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 $E$)

$A$ block of mass $M$ moving on a frictionless horizontal surface collides with a spring of spring constant $K$ and compresses it by a length $L$. The maximum momentum of the block during the collision process is

$A$ car of mass $1000 \,kg$ having a velocity of $10 \,ms^{-1}$ collides with a horizontally mounted spring. If the spring constant is $4000 \,Nm^{-1}$, then the maximum compression of the spring is: (in $\,m$)

$Assertion$ : 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 ....... $J$.

When equal forces are applied to two springs with spring constants $1500 \, N/m$ and $3000 \, N/m$,what is the ratio of their stored potential energies?