Elastic Potential Energy in a Stretched Wire

When a wire is put under a tensile stress, work is done against the inter-atomic forces. This work is stored in the wire in the form of elastic potential energy. When a wire of original length L and area of cross-section A is subjected to a deforming force F along the length of the wire, let the length of the wire be elongated by l. 

Then from Eq. (9.8), we have F = YA × (l/L). 

Here Y is the Young’s modulus of the material of the wire. 

Now for a further elongation of infinitesimal small length dl, work done dW is 

F × dl or $\frac{YAldl}{L}$. 

Therefore, the amount of work done (W) in increasing the length of the wire from L to L + l, that is from l = 0 to l = l is

W = $\int_{0}^{l} \frac{YAl}{L}dl$

W = $\frac{YA}{2}\times \frac{l^2}{L}$

W = $\frac{1}{2}\times Y \times \left(\frac{l}{L}\right)^2 \times AL$

W = $\frac{1}{2}$ × Young’s modulus × strain $^2$  × volume of the wire

W = $\frac{1}{2}$ × stress × strain × volume of the wire

This work is stored in the wire in the form of elastic potential energy (u). Therefore the elastic potential energy per unit volume of the wire (u) is

$u=\frac{1}{2}\sigma \epsilon$

Example 1

If in a wire of Young's modulus Y, longitudinal strain X is produced then the potential energy stored in its unit volume will be :

A. 0.5YX$^2$
B. 0.5Y$^2X$
C. 2YX$^2$
D. YX$^2$

Answer:

We know that, the potential energy stored per unit volume is

W = $\frac{1}{2}$ × stress × strain

and

Y = Stress/strain

So, stress = Y x strain

W = $\frac{1}{2}$ × Young’s modulus × strain $^2$   

W = $\frac{1}{2}$ × Y × X $^2$ = 0.5YX$^2$

Example 2:

Find the elastic potential energy in a system shown below if the material of wires is same (Y = Young’s modulus)

Answer:

Elastic potential energy,

E = $\frac{1}{2}$ × stress × strain × volume of the wire

E = $\frac{1}{2}$ × stress × $\frac{stress}{Young's modulus}$ × volume of the wire

$E_1$ = $\frac{1}{2} \times \frac{w}{\pi R^2} \times \frac{w}{\pi R^2 Y} \times (\pi R^2)(\frac{L}{2})$

$E_1$ = $\frac{1}{2} \times \frac{w^2}{Y} \times \frac{L}{2\pi R^2}$

$E_1$ = $\frac{w^2L}{4Y\pi R^2}$

and

$E_2$ = $\frac{1}{2} \times \frac{w}{\pi R^2/4} \times \frac{w}{\pi YR^2/4} \times (\pi R^2/4) (2L)$

$E_2$ = $\frac{4w^2L}{Y\pi R^2}$

the elastic potential energy in a system is

$E_1 + E_2=\frac{w^2L}{4Y\pi R^2} + \frac{4w^2L}{Y\pi R^2}$

$E_{system}=\frac{17w^2L}{4Y\pi R^2}$

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