Determine the elastic potrntial energy stored in stretched wire.

Explanation

When a wire is stretched, work is done against the internal restoring forces acting between particles of the wire. This work done appears as elastic potential energy in the wire.
Consider a wire of length L and area of cross section A. Let deforming force F be the applied on the wire and l be the increase in length of the wire.
`Y = (FL)/(ADeltaL)` here U is the Young.s modulus of wire.
`Delta l = l`
Let work dW is needed to increase in small length dl.
`therefore dW =F xx dl OR Yal (dl)/(L)`
Work W is done for increae the length of wire from L to `L + l.` It is work for `l =0` to `l =l`
`W = int _(0)^(l) (YAl)/(L) dl (YA )/(2) (l ^(2))/(L)`
`W = 1/2 xx Y xx ((l )/(L)) ^(2) xx AL`
`1/2xx` Young modulus `xx("Strain")^(2) xx` Volume of wire
`W =1/2 xx ` Strees `xx ` Stress ` xx` Volume of wire
Hence, work stored in wire it the elastic potential energy. So the stored elastic potential energy per unit is obtained from `u = 1/2 sigma epsi`
SI unit of elastic potential density is `Nm ^(-12) or `Pa and dimensional formula `[M^(1) L ^(-1) T ^(-2)].`
Elastic potential energy in eleastic body in equal to the enclosed area under the applied force F to the body versus change in length. But enclosed area under the graph of stress versus strain is equal to the elastic potnetial energy per unit voluem (energy) (density) of the body.