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Mechanics
Elastic Potential Energy
Elastic potential energy is the potential energy stored within an elastic body as a result of deformation, such as stretching or compressing. It arises from the conservative internal forces that oppose the deformation.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Definition of elastic energy, work done in stretching a wire, energy density, and basic numerical problems based on Hooke's law and Young's modulus.
PGT extension
Detailed derivation of elastic potential energy per unit volume, strain energy in terms of elastic constants, and energy stored under complex stress-strain conditions.
Prerequisites
Hooke's Law and Elastic Constants
Work-Energy Theorem
Stress, Strain, and Young's Modulus
Origin of Elastic Energy
When an external force deforms an elastic body, work is done against the interatomic forces.
This work is stored within the material in the form of potential energy, termed elastic potential energy or strain energy.
Upon removal of the deforming force, this stored energy is completely recovered if the deformation remains within the elastic limit.
Energy Density in a Stretched Wire
Energy density is defined as the elastic potential energy stored per unit volume of the deformed material.
It depends quadratically on the longitudinal strain or linearly on the square of stress divided by Young's modulus.
Expressed in SI units as joules per cubic metre (J/m³).
Behavior under Cyclic Loading and Hysteresis
During loading and unloading cycles within the elastic limit, the energy stored is fully recovered.
If the deformation exceeds the elastic limit, permanent plastic deformation occurs, and some energy is dissipated as heat.
The area of the stress-strain hysteresis loop represents the thermal energy dissipated per unit volume per cycle.
Work Done in Stretching a Spring vs. Wire
For a helical spring, the displacement is macroscopic and measured in meters, governed by spring constant k.
For a uniform wire of length L and cross-sectional area A, the displacement is distributed throughout, governed by Young's modulus Y.
Both follow the fundamental quadratic relation $U = ½ Fx$ where F is the restoring force and x is the extension.
Derivations
Derivation of Elastic Potential Energy of a Stretched Wire
Conditions: Uniform wire of original length L, cross-sectional area A, Young's modulus Y, stretched by an incremental extension x.
According to Hooke's law, the tension F required to produce an extension x is given by F = \frac{YA}{L} x.
The small amount of work dW done in producing an additional infinitesimal extension dx is dW = F dx = \frac{YA}{L} x dx.
Integrating this work from zero to total extension l yields the total elastic potential energy $U = \int_{0}^{l} \frac{YA}{L} x dx$.
Solving the integral gives $U = \frac{YA}{2L} l^2$.
Result: U = \frac{1}{2} \left( \frac{YA}{L} l \right) l = \frac{1}{2} \times \text{Load} \times \text{Extension}
Derivation of Elastic Energy Density (Energy per Unit Volume)
Conditions: Isotropic elastic medium undergoing longitudinal strain within the elastic limit.
Start with the expression for total potential energy $U = \frac{1}{2} \frac{YA}{L} l^2$.
Divide the total energy by the volume of the wire V = A \times L to obtain energy density u.
Y = Young's modulus, A = Area, l = Extension, L = Original length Joule (J)
Within elastic limit
Energy in terms of Load and Extension
$$U = \frac{1}{2} F l$$
F = Stretching force, l = Extension Joule (J)
Linear force-extension relationship
Energy Density (Stress-Strain Form)
$$u = \frac{1}{2} \sigma \epsilon$$
$\sigma = Tensile stress$, $\epsilon = Tensile strain$ Joule per cubic metre (J/m³)
Homogeneous and isotropic material
Energy Density in terms of Young's Modulus
$$u = \frac{1}{2} Y \epsilon^2$$
Y = Young's modulus, $\epsilon = Strain$ Joule per cubic metre (J/m³)
Hooke's law applies
Energy Density in terms of Stress
$$u = \frac{\sigma^2}{2Y}$$
$\sigma = Stress$, Y = Young's modulus Joule per cubic metre (J/m³)
Linear elastic regime
Elastic Potential Energy of a Spring
$$U = \frac{1}{2} k x^2$$
k = Spring constant, x = Displacement Joule (J)
Ideal spring obeying Hooke's law
Bulk Elastic Energy Density
$$u = \frac{1}{2} B \Delta V^2$$
B = Bulk modulus, ΔV = Volume change fraction Joule per cubic metre (J/m³)
Uniform volumetric compression
Total Energy Stored in Stretched Rod
$$U = \frac{1}{2} \frac{F^2 L}{YA}$$
F = Force, L = Length, Y = Young's modulus, A = Area Joule (J)
Static equilibrium
Graphs and experiments
Force versus Extension Graph
Draw: A straight line passing through the origin on a F-x Cartesian plot, ending within the elastic limit.
