The force constants of two springs are $K_1$ and $K_2$. Both are stretched until their elastic potential energies are equal. If the stretching forces applied are $F_1$ and $F_2$,then the ratio $F_1:F_2$ is:

The potential energy of a long spring when stretched by $2\,cm$ is $U$. If the spring is stretched by $8\,cm$,the potential energy stored in it will be $.......\,U$.

$A$ spring $40 \, mm$ long is stretched by the application of a force. If $10 \, N$ force is required to stretch the spring by $1 \, mm$,then the work done in stretching the spring by $40 \, mm$ is ............. $J$.

$A$ light spring of length $10 \ cm$ is stretched by $2 \ cm$ when a mass of $20 \ g$ is attached to its end. The mass is pulled until the total length of the spring becomes $14 \ cm$. What is the elastic potential energy stored in the spring (in Joules)?

Two springs have spring constants $k_{A}$ and $k_{B}$ such that $k_{A} > k_{B}$. If both springs are stretched by the same extension $x$,the work required will be:

$A$ block of mass $m=25 \ kg$ sliding on a smooth horizontal surface with a velocity $v=3 \ ms^{-1}$ meets a spring of spring constant $k=100 \ N/m$ fixed at one end as shown in the figure. The maximum compression of the spring and the velocity of the block as it returns to the original position are,respectively: