What is called elastic potential energy? Write its different formulas.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Elastic potential energy is the energy stored in a body due to its deformation (change in shape or size) within its elastic limit. When a deforming force is applied to an elastic body,work is done against the internal restoring forces. This work is stored in the body as elastic potential energy.
Let a wire of length $L$,area of cross-section $A$,and Young's modulus $Y$ be stretched by a force $F$ such that its extension is $\Delta L$.
The different formulas for elastic potential energy $(U)$ are:
$1$. In terms of force and extension: $U = \frac{1}{2} F \Delta L$
$2$. In terms of stress and strain: $U = \frac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume}$
$3$. In terms of Young's modulus: $U = \frac{1}{2} Y \times (\text{strain})^2 \times \text{volume}$
$4$. Energy density (Energy per unit volume): $u = \frac{U}{V} = \frac{1}{2} \times \text{stress} \times \text{strain} = \frac{1}{2} \frac{(\text{stress})^2}{Y}$

Explore More

Similar Questions

The work done per unit volume to stretch a wire of cross-sectional area $2 \, mm^2$ by $2 \%$ is ....... $MJ/m^3$. Given: Young's modulus $Y = 8 \times 10^{10} \, N/m^2$.

$A$ rubber band catapult has an initial length of $2 \, cm$ and a cross-sectional area of $5 \, mm^2$. It is stretched by an additional $2 \, cm$ and then released to project a stone of mass $20 \, g$. The velocity of the projected stone is (Young's modulus of rubber $= 5 \times 10^8 \, N/m^2$) (in $ \, m/s$)

When a weight $Mg$ is suspended from a wire of length $L$,the increase in length is $l$ meters. The energy stored in the wire is:

$A$ steel rod of length $\ell$,cross-sectional area $A$,Young's modulus of elasticity $Y$,and thermal coefficient of linear expansion $\alpha$ is heated so that its temperature increases by $t\,^\circ C$. The work that can be done by the rod on heating is:

Difficult
View Solution

When a force is applied on a wire of uniform cross-sectional area $3 \times 10^{-6} \, m^2$ and length $4 \, m$,the increase in length is $1 \, mm$. The energy stored in it will be $(Y = 2 \times 10^{11} \, N/m^2)$. (in $, J$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo