If a spring extends by $x$ on loading,then the energy stored in the spring is (where $T$ is the tension in the spring and $K$ is the spring constant):

$A$ spring with force constant $k$ is initially stretched by a distance $x$. If it is further stretched by a distance $y$,the work done during the second stretching 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:

$A$ body of mass $0.5 \ kg$ moves with a speed of $1.5 \ m/s$ on a smooth horizontal surface. It hits a massless spring with a force constant $k = 50 \ N/m$. What will be the maximum compression of the spring in $m$?

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

Two particles with mass $m_1 = 16 \ kg$ and $m_2 = 2 \ kg$ slide as a unit with a common velocity of $12 \ ms^{-1}$ on a level frictionless surface. Between them is a compressed massless spring with spring constant $k = 100 \ Nm^{-1}$. The spring,originally compressed by $25 \ cm$,is suddenly released,sending the two masses,which are connected to the spring,flying apart from each other. The orientation of the spring w.r.t. the initial velocity is shown in the diagram. What is the relative velocity of separation in $ms^{-1}$,after the particles lose contact?