To simulate car accidents,auto manufacturers study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car of mass $1000 \; kg$ moving with a speed $18.0 \; km/h$ on a smooth road and colliding with a horizontally mounted spring of spring constant $6.25 \times 10^{3} \; N m^{-1}$. What is the maximum compression of the spring in $m$?

$A$ spring has a spring constant $200 \,N/m$. If it is stretched by $1 \,cm$, then the potential energy stored in it is: (in $\,J$)

One end of a spring of spring constant $80 \ Nm^{-1}$ and unstretched length of $30 \ cm$ is fixed at point $A$ and the other end of the spring is fitted with a smooth ring of mass $300 \ g$ as shown in the figure. The ring is allowed to slide on a horizontal rod fixed at a height of $40 \ cm$. Initially the spring makes an angle of $60^{\circ}$ with the vertical and the system of spring and ring is released from rest. The speed of the ring when the spring becomes vertical is . . . . . . $ms^{-1}$.

$A$ spring of spring constant $5 \times 10^3 \, N/m$ is stretched initially by $5 \, cm$ from the unstretched position. Then the work required to stretch it further by another $5 \, cm$ is .............. $J$.

$Assertion$ : Graph between potential energy of a spring versus the extension or compression of the spring is a straight line.
$Reason$ : Potential energy of a stretched or compressed spring is proportional to the square of extension or compression.

When a spring is stretched by $2 \ cm$,it stores $100 \ J$ of energy. If it is stretched by another $2 \ cm$,the additional energy stored is ....... $J$.