $10 \, kg$ and $20 \, kg$ respectively are connected by a massless spring as shown in the figure. $A$ force of $200 \, N$ acts on the $20 \, kg$ mass. At the instant shown,the $10 \, kg$ mass has an acceleration of $12 \, m/s^2$ towards the right. The acceleration of the $20 \, kg$ mass at this instant is ........ $m/s^2$.