If a spring extends by $x$ on loading,then the energy stored by the spring is (if $T$ is tension in the spring and $k$ is spring constant).

An elastic string of unstretched length $L$ and force constant $k$ is stretched by a small length $x$. It is further stretched by another small length $y$. The work done in the second stretching is

$A$ block of mass $m$ slides down an inclined plane as shown in the figure and hits a spring at the bottom,causing it to compress. If the spring constant is $K$,what is the maximum compression of the spring?

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:

$A$ block of mass $M$ moving on a frictionless horizontal surface collides with a spring of spring constant $K$, as shown in the figure. If the spring compresses by a length $L$, then the maximum momentum of the block is:

$A$ ring of mass $m$ is attached to a horizontal spring of spring constant $k$ and natural length $l_0$. The other end of the spring is fixed,and the ring can slide on a smooth horizontal rod as shown. Now,the ring is shifted to position $B$ and released. The speed of the ring when the spring attains its natural length is: