As a metal wire elongates due to a load suspended from it, the...

When a load of mass $M$ is suspended from a metal wire, the wire elongates by a distance $x$ under the constant force $F = Mg$.

The decrease in the gravitational potential energy of the load is $\Delta U = Mgx$.

The elastic potential energy stored in the stretched wire is $U_{elastic} = \dfrac{1}{2} F x = \dfrac{1}{2} Mgx$.

The remaining energy is $\Delta U - U_{elastic} = Mgx - \dfrac{1}{2}Mgx = \dfrac{1}{2}Mgx$.

This remaining energy is dissipated as heat. Thus, the decrease in gravitational potential energy is shared between the increased elastic potential energy of the wire and heat.