Notes : Elastic Potential Energy Stored in a Stretched Wire: Derivation, Formula, Energy Density, Numericals, MCQs , FAQs

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Notes : Elastic Potential Energy Stored in a Stretched Wire: Derivation, Formula, Energy Density, Numericals, MCQs , FAQs Class 11 physics chapter 8 mechanical - Physicskund 

Introduction

When an external force stretches a wire, work is done against the internal restoring forces of the material. This work gets stored in the wire as Elastic Potential Energy (EPE). If the wire is within its elastic limit, the stored energy can be completely recovered when the force is removed.

Basic Terms

Stress

Stress is defined as the restoring force acting per unit area.

$$ \text{Stress}=\frac{F}{A} $$

Where:

  • \(F\) = Applied Force
  • \(A\) = Area of Cross-section

SI Unit: Pascal (Pa)

Strain

Strain is the ratio of change in length to the original length.

$$ \text{Strain}=\frac{\Delta L}{L} $$

or

$$ \text{Strain}=\frac{l}{L} $$

Where:

  • \(l\) = Extension Produced
  • \(L\) = Original Length

SI Unit: No Unit (Dimensionless)

Young's Modulus

Young's modulus is the ratio of stress to strain.

$$ Y=\frac{\text{Stress}}{\text{Strain}} $$

Substituting stress and strain:

$$ Y=\frac{F/A}{l/L} $$

$$ Y=\frac{FL}{Al} $$

Therefore,

$$ F=\frac{YAl}{L} $$

Derivation of Elastic Potential Energy Stored in a Stretched Wire

Consider a wire having:

  • Length = \(L\)
  • Cross-sectional Area = \(A\)
  • Extension Produced = \(l\)

A force \(F\) is applied to stretch the wire.

From Young's modulus,

$$ F=\frac{YAl}{L} $$

Small Work Done

If the wire is stretched by a small amount \(dl\),

$$ dW = F\,dl $$

Substituting \(F\),

$$ dW=\frac{YAl}{L}dl $$

Total Work Done

To stretch the wire from \(0\) to \(l\),

$$ W=\int_{0}^{l} dW $$

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

Taking constants outside the integral,

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

Integrating,

$$ W=\frac{YA}{L}\left[\frac{l^2}{2}\right]_{0}^{l} $$

$$ W=\frac{YAl^2}{2L} $$

Using

$$ F=\frac{YAl}{L} $$

Therefore,

$$ W=\frac{1}{2}Fl $$

Elastic Potential Energy

The work done is stored as elastic potential energy.

$$ U=\frac{1}{2}Fl $$

or

$$ U=\frac{YAl^2}{2L} $$

Why Does the Factor \( \frac{1}{2} \) Appear?

The stretching force increases gradually from zero to its maximum value \(F\).

Average force:

$$ F_{\text{avg}}=\frac{0+F}{2} $$

$$ F_{\text{avg}}=\frac{F}{2} $$

Work done:

$$ W=F_{\text{avg}}\times l $$

$$ W=\frac{F}{2}\times l $$

$$ W=\frac{1}{2}Fl $$

Force–Extension Graph

According to Hooke's law,

$$ F\propto l $$

The graph between force and extension is a straight line passing through the origin. The area under the graph represents the elastic potential energy stored in the wire.

$$ \text{Area}=\frac{1}{2}\times \text{Base}\times \text{Height} $$

$$ \text{Area}=\frac{1}{2}\times l\times F $$

$$ \text{Area}=\frac{1}{2}Fl $$

Potential Energy Per Unit Volume (Energy Density)