Interpret: The slope of the line represents the stiffness constant (YA/L or k). The area enclosed under the line with the displacement axis represents the total elastic potential energy stored.
Exam use: Directly tested for calculating work done or comparing energy storage capacities of different wires or springs.
Stress-Strain Curve and Elastic Energy Storage
Draw: A standard stress-strain curve showing the proportional limit, elastic limit, yield point, and fracture point.
Interpret: The area under the curve up to the elastic limit gives the resilience or maximum elastic energy density the material can store without permanent damage.
Exam use: Used to distinguish between resilience, toughness, and proof stress in conceptual PGT questions.
Searle's Apparatus / Young's Modulus Energy Experiment
Draw: Experimental setup showing reference and experimental wires, spirit level, micrometer screw gauge, and added load weights.
Interpret: Measurement of extension corresponding to known incremental loads allows experimental verification of the work-energy relations in stretched metallic wires.
A steel wire of length 2 m and cross-sectional area 1.0 mm² is stretched by 1 mm. If Young's modulus of steel is 2 × 10¹¹ N/m², calculate the elastic potential energy stored in the wire.
Identify given parameters: L = 2 m, $A = 1.0 mm^{2} = 1.0 \times 10^{-6} m^{2}$, $l = 1 mm = 1.0 \times 10^{-3} m$, $Y = 2 \times 10^{11} N/m^{2}$.
Substitute values into the formula $U = \frac{YA l^2}{2L}$.
Find the energy density stored in a copper wire subject to a longitudinal stress of 4 × 10&sup7; N/m². Given Young's modulus for copper is 1.2 × 10¹¹ N/m².
Valid strictly within the elastic limit where Hooke's law is obeyed and no permanent plastic deformation occurs.
Assumes the material is homogeneous, isotropic, and its cross-sectional area remains nearly constant during stretching (neglecting lateral contraction governed by Poisson's ratio for basic calculations).
Common misconceptions
Common error: Elastic potential energy stored in a wire is directly proportional to the total length of the wire.
Correct: Energy density is independent of length, and total energy for a given extension is inversely proportional to the original length (U ∝ 1/L).
Longer wires offer less stiffness ($k = YA/L)$ for the same cross-sectional area, reducing the restoring force for a given extension l.
Common error: Work done in stretching a wire equals Force times Extension (F × l).
Correct: Work done equals one-half of the product of Force and Extension (½ F l).
The force increases linearly from zero to F as the extension increases from zero to l, requiring integration or the average force.
Quick revision
Elastic energy formula: $U = ½ F l = \frac{YA l^2}{2L}$.
Energy density formula: $u = ½ \sigma\epsilon = \frac{\sigma^2}{2Y} = \frac{1}{2} Y \epsilon^{2}$.
Elastic Potential Energy of a Wire: $U = \frac{YA l^2}{2L}$
Must remember
Elastic potential energy stored in a stretched wire is given by $U = \frac{YA l^2}{2L}$ or $U = \frac{1}{2}Fl$.
Energy density (energy per unit volume) in terms of stress $\sigma$ and strain $\epsilon$ is $u = \frac{1}{2} \sigma \epsilon = \frac{1}{2} Y \epsilon^2 = \frac{\sigma^2}{2Y}$.
The area under the stress-strain hysteresis loop during cyclic loading represents thermal energy dissipated per unit volume per cycle.
For a given extension $l$, the elastic potential energy of a wire is inversely proportional to its original length ($U \propto \frac{1}{L}$).
Elastic potential energy of an ideal spring obeying Hooke's law is given by $U = \frac{1}{2} k x^2$.
Glossary
Elastic Potential EnergyWork done against interatomic forces in deforming an elastic body, stored within the material.
Energy Density ($u$)Elastic potential energy stored per unit volume of the deformed material, measured in $\text{J/m}^3$.
Young's Modulus ($Y$)Ratio of tensile stress to tensile strain within the linear elastic regime.
Hysteresis LoopClosed loop formed on a stress-strain graph during cyclic loading and unloading beyond the elastic limit, indicating energy dissipation.
Spring Constant ($k$)Restoring force per unit macroscopic displacement for a helical spring.
Stretching a Wire vs. Helical Spring
Parameter
Uniform Wire
Helical Spring
Displacement nature
Distributed throughout length ($l$)
Macroscopic coil extension ($x$)
Governing elastic property
Young's Modulus ($Y$) and Area ($A$)
Spring Constant ($k$)
Energy relation
$U = \frac{YA l^2}{2L}$
$U = \frac{1}{2} k x^2$
Forms of Energy Density
In Terms Of
Equation
Applicability
Stress & Strain
$u = \frac{1}{2} \sigma \epsilon$
Homogeneous and isotropic material
Strain & Young's Modulus
$u = \frac{1}{2} Y \epsilon^2$
Hooke's law regime
Stress & Young's Modulus
$u = \frac{\sigma^2}{2Y}$
Linear elastic regime
Practice Quiz
Choose an option to reveal the explanation.
