## Elasticity Work Done In Stretching A Wire Energy Density

When a body gets strained, an internal reaction force develops inside the body. As a result, the applied external force does some work against this reaction force to produce deformation in the body.

- This work remains stored inside the body as potential energy, known as elastic potential energy. If the external force is withdrawn, then the internal reaction force, i.e., stress, ceases and this potential energy is converted into heat energy.
- Let us consider a wire of length L and cross-sectional area whose one end is fixed to a rigid support and a force F is applied at its other end. Let the total elongation of the wire be l. Within the elastic limit, the elongation is directly proportional to the applied force.
- When the applied force is zero, the elongation is zero; and when the applied force is F, then the elongation of the wire becomes l. So we can say that during the elongation l of the wire, an average force \(\frac{0+F}{2} \text { or } \frac{F}{2}\) actually acts on the wire.

**Read and Learn More: Class 11 Physics Notes**

Therefore, work done, W = average force x displacement = 1/2 Fl………..(1)

The Young’s modulus for the material of the wire is,

Y = \(\frac{F / \alpha}{l / L} \quad \text { or, } F=\frac{Y \alpha l}{L} \quad therefore W=\frac{1}{2} F l=\frac{1}{2} \frac{Y \alpha l^2}{L}\) …….(2)

This work remains stored in the wire as potential energy.

Volume of the wire = La.

Therefore, the work done per unit volume of the wire, or the potential energy stored per unit volume of the wire (energy density), is

w = \(\frac{W}{L \alpha}=\frac{1}{2} \frac{Y \alpha l^2}{L \cdot L \alpha}=\frac{1}{2} \frac{Y l^2}{L^2}\) ..(3)

= \(\frac{1}{2} \times \frac{Y l}{L} \times \frac{l}{L}=\frac{1}{2} \times \text { stress } \times \text { strain }\)………..(4)

[stress = \(\frac{F}{\alpha}=\frac{Y l}{L}\) = Q strain = \(\frac{l}{L}\)

So, elastic energy density = 1/2 x stress x strain.

**Calculation with the help of calculus:** if the total elongation of the wire is supposed to be an aggregate of infinitesimal elongations, then for an elongation dl of the wire, the work done, dW = Fdl

We know that, F = \(\frac{Y \alpha l}{L} \quad therefore d W=\frac{Y \alpha l}{L} \cdot d l\)

Therefore, the work done for the total elongation l becomes,

W = \(\int d W=\int_0^l \frac{Y \alpha l}{L} d l=\frac{Y \alpha}{L} \int_0^l l d l=\frac{1}{2} \frac{Y \alpha l^2}{L}\)

This is exactly same as equation (2).

## Elasticity Work Done In Stretching A Wire Energy Density Numerical Examples

**Example 1. What amount of work must be done In stretching a wire, of length 1 m and of cross-sectional area mm², by 0.1mm? Young’s modulus for the material = 2 x 10 ^{-11} N • m^{-2}.**

**Solution:**

length 1 m and of cross-sectional area mm², by 0.1mm

We know that, work’s done, W = \(\frac{1}{2} \frac{Y \alpha l^2}{L}\)

Here, Y = 2 x 10^{11} N • m^{-2}, a=1 mm^{2} = 10^{-6} m^{2},

l = 0.1 mm = 10^{-4} m, L = 1 m

∴ W = \(\frac{1}{2} \times \frac{2 \times 10^{11} \times 10^{-6} \times\left(10^{-4}\right)^2}{1}=10^{-3} \mathrm{~J} .\)

**Example 2. When the load on a wire Is increased from 3 kg to 5 kg, the elongation of the wire increases from 0. 6 mm to 1 mm. How much work is done during this extension of the wire?**

**Solution:**

When the load on a wire Is increased from 3 kg to 5 kg, the elongation of the wire increases from 0. 6 mm to 1 mm.

We know that, work done, W= 1/2 Fl.

Let the work done in stretching the wire through 0.6 mm be W_{1}

Here, F = 3 kgf = (3×9.8) N and l = 0.6 mm = 6 x 10^{-4} m

∴ W_{1} = 1/2 x 3 x 9.8 x 6 x 10^{-4} = 8.82 x 10^{-3} J

Let the work done in stretching the wire through 1 mm be W_{2}

In this case, F = 5 kgf = 5 x 9.8 N and l = 1 mm = 10^{-3} m

∴ W_{2} = 1/2 x 5 x 9.8 x 10^{-3} = 2.45 x 10^{-2}

So, during the extension of the wire from 0.6 mm to 1 mm, the net work done = W_{2} – W_{1} = (2.45 – 0.882) x 10^{-2}

= 1.568 x 10^{-2} J

**Example 3. For a uniform wire of length 3 m and cross-sectional area 1 mm ^{2}, 0.021 J of work is necessary to stretch it through 1 mm. Calculate the Young’s modulus for its material.**

**Solution:**

For a uniform wire of length 3 m and cross-sectional area 1 mm^{2}, 0.021 J of work is necessary to stretch it through 1 mm.

Work done, W = \(\frac{1}{2} \frac{Y \alpha l^2}{L}\)

or, \(Y=\frac{2 W L}{\alpha l^2}\)

Here, W = 0.021 J, α = 1 mm^{2} = 10^{-6} m^{2}, L = 3 m, l = 1 mm = 10^{-3} m

∴ Y = \(\frac{2 \times 0.021 \times 3}{10^{-6} \times\left(10^{-3}\right)^2}=1.26 \times 10^{11} \mathrm{~N} \cdot \mathrm{m}^{-2}\)

**Example 4. Two uniform wires of length 3 m and 4 m respectively are made of the same material. To stretch both these wires by the same length, 0.031 and 0.05 J of work are necessary. Calculate the ratio of the cross sectional areas of the two wires.**

**Solution:**

Two uniform wires of length 3 m and 4 m respectively are made of the same material. To stretch both these wires by the same length, 0.031 and 0.05 J of work are necessary.

Let the work done the case of the first antj t^e second wires respectively be,

⇒ \(W_1=\frac{1}{2} \frac{Y \alpha_1 l^2}{L_1} \text { and } w_2=\frac{1}{2} \frac{Y \alpha_2 l^2}{L_2}\)

∴\(\frac{W_1}{W_2}=\frac{a_1 L_2}{a_2 L_1} \text { or, } \frac{\alpha_1}{\alpha_2}=\frac{W_1 L_1}{W_2 L_2}\)

Here, \(W_1=0.03 \mathrm{~J}, W_2=0.05 \mathrm{~J}, L_1=3 \mathrm{~m}\) and \(L_2=4 \mathrm{~m}\)

∴ \(\frac{\alpha_1}{\alpha_2}=\frac{0.03 \times 3}{0.05 \times 4} \text { or, } \alpha_1: \alpha_2=9: 20\)