Q.A spring of force constant $k$ is cut into two pieces such that one piece is double the length of the other. Then, the long piece will have a force constant of

A stretched or compressed spring stores elastic potential energy. For a spring of force constant $k$ obeying Hooke's law, the internal restoring force is $\vec F_s=-k\vec x$ (the negative sign showing it always opposes the displacement $x$ from the natural, equilibrium length), so the external force needed to produce and hold a stretch $x$ is the equal-and-opposite, position-dependent force $F_a=kx$.

Because this applied force is variable (proportional to $x$, unlike the constant gravitational force), the stored energy must be found by integrating it from the equilibrium position (where $U=0$) out to the final extension $x$:

$U = \int_0^x kx\,dx = \tfrac{1}{2}kx^2$ …