krit.club logo

Mechanical Properties of Solids - Elastic Potential Energy

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

When a solid body like a wire is stretched, work is done against the internal restoring forces. This work is stored in the body in the form of Elastic Potential Energy (UU).

•

The energy stored is equal to the work done by the stretching force: W=∫0ΔlFdlW = \int_{0}^{\Delta l} F dl.

•

For a wire of length LL and area of cross-section AA, the internal restoring force FF for an extension ll is given by F=YAlLF = \frac{Y A l}{L}, where YY is the Young's Modulus of the material.

•

Energy Density (uu) is defined as the elastic potential energy stored per unit volume of the material. Its SI unit is J m−3\text{J m}^{-3} or N m−2\text{N m}^{-2}.

•

The relationship between energy density, stress (σ\sigma), and strain (ϵ\epsilon) can be expressed using Young's Modulus (YY) since Y=σϵY = \frac{\sigma}{\epsilon}.

📐Formulae

W=12×Load×Extension=12FΔlW = \frac{1}{2} \times \text{Load} \times \text{Extension} = \frac{1}{2} F \Delta l

U=12×Stress×Strain×VolumeU = \frac{1}{2} \times \text{Stress} \times \text{Strain} \times \text{Volume}

u=UV=12×Stress×Strainu = \frac{U}{V} = \frac{1}{2} \times \text{Stress} \times \text{Strain}

u=12×Y×(Strain)2u = \frac{1}{2} \times Y \times (\text{Strain})^2

u=(Stress)22Yu = \frac{(\text{Stress})^2}{2Y}

💡Examples

Problem 1:

A steel wire of length 2.0 m2.0\text{ m} and cross-sectional area 0.8×10−6 m20.8 \times 10^{-6}\text{ m}^2 is stretched by 2.0 mm2.0\text{ mm}. Calculate the elastic potential energy stored in the wire. (Given: YY for steel =2.0×1011 N m−2= 2.0 \times 10^{11}\text{ N m}^{-2})

Solution:

L=2.0 m,A=0.8×10−6 m2,Δl=2.0×10−3 m,Y=2.0×1011 N m−2L = 2.0\text{ m}, A = 0.8 \times 10^{-6}\text{ m}^2, \Delta l = 2.0 \times 10^{-3}\text{ m}, Y = 2.0 \times 10^{11}\text{ N m}^{-2} First, find the stretching force FF: F=YAΔlL=(2.0×1011)×(0.8×10−6)×(2.0×10−3)2.0=160 NF = \frac{Y A \Delta l}{L} = \frac{(2.0 \times 10^{11}) \times (0.8 \times 10^{-6}) \times (2.0 \times 10^{-3})}{2.0} = 160\text{ N} Now, calculate energy UU: U=12FΔl=12×160×(2.0×10−3)=0.16 JU = \frac{1}{2} F \Delta l = \frac{1}{2} \times 160 \times (2.0 \times 10^{-3}) = 0.16\text{ J}

Explanation:

The energy is calculated by first determining the force required to produce the given extension using Young's modulus, then applying the work-done formula for a stretched string.

Problem 2:

Calculate the energy density of a metal wire which is under a stress of 108 N m−210^8\text{ N m}^{-2}. The Young's modulus of the metal is 2×1011 N m−22 \times 10^{11}\text{ N m}^{-2}.

Solution:

Stress (σ)=108 N m−2\text{Stress } (\sigma) = 10^8\text{ N m}^{-2} Y=2×1011 N m−2Y = 2 \times 10^{11}\text{ N m}^{-2} Using the formula for energy density in terms of stress: u=σ22Y=(108)22×2×1011=10164×1011=0.25×105 J m−3=2.5×104 J m−3u = \frac{\sigma^2}{2Y} = \frac{(10^8)^2}{2 \times 2 \times 10^{11}} = \frac{10^{16}}{4 \times 10^{11}} = 0.25 \times 10^5\text{ J m}^{-3} = 2.5 \times 10^4\text{ J m}^{-3}

Explanation:

Energy density is the energy per unit volume, which can be found directly from the applied stress and the material's Young's modulus.

Elastic Potential Energy - Revision Notes & Key Formulas | CBSE Class 11 Physics