Volume of wire:

$$ V=AL $$

Energy density:

$$ \text{Energy Density}=\frac{U}{V} $$

Substituting \(U=\frac{YAl^2}{2L}\),

$$ \frac{U}{V} = \frac{\frac{YAl^2}{2L}}{AL} $$

$$ = \frac{1}{2}Y\frac{l^2}{L^2} $$

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

Since

$$ \frac{l}{L}=\text{Strain} $$

Therefore,

$$ \text{Energy Density} = \frac{1}{2}Y(\text{Strain})^2 $$

Alternative Form

Since

$$ Y=\frac{\text{Stress}}{\text{Strain}} $$

Substituting,

$$ \text{Energy Density} = \frac{1}{2} \left( \frac{\text{Stress}}{\text{Strain}} \right) (\text{Strain})^2 $$

$$ \text{Energy Density} = \frac{1}{2} (\text{Stress}\times\text{Strain}) $$

Important Formulae

$$ \text{Stress}=\frac{F}{A} $$

$$ \text{Strain}=\frac{l}{L} $$

$$ Y=\frac{\text{Stress}}{\text{Strain}} $$

$$ U=\frac{1}{2}Fl $$

$$ U=\frac{YAl^2}{2L} $$

$$ \frac{U}{V} = \frac{1}{2}Y(\text{Strain})^2 $$

$$ \frac{U}{V} = \frac{1}{2} (\text{Stress}\times\text{Strain}) $$

$$ U= \frac{1}{2} (\text{Stress}) (\text{Strain}) (\text{Volume}) $$

Solved Numericals

Numerical 1

A wire is stretched by a force of \(200N\) producing an extension of \(0.02m\). Calculate the elastic potential energy stored.

$ U=\frac{1}{2}Fl $

$ U=\frac{1}{2}\times200\times0.02 $

$ U=2J $

Answer: \(2J\)

Numerical 2

A wire stores \(10J\) energy when stretched by \(0.1m\). Find the stretching force.

$ U=\frac{1}{2}Fl $

$ 10=\frac{1}{2}\times F\times0.1 $

$ 10=0.05F $

$ F=200N $

Answer: \(200N\)

Very Short Answer Questions

  1. What is elastic potential energy?
    Answer: Energy stored due to elastic deformation.
  2. What is the SI unit of elastic potential energy?
    Answer: Joule (J).
  3. What is stress?
    Answer: Force per unit area.
  4. What is strain?
    Answer: Change in length divided by original length.
  5. Define Young's modulus.
    Answer: Ratio of stress to strain.

Short Answer Questions

  1. Why is energy stored in a stretched wire?
  2. What does the area under the force-extension graph represent?
  3. Why does the factor \( \frac{1}{2} \) appear in the energy formula?
  4. Define energy density.
  5. State Hooke's law.

Long Answer Questions

  1. Derive the expression \(U=\frac{1}{2}Fl\).
  2. Derive the expression for energy density.
  3. Explain the force-extension graph and prove that its area represents elastic potential energy.

Fill in the Blanks

  1. Stress is force per unit area.
  2. Strain is change in length divided by original length.
  3. Energy stored in a stretched wire is called elastic potential energy.
  4. Young's modulus is stress divided by strain.
  5. Energy density means energy per unit volume.

True or False

  1. Elastic potential energy is stored during stretching. — True
  2. Strain has units. — False
  3. Stress is measured in pascal. — True
  4. Energy density is energy per unit mass. — False
  5. Hooke's law is valid beyond elastic limit. — False

Multiple Choice Questions (MCQs)

  1. Elastic potential energy stored in a stretched wire is:
    • A. \(Fl\)
    • B. \(\frac{1}{2}Fl\) ✓
    • C. \(2Fl\)
    • D. \(F/l\)
  2. SI unit of Young's modulus is:
    • A. Joule
    • B. Newton
    • C. Pascal ✓
    • D. Meter
  3. Strain is:
    • A. Vector
    • B. Dimensionless ✓
    • C. Force
    • D. Energy

Frequently Asked Questions (FAQ)

What is elastic potential energy?

Energy stored in a body due to elastic deformation.

What is the SI unit of elastic potential energy?

Joule (J).

What is energy density?

Energy stored per unit volume.

What is Hooke's law?

Within the elastic limit, stress is directly proportional to strain.

What does the area under the force-extension graph represent?

The elastic potential energy stored in the wire.

Quick Revision

$ U=\frac{1}{2}Fl $

$ U=\frac{YAl^2}{2L} $

$ \text{Energy Density} = \frac{1}{2}Y(\text{Strain})^2 $

$ \text{Energy Density} = \frac{1}{2} (\text{Stress}\times\text{Strain}) $

Area under Force-Extension Graph = Elastic Potential Energy

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