# Work done in Stretching a wire

## Work done in Stretching a wire

In stretching a wire work is done against internal restoring forces. This work is stored as elastic potential energy or strain energy.

If a force F acts along the length L of the wire or cross section A and stretches it by x then:

$$Young’s\,\,Modulus\left( Y \right)=\frac{Stress}{Strain}=\frac{F/A}{x/L}=\frac{FL}{Ax}\Rightarrow F=\frac{YA}{L}x$$.

So, the work done for an additional small increase dx in length,

$$dW=F\,dx=\frac{YA}{L}x.dx$$.

Hence, total work done in increasing the length (l):

$$W=\int\limits_{0}^{l}{dW}=\int\limits_{0}^{l}{F.dx}=\frac{1}{2}\frac{YA}{L}{{l}^{2}}$$, this work done is stored in the wire

Therefore, Energy stored in wire, $$U=\frac{1}{2}\frac{YA{{l}^{2}}}{L}=\frac{1}{2}Fl\,\,\,\left( As,\,\,F=\frac{YAl}{L} \right)$$.

Now, dividing both sides by volume of the wire we get energy stored in unit volume of wire.

$${{U}_{V}}=\frac{1}{2}\frac{Fl}{AL}$$ = ½ x Stress x Strain = ½ x Y x (Strain)²

$${{U}_{V}}=\frac{1}{2Y}{{\left( Stress \right)}^{2}}$$ (As AL = Volume of Wire)

If the force on the wire is increased from F₁ to F₂ and the elongation in wire is l then energy stored in the wire, $$U=\frac{1}{2}\left( \frac{{{F}_{1}}+{{F}_{2}}}{2} \right)l$$.

Thermal energy density = Thermal energy per unit volume = ½ x Thermal Stress x Strain

$$=\frac{1}{2}\frac{Fl}{AL}$$ = ½ (Y α Δθ) (α Δθ) = ½ Y α² (Δθ)².