- Consider a wire of length ‘L’ and area of cross-section ‘A’ be fixed at one end and stretched by suspending a load ‘M’ from the other end. The extension in the wire takes place so slowly that it can be treated as quasi-static change; because internal elastic force in the wire is balanced by the external applied force and hence acceleration is zero. Let at some instant during stretching the internal elastic force be ‘f’ and the extension produced be ‘x’. Then,

- Since at any instant, the external applied force is equal and opposite to the internal elastic force, we can say that the work done by the external applied force in producing a further infinitesimal dx is

- Let ‘ l ‘ be the total extension produced in the wire, and work done during the total extension can be found by integrating above equation.

This is an expression for the work done in stretching wire.

**Strain Energy:**

- The work done by the external applied force during stretching is stored as potential energy (U) in the wire and is called as strain energy in the wire.
- Thus the strain energy is given by

- Its S.I. unit is J (joule) and its dimensions are [L
^{2}M^{1}T^{-2}].

**Strain Energy Per Unit Volume of a Wire (Different Forms of Expression):**

The work done by external applied force during stretching is stored as potential energy (U) in the wire and is called as strain energy in the wire.

- This is an expression for strain energy or potential energy per unit volume of stretched wire. This is also called as the energy density of the strained wire. Its S.I. unit is J m
^{-3}and its dimensions are [L^{-1}M^{1}T^{-2}].

**Note:**

- More work is to be done for stretching a steel wire than stretching a copper wire because steel is more elastic than copper. Due to which more restoring force is produced in the steel, hence we have to do more work to overcome these larger restoring forces.

### Problems Based on Work Done in Stretching a Wire and Strain energy

- Find the work done in stretching a wire of length 2 m and of sectional area 1 mm² through 1 mm if Young’s modulus of the material of the wire is 2 × 10
^{11}N/m².**Ans:**(0.05 J) - Calculate the work done in stretching a wire of length 3 m and cross-sectional area 4 mm² when it is suspended from a rigid support at one end and a load of 8 kg is attached at the free end. Y = 12 × 10
^{10}N/m² and g = 9.8 m/s² .**Ans:**(0.0192 J) - When the load on a wire is increased slowly from 3 to 5 kg wt, the elongation increases from 0.6 to 1 mm. How much work is done during the extension? g = 9.8 m/s² .
**Ans :**(0.01568 J) - A spring is compressed by 1 cm by a force of 3.92 N. What force is required to compress it by 5 cm? What is the work done in this case? Assume the Hooke’s Law.
**Ans :**(9.8 N; 0.49 J) - A wire 4m long and 0.3 mm in diameter is stretched by a load of 0.8 kg. If the extension caused in the wire is 1.5 mm, find the strain energy per unit volume of the wire.g = 9.8 m/s².
**Ans :**(2.08 × 10^{4 }J/m³) - Find the energy stored in a stretched brass wire of 1 mm² cross-section and of an unstretched length 1 m when loaded by 2 kg wt. What happens to this energy when the load is removed. Y = 10
^{11 }N/m².**Ans:**(1.921 × 10^{-3}J) - A metal wire of length 2.5 m and are of cross section 1.5 × 10
^{-6}m² is stretched through 2 mm. Calculate work done during stretching. Y = 1.25 × 10^{11}N/m².**Ans:**(150 J) - A copper wire is stretched by 0.5% of its length. Calculate the energy stored per unit volume in the wire. Y = 1.2 × 10
^{11}N/m².**Ans:**(1.5 × 10^{6 }J/m³) - Calculate the strain energy per unit volume in a brass wire of length 2.0 m and cross-sectional area 0.5 mm2, when it is stretched by 2mm and a force of 5 kg-wt is applied to its free end.
**Ans:**(4.9 × 10^{4 }J/m³) - Calculate the work done in stretching a wire of length 2 m and cross-sectional area 0.0225 mm² when a load of 100 N is applied slowly to its free end. Young’s modulus of elasticity = 20 × 10
^{10 }N/m².**Ans:**(2.222 J) - A uniform steel wire of length 3 m and area of cross-section 2 mm² is extended through 3mm. Calculate the energy stored in the wire, if the elastic limit is not exceeded. Young’s modulus of elasticity = Y = 20 × 10
^{10 }**Ans:**(0.6 J)

### For Solutions of Above Problems Click Here

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