JEE Main Physics Elastic Potential Energy

What is Potential Energy?

Potential energy is the energy that an object possesses due to its pressure, electric charge, or relatively stationary position in space. It is the body's innate energy about its location relative to other objects. One of the two primary forms of energy is potential energy, the other being kinetic energy. Elastic potential energy and gravitational potential energy are the two different types of potential energy.

What is Elastic Potential Energy?

The energy that is stored when a force is used to deform an elastic object is known as elastic potential energy. Until the force is released and the object springs back to its original shape, doing labour in the process, the energy is retained. The object may be compressed, stretched, or twisted during the deformation.

In simple words, Elastic potential energy is the energy stored in an elastic object, such as a spring or a stretched wire, due to its deformation. The elastic potential energy stored in a stretched wire depends on the extent of its stretching or compression and the material's elastic properties.

Elastic Energy Examples

Numerous items are made to hold elastic potential energy, such as:

• The coil spring of a wind-up clock

• An archer's stretched bow

• A bent diving board, just before a diver’s jump

• The twisted rubber band that powers a toy aeroplane

• A bouncy ball, compressed at the moment, bounces off a brick wall.

Hooke's Law:

Hooke's Law is named after Robert Hooke, a physicist from the 17th century. It explains that small changes in an object's size or shape are directly related to the force applied to it. The formula for this law is F = kx, where F is the applied force, k is the spring constant, and x is the displacement of the object.

Deformation can happen through stretching, bending, squeezing, or twisting. Hooke's Law is not just about rubber bands; it applies to things like breathing, car suspensions, and even mattresses. However, the law works only for elastic objects and in cases where the deformation is small, so it’s not universal.

Definition of Stress, Strain, and Young's Modulus 

Stress: The definition of stress is force per unit area, where F is the applied force, and A is the cross-sectional area.

• Tensile Stress: Stress that has a propensity to extend or stretch the material - behaves normally in the stressed area.

• Compressive Stress: Stress that has a propensity to shorten or compress the material - behaves normally in the stressed area.

• Shearing Stress: Stress that has a propensity to shear the material - compressive or tensile stress acts perpendicular to the strained area at right angles.

Strain: Strain is defined as the deformation of a solid due to stress.

• Normal strain: lengthening or shortening of a line segment.

• Shear strain: Angle difference between two once perpendicular lines.

Young's Modulus: Over a range of loads, most metals deform in proportion to the applied force. According to Hooke's Law, strain is proportional to deformation, and stress is proportional to load.

Elastic Potential Energy Formula using Stretched Wire 

A specific amount of labour must be done against restoring force to stretch a wire. The potential energy (U) of the stretched wire is where this work is stored.

Consider a wire that is being bent by a deforming force of length "L" and cross-section area "A". The subsequent modest task performed by "dw" to be extended by "dl" is given by:

$dW=F~dl$                                                    ----(1)

Thus, by integrating dw between the limits of 0 and 'L', the total work 'W' required to extend the wire by length 'l' is determined.

Therefore,

$W=\int dw$

$W=\int_{0}^{l}F~dl$                                       ---(2)

Here, ‘F’ is elongation-dependent. Therefore, the definition of Young's Modulus (Y) when the elongation is ‘l’ yields

$Y=\dfrac{stress}{strain}=\dfrac{\dfrac{F}{A}}{\dfrac{l}{L}}$

$\therefore F=\dfrac{YAl}{L}$                            ----(3)

Using this in equation (2) and integrating, we get

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

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

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

$=\dfrac{1}{2}\dfrac{YAl}{L}{{l}^{2}}$

$\therefore W=\dfrac{1}{2}F\times l$ 

This work is equivalent to the stored potential energy (U).

Energy stored (U) = work done (w)

$U=\dfrac{1}{2}force\times elongation$

Elastic Potential Energy Formula in terms of Stress and Strain

The elastic potential energy of the stretched wire is the amount of energy it can hold relative to its volume.

