Question

derive expression for elastic potential energy in a stretched wire ?

Answer 1

some work has to be done against the internal restoring force in streching a wire. this is known as E pfwell ,, averafe force = F/2we know that work = force x displacementW = F/2 * dlwork = 1/2 F/a*dl/l * al1/2 ( stress)( strain ) ( Volume )elastic potential energy per unit volume or Energy density of wire ....u = U/al= 1/2 F/a*dl/l * al/al1/2 stress X strain= 1/2 (young's modulus x strain ) strain1/2 (young's modulus) ( strain)2

Answer 2

$\huge\textbf\red{Answer :}$

Consider a wire of length (L) and area of cross section (A) is stretched by force (F). As the length increases from L to L + ∆L. The restoring force increases from 0 to F. Thus, average restoring force is..

$F_{av}$ = $\frac{0\:+\:F}{2}$

$F_{av}$ = $\frac{F}{2}$

Thus work done,

$W$ = $F_{av}$ × $\triangle\:L$

$W$ = $\frac{F}{2}$ × $\frac{\triangle\:L}{AL}$ × $AL$

$\frac{W}{AL}$ = $\frac{1}{2}$ × $\frac{F}{A}$ × $\frac{\triangle\:L}{L}$

$W$ = $\frac{1}{2}$ × $stress\:\times\:strain$

$U$ = $\frac{1}{2}$ × $stress\:\times\:strain$

Here; U = energy stored per unit volume.