$A$ long spring, when stretched by a distance $x$, has the potential energy $u$. On increasing the stretching to $nx$, the potential energy of the spring will be:

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

$A$ spring of force constant $10\, N/m$ has an initial stretch $0.20\, m$. In changing the stretch to $0.25\, m$,the increase in potential energy is about.....$joule$.

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?

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):

Explain the elastic potential energy of a spring and obtain an expression for this energy.