$U=\dfrac{energy~stored}{volume} $

$=\dfrac{YF\times l}{A\times L}$

$ =\dfrac{1}{2}\dfrac{F}{A}\times \dfrac{l}{L}$

Elastic Potential Energy per unit volume is $\dfrac{1}{2}stress\times strain  $

Elastic Potential Energy of a Spring

Among the three springs, which one has the most potential energy? It’s the third spring since it is stretched the most (2x) compared to the others (0 and x). This greater stretch means the third spring stores the highest elastic potential energy. Think of it this way: the work done in stretching the spring is converted into stored energy.

Here is the Elastic Potential Energy Derivation,

According to Hooke’s Law, the force applied to an elastic material is directly proportional to its displacement. This is expressed as:  

F = kx,  

Where:  

• F is the applied force,  

• x is the displacement (stretch or compression),  

• k is the spring constant.

At equilibrium (where no external force is acting), the system has no potential energy. To find the potential energy stored in a spring when it is displaced, we use the formula:  

$\dfrac{1}{2}$ kx²,  

Where:  

U is the elastic potential energy of the system.  

Elastic After Effect 

We are aware that when the deforming forces are removed from elastic bodies, they return to their original shape. The original configurations of some elastic bodies can be recovered right away, while for others, it takes some time. Elastic after-effect is the transient delay experienced by an elastic body in returning to its initial shape following the removal of a deforming force.

Elastic Fatigue 

Elasticity is lost when a body is repeatedly deformed and released. Elastic fatigue is the term for this. After repeatedly deforming, an iron nail may break due to elastic fatigue.

Elastic Hysteresis 

Elastic hysteresis is the phenomenon where the strain produced in elastic lags behind the tension to which it is subjected as a result of elastic after-effect. The strain created when a body is subjected to the road cycle of growing and reducing stress is greater when the body is empty than when it is loaded.

Elastic Potential Energy Examples for Better Understanding

Example 1: A spring is stretched such that it stores 25 J of potential energy, and its spring constant is 250 N/m. Find the displacement of the spring.  

Answer:  Given: Potential Energy (P.E) = 25 J  

Spring Constant (k) = 250 N/m  

The potential energy formula is:  

P.E = $\frac{1}{2}kx^2 $

Rearranging to find displacement:  

$x = \sqrt{\frac{2P.E}{k}} $

Substitute the values:  

$x = \sqrt{\frac{2 \times 25}{250}} $

$x = \sqrt{\frac{50}{250}} $

$x = \sqrt{0.2} $

$ x = 0.447\text{m} $

The displacement of the spring is 0.447 m.  

Example 2: A compressed spring has a spring constant of 150 N/m and is displaced by 0.5 m. Calculate the elastic potential energy stored in the spring.  

Answer: Given: 

Spring Constant (k) = 150 N/m  

Displacement (x) = 0.5 m  

The potential energy formula is:  

$P.E = \frac{1}{2}kx^2 $

Substitute the values:  

$P.E = \frac{1}{2} \times 150 \times (0.5)^2 $

$P.E = \frac{1}{2} \times 150 \times 0.25 $

$P.E = 75 \times 0.25 $

$P.E = 18.75 \text{J} $

The elastic potential energy stored is 18.75 J.

Example 3: A spring with a spring constant of 300 N/m is compressed by 0.4 m. Calculate the elastic potential energy stored in the spring.  

Answer: Given:  

Spring Constant (k) = 300 N/m  

Displacement (x) = 0.4 m  

The potential energy formula is:  

$P.E = \frac{1}{2}kx^2 $

Substitute the values:  

$P.E = \frac{1}{2} \times 300 \times (0.4)^2 $

$P.E = \frac{1}{2} \times 300 \times 0.16 $

$P.E = 150 \times 0.16 $

$P.E = 24  \text{J} $

The elastic potential energy stored in the spring is 24 J.

Conclusion

The potential energy that is held when an elastic item is stretched or compressed by an external force, such as the stretching of a spring, is known as elastic potential energy. It is equivalent to the work required to stretch the spring, which is dependent on both the length of the stretch and the spring constant k.

The elastic limit of an object intended to store elastic potential energy will normally be high. Still, the maximum load an elastic object can support is a property shared by all elastic objects. The elastic limit is the point at which an object can be deformed without losing its original shape.

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