1. What is the physical significance of the area of the stress-strain hysteresis loop during cyclic loading of a material?
Explanation: If deformation exceeds the elastic limit, permanent plastic deformation occurs, and the area of the stress-strain hysteresis loop represents the thermal energy dissipated per unit volume per cycle.
2. The elastic potential energy density $u$ in terms of tensile stress $\sigma$ and Young's modulus $Y$ is expressed as:
Explanation: Energy density in terms of stress is given by $u = \frac{\sigma^2}{2Y}$, derived by substituting $\sigma = Y\epsilon$ into $u = \frac{1}{2}Y\epsilon^2$.
3. How does the total elastic potential energy $U$ of a uniform wire vary with its original length $L$ for a given extension $l$ and cross-sectional area $A$?
Explanation: From the formula $U = \frac{YA l^2}{2L}$, for a fixed extension $l$, the energy $U$ is inversely proportional to the original length $L$ because longer wires offer less stiffness.
4. What is the correct expression for the work done in stretching a uniform wire of length $L$, cross-sectional area $A$, and Young's modulus $Y$ by an extension $l$?
Explanation: The work done is integrated from zero extension to $l$, resulting in $U = \frac{1}{2} F l = \frac{1}{2} \left(\frac{YA l}{L}\right) l = \frac{YA l^2}{2L}$.
5. What is the SI unit of elastic potential energy density?
Explanation: Energy density is defined as the elastic potential energy stored per unit volume, making its SI unit $\text{J/m}^3$.
6. If a force $F$ produces an extension $l$ in a wire obeying Hooke's law, the elastic potential energy stored is:
Explanation: Since force increases linearly from zero to $F$, the average force is $\frac{1}{2}F$, so the work done (energy stored) is $U = \frac{1}{2} F l$.
7. Which of the following conditions is mandatory for the complete recovery of elastic potential energy upon removal of the deforming force?
Explanation: Upon removal of the deforming force, stored energy is completely recovered only if the deformation remains within the elastic limit.
8. The energy density in terms of strain $\epsilon$ and Young's modulus $Y$ is given by:
Explanation: Energy density depends quadratically on longitudinal strain, expressed as $u = \frac{1}{2} Y \epsilon^2$.
9. What governs the macroscopic displacement in a helical spring compared to a uniform wire?
Explanation: For a helical spring, the displacement is macroscopic and measured in meters, governed by the spring constant $k$.
10. Why is the work done in stretching a wire equal to $\frac{1}{2}Fl$ rather than $Fl$?
Explanation: The force starts at zero and increases linearly to $F$, so the average force during extension is $\frac{1}{2}F$.
11. What form of energy is stored within an elastic material when it is deformed by an external force?
Explanation: Work done against interatomic forces during deformation is stored as elastic potential energy or strain energy.
12. If the tensile stress in a wire is $\sigma$ and tensile strain is $\epsilon$, the elastic energy density is:
Explanation: Energy density expressed in terms of stress and strain is $u = \frac{1}{2} \sigma \epsilon$.
13. What happens to the stored elastic energy if deformation exceeds the elastic limit?
Explanation: Exceeding the elastic limit causes permanent plastic deformation and thermal energy dissipation.
14. The total energy stored in a stretched rod subjected to a tensile force $F$ is given by:
Explanation: Using $l = \frac{FL}{YA}$ in $U = \frac{1}{2}Fl$ gives $U = \frac{1}{2} F \left(\frac{FL}{YA}\right) = \frac{1}{2} \frac{F^2 L}{YA}$.
15. Bulk elastic energy density under uniform volumetric compression involving bulk modulus $B$ and volume change fraction $\Delta V$ is:
Explanation: Bulk elastic energy density is given by $u = \frac{1}{2} B \Delta V^2$ under uniform volumetric compression.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. A wire of length $L$ and cross-sectional area $A$ is stretched by an amount $l$. If $Y$ is Young's modulus of the material, the potential energy stored per unit volume is:
Explanation: Strain $\epsilon = \frac{l}{L}$. Energy density $u = \frac{1}{2} Y \epsilon^2 = \frac{1}{2} Y \left(\frac{l}{L}\right)^2$.
