When a spring is stretched by $10 \ cm$, the potential energy stored is $E$. When the spring is stretched by $10 \ cm$ more, the potential energy stored in the spring becomes (in $E$)

$Assertion$ : 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.

$A$ spring of spring constant $200 \, Nm^{-1}$ is initially stretched by $10 \, cm$ from the unstretched position. The work to be done to stretch the spring further by another $10 \, cm$ is (in $J$)

$A$ spring with spring constant $k$ is extended from $x = 0$ to $x = x_1$. The work done will be

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 with spring constant $k$ when stretched through $1 \, cm$,the potential energy is $U$. If it is stretched by $4 \, cm$,the potential energy will be (in $U$)