Question

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 ​

Answer 1

Here we're given to find an expression for elastic potential energy which is given by,

$\mathrm {Elastic\ Potential\ Energy=\dfrac {1}{2}\times\ Load\ \times\ Extension}\\\\\\\mathrm {U=\dfrac {1}{2}\ F\ \Delta L}$

But here, extension, $\mathrm {\Delta L=x}$

The tension in the spring is equal to the load hung to the spring, i.e., $\mathrm {T=F}$

Hence the potential energy is,

$\mathbf {U=\dfrac {1}{2}Tx}$

Well, $\mathrm {T=kx}.$ Then,

$\mathbf {U=\dfrac {1}{2}kx^2}$