2. Two wires $A$ and $B$ of the same material and same length have radii $r_1$ and $r_2$ respectively. When stretched by the same force, the strain energy stored per unit volume is in the ratio:
Explanation: Energy density $u = \frac{\sigma^2}{2Y}$. Since both wires are made of the same material ($Y$ is same) and subjected to the same force ($F$ and $A$ give same stress $\sigma$), the energy density is identical for both, ratio is $1:1$.
3. If the potential energy stored in a spring is $U$ when stretched by $x$, what is the potential energy stored when it is stretched by $2x$?
Explanation: Elastic potential energy of a spring is $U = \frac{1}{2} k x^2$. When extension becomes $2x$, $U' = \frac{1}{2} k (2x)^2 = 4 \left(\frac{1}{2} k x^2\right) = 4U$.
4. A steel wire of length $2\text{ m}$ and cross-sectional area $10^{-6}\text{ m}^2$ is stretched by $1\text{ mm}$. If Young's modulus of steel is $2 \times 10^{11}\text{ N/m}^2$, the work done in stretching the wire is:
5. Which of the following graphs correctly represents the elastic potential energy $U$ versus extension $x$ for an ideal spring?
Explanation: The relation $U = \frac{1}{2}kx^2$ is quadratic, plotting as a parabola opening upwards.
6. The work done in pulling a spring from extension $0$ to $l$ is $W_1$, and from $l$ to $2l$ is $W_2$. The ratio $W_1 : W_2$ is:
Explanation: $W_1 = \frac{1}{2}kl^2$. Work from $l$ to $2l$ is $W_2 = \frac{1}{2}k(2l)^2 - \frac{1}{2}kl^2 = \frac{3}{2}kl^2$. Thus, $W_1 : W_2 = 1 : 3$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. If a compressive force causes a volume contraction of $\Delta V$ in a material with bulk modulus $B$, the stored elastic potential energy per unit volume is:
Explanation: Bulk elastic energy density is given by $u = \frac{1}{2} B \Delta V^2$.
2. A force $F$ stretches a wire of length $L$ and radius $r$ by an amount $l$. If both length and radius are doubled, keeping the force same, the elastic potential energy stored in the wire becomes:
Explanation: $U = \frac{1}{2} \frac{F^2 L}{YA} = \frac{F^2 L}{2 Y \pi r^2}$. If $L' = 2L$ and $r' = 2r$, then $U' = \frac{F^2 (2L)}{2 Y \pi (2r)^2} = \frac{2}{4} U = \frac{1}{2} U$. Stored energy is halved.
3. During a cyclic loading experiment, a stress-strain hysteresis loop is formed. The area of this loop is $150\text{ J/m}^3$. If the volume of the wire is $2\times 10^{-3}\text{ m}^3$, the total thermal energy dissipated per cycle is:
Explanation: Thermal energy dissipated = Loop area $\times$ Volume = $150 \times (2 \times 10^{-3}) = 0.30\text{ J}$.
4. What is the elastic potential energy stored per unit volume in a copper wire whose tensile stress is $3\times 10^7\text{ N/m}^2$ and Young's modulus is $1.2 \times 10^{11}\text{ N/m}^2$?
5. When a spring is stretched by $10\text{ cm}$, its potential energy is $U$. If it is compressed by $5\text{ cm}$ instead, its potential energy will be:
Explanation: Potential energy $U \propto x^2$. Compression by $5\text{ cm}$ ($x/2$) yields energy $\left(\frac{1}{2}\right)^2 U = \frac{U}{4}$.
6. Which of the following correctly defines energy density in a deformed elastic material?
Explanation: Energy density is defined as the elastic potential energy stored per unit volume of the deformed material.
7. If a wire is stretched such that the stress developed is equal to Young's modulus ($Y$), the energy density $u$ in terms of $Y$ is:
Explanation: Substitute $\sigma = Y$ into $u = \frac{\sigma^2}{2Y}$ to get $u = \frac{Y^2}{2Y} = \frac{Y}{2}$.
8. Two springs of spring constants $k_1$ and $k_2$ ($k_1 > k_2$) are stretched by the same force. Which spring stores more elastic potential energy?
Explanation: Energy in terms of force is $U = \frac{F^2}{2k}$. Since $k_2 < k_1$, Spring 2 stores more potential energy for the same applied force $F$.
9. The work done in stretching an elastic wire within its elastic limit is stored as:
Explanation: Work done against interatomic forces is stored within the material as elastic potential energy.
10. What is the primary factor distinguishing macroscopic displacement energy formulas in springs from microscopic/distributed strain energy formulas in wires?
Explanation: Helical springs feature macroscopic displacement governed by spring constant $k$, while wires have distributed displacement governed by Young's modulus $Y$.