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Today, we'll learn about elastic potential energy in a wire. Can anyone explain what happens when we stretch a wire?
I think stretching a wire requires a force, and the wire gets longer.
That's correct! The work done by the force is stored in the wire as elastic potential energy. Why do you think this energy is important?
Because the wire can return to its original shape after we stop applying the force!
Exactly! This is crucial in applications like springs. Remember, the energy stored is a function of stress and strain.
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Let's dive into the equations. When we stretch the wire, we apply a force, right? Can someone write the formula for the force acting in terms of Young's modulus?
It should be F = YA Γ (l/L)!
Great! Now, if we want to find the work done as we stretch the wire from L to L + l, what equation do we use?
We can integrate, right? So W would be the integral of that force over the length.
Exactly! Integrating gives us the total work done, or the elastic potential energy stored, represented as u = (1/2) Γ Ο Γ Ξ΅. This connection is key for engineers!
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Why do we care about the elastic potential energy in designs like bridges or cranes?
Because we need materials that can handle stress without permanent deformation!
Correct! It's risky to exceed the elastic limit of a material. Can you visualize a scenario where knowing this might save lives?
If a cable in a crane stretched too much and broke?
Exactly! Understanding these concepts helps us design safer and more efficient structures.
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Let's do a quick calculation together. If we have a wire with L = 2 m and A = 1 Γ 10^-4 mΒ², with a Youngβs modulus of 200 GPa, what would the elastic potential energy be if the wire is stretched by 0.01 m?
I think we use the formula for work done!
Yes, we start with calculating stress and strain first. What is the stress here?
Stress would be 200 GPa Γ (0.01/2) = 1 GPa!
Good! And the energy would then be calculated using u = (1/2) Γ Ο Γ Ξ΅. Youβre getting it!
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When a wire is stretched due to an applied force, work is performed against inter-atomic forces, resulting in the storage of elastic potential energy. The mathematical relation depicts the dependence on Young's modulus, stress, strain, and the wire's volume.
When a wire experiences tensile stress, work must be done against the inter-atomic forces to elongate the wire. The work done on the wire is stored as elastic potential energy, which can be released when the wire returns to its original length after the force is removed. For a wire of original length L and cross-sectional area A subjected to a deforming force F, if the length is elongated by a small amount l, we can establish the relationship:
Using Young's modulus (Y) which relates stress (Ο) and strain (Ξ΅), we can express the work done (W) in increasing the length of the wire:
$$F = YA \frac{l}{L}$$
$$dW = F \cdot dl = YA \frac{l}{L} dl$$
$$W = \int_0^{l} \frac{1}{2} YA \frac{l}{L} dl = \frac{1}{2} Y \cdot \text{strain}^2 \cdot \text{volume of the wire}$$
$$u = \frac{1}{2} \sigma \epsilon$$
This relationship illustrates that the stored elastic potential energy is not just a function of the material properties but also of the geometry and the deformation applied.
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When a wire is put under a tensile stress, work is done against the inter-atomic forces.
When a wire is stretched, atomic forces attempt to return it to its original shape. To elongate the wire, external force must be applied, overcoming these forces. This work done is stored as elastic potential energy in the wire.
Think of stretching a rubber band. When you pull on it, you're doing work against the forces that try to keep it at its original length. The more you stretch it, the more energy you store in the band.
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When a wire of original length L and area of cross-section A is subjected to a deforming force F along the length of the wire, let the length of the wire be elongated by l. Then from Eq. (8.8), we have F = YA Γ (l/L).
This equation derives from the definition of Young's modulus (Y), which describes how much a wire will elongate (l) when a certain force (F) is applied. The relationship indicates that the force required to stretch the wire is proportional to its cross-sectional area and inversely proportional to its original length.
Imagine trying to stretch two different lengths of wire; a longer wire takes less force to stretch a certain distance than a shorter one. This is because the longer wire can distribute the stress more evenly.
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Now for a further elongation of infinitesimal small length d l, work done d W is F Γ dl or YAldl/L.
The work done (dW) to stretch the wire by an infinitesimal length (dl) can be calculated by multiplying the force (F) by the small change in length (dl). This relationship integrates the force applied over the length to find the total work done (W) during the entire elongation.
Think of pushing a swing. The further you push it (elongate it), the more effort (work) you exert to keep it movingβsimilarly, the effort increases as you stretch a wire more.
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The amount of work done ( W) in increasing the length of the wire from L to L + l, that is from l = 0 to l = l is W = (1/2) Γ Youngβs modulus Γ strainΒ² Γ volume of the wire.
This formula reveals that the total elastic potential energy (W) stored in the wire equals half the product of Youngβs modulus (Y), the square of the strain, and the volume of the wire. This indicates that stronger materials and greater deformations lead to more stored energy.
Consider a charged battery. The more you charge it (like the more you stretch the wire), the more energy it stores. Similarly, the wire retains energy based on how much it is stretched and its material properties.
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Therefore the elastic potential energy per unit volume of the wire (u) is u = (1/2) Γ Ο Γ Ξ΅.
The formula u = (1/2) Γ Ο Γ Ξ΅ tells us how much elastic potential energy is stored in a given volume of material. Here, Ο is stress, and Ξ΅ is strain. This signifies that the energy density increases with stress and strain, consolidating the relationship between deformation and energy storage.
Think of a sponge. The denser the sponge, the more energy (or force) it takes to compress it, which relates to how much potential energy it can store when compressed.
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Key Concepts
Elastic Potential Energy: Energy stored when a wire is stretched, derived from work done against inter-atomic forces.
Young's Modulus: Relates stress with strain; crucial for understanding how materials behave under load.
Work Done: Energy required to stretch a wire, calculated through the force applied over a distance.
See how the concepts apply in real-world scenarios to understand their practical implications.
Example 1: A steel wire of 2 m length with a cross-section of 1 cmΒ² stretched by 1 cm stores elastic potential energy that can be calculated using the established formulas.
Example 2: A rubber band when stretched stores elastic potential energy, and this energy can be used for propulsion when released.
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
Stretch the wire, feel it grow, Energy stored, now you know!
Imagine a rubber band, stretched tight, holds energy like a superhero ready for flight!
W = 1/2 Γ stress Γ strain, remember: that's how energy is gained!
Review key concepts with flashcards.
Review the Definitions for terms.
Term: Elastic Potential Energy
Definition:
The energy stored in a body when it is deformed elastically, proportional to the stress and strain applied.
Term: Tensile Stress
Definition:
The stress that occurs when forces are applied to stretch a material.
Term: Young's Modulus
Definition:
A measure of the stiffness of a solid material and a factor that relates stress to strain.
Term: Strain
Definition:
The measure of deformation representing the displacement between particles in a material body.
Term: Work Done
Definition:
The energy transferred by a force acting over a distance, in this case, the force stretching the wire.