# Elastic potential energy

##### Elastic potential energy in a stretched wire :

When a wire is stretched, some 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 F be the stretching forces applied on the wire and Δl be the increase in length of the wire.

Initially, the internal restoring force was zero but when length is increased by Δl ,

the internal force for an increase in length Δl of the wire

$\displaystyle = \frac{(0 + F)}{2} = \frac{F}{2}$

Hence, work done on the wire, w = average force × increase in length

$\displaystyle \frac{F}{2} \times \Delta l$

This is stored as elastic potential energy U in the wire.

$\displaystyle U = \frac{1}{2}\times F \times \Delta l$

$\displaystyle U=\frac{1}{2} \times \frac{F}{a}\times \frac{\Delta l}{l} \times a l$

$\displaystyle U = \frac{1}{2}Stress \times Strain \times Volume \, of \, wire$

∴ elastic potential energy per unit volume of the wire

$\displaystyle \frac{U}{V}= u = \frac{1}{2}Stress \times Strain$

$\displaystyle u = \frac{1}{2}(Young’s \, modulus \times Strain) \times Strain$

( Young’s modulus = stress / strain)

$\displaystyle u = \frac{1}{2}(Young’s \, modulus) \times( Strain)^2$

Illustration : A steel wire of 4.0 m in length is stretched through 2.0 mm. The cross-sectional area of the wire is 2.0 mm2. If Young’s modulus of steel is 2.0 × 1011 N/m2 find (I) the energy density of wire (ii) the elastic potential energy stored in the wire.

Solution: Here, l = 4.0 m ; Δl = 2 × 10-3 m ; a = 2.0 × 10-6 m2

Y = 2.0 × 1011 N/m2

(i) The energy density of stretched wire

u = (1/2)× stress × strain = (1/2) × Y × (strain)2

= (1/2) × 2.0 × 1011 × (2 × 10-3) / 4)2

= 0.25 × 105 = 2.5 × 104 J/m3.

(ii) Elastic potential energy = energy density × volume

= 2.5 × 104 × (2.0 × 10-6 ) × 4.0 J = 20 × 10-2 = 0.20 J.

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