Calculating Energy Stored in a Wire: Young's Modulus, F, L, & a

In summary, Sir, Hooke's law relates tension (T) to extension (x) in a wire in relative to it's origional length(L) and young's modulus (\lambda). If \lambda is Young's modulus, then T = A \frac{\lambda x}{L}.
  • #1
Amith2006
427
2
Sir,
Please help me with this problem.
# A wire of cross sectional area a, length L and young’s modulus Y is extended by an external force F. What is the total energy stored in the wire?
I solved it in the following way:

Energy stored = ½ x stress x strain
= (½) x (F/a) x (dL/L)
= (½) x (F/a) x (F/aY)
= (1/2) x (F^2/a^2Y)

Is it right? But the answer given in my book is ½(YF^2L/a).
Here the symbol ^ represents power and x represents multiplication.
 
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  • #2
Hooke's law relates tension (T) to extension (x) in a wire in relative to it's origional length(L) and young's modulus ([itex]\lambda[/itex]);
[tex]T = \frac{\lambda x}{L}[/tex]
Energy stored is simply work done by stretching the wire, which is force multiplied by distance moved, which is given by integrating Hooke's law between the limits of zero and maximum extension (e);
[tex]E_p = \int_{0}^{e} \frac{\lambda x}{L} \;\; dx = \frac{\lambda e^2}{2L}[/tex]
 
  • #3
Hootenanny said:
Hooke's law relates tension (T) to extension (x) in a wire in relative to it's origional length(L) and young's modulus ([itex]\lambda[/itex]);
[tex]T = \frac{\lambda x}{L}[/tex]
You're missing the area. If [itex]\lambda[/itex] is Young's modulus, then:
[tex]T = A \frac{\lambda x}{L}[/tex]
 
  • #4
Doc Al said:
You're missing the area. If [itex]\lambda[/itex] is Young's modulus, then:
[tex]T = A \frac{\lambda x}{L}[/tex]

Indeed, I stand corrected, I have just work through the derivation using stress and strain. However, glancing through my textbooks it appears that they make no mention of area and states the [itex]\lambda[/itex] is simply the modulus of elasticity, which is the same as young's modulus. I am now rather confused and worried with regards to my upcomming exam :confused: Could you enlighten my Doc Al?
 
  • #5
Not sure about bringing enlightenment before having coffee, but the standard definition of Young's modulus is Stress (F/A) over Strain ([itex]\Delta L / L[/itex]). How does your text define it?

See here: http://hyperphysics.phy-astr.gsu.edu/HBASE/permot3.html#c2

Except for leaving out the area, your analysis is perfectly correct.

Note to Amith2006:
Amith2006 said:
But the answer given in my book is ½(YF^2L/a).
Check the units of that answer; the correct answer must have units of energy.
 
  • #6
Doc Al said:
Not sure about bringing enlightenment before having coffee

I've just had a big cup :tongue2:

Doc Al said:

Yeah, I've just been reading through that and agree with it totally.

Just reading through my text(applied mathematics textbook) it says that;
The units of [itex]\lambda[/itex] are Newtons
However, I know from my physics that youngs modulus is defined as;
[tex]\lambda = \frac{FL}{Ax}[/tex]
which should leave units as [itex]N\cdot m^{-3}[/itex]. Perhaps the applied mathematics textbook is using a different constant and incorrectly naming it the 'modulus of elasticity'?
 
  • #7
Hootenanny said:
Perhaps the applied mathematics textbook is using a different constant and incorrectly naming it the 'modulus of elasticity'?
That must be it. No problem as long as you use the definition consistently. (But it looks like the OP is using the standard definition--so be careful!)

Funny, I just saw another problem where this same issue came up (https://www.physicsforums.com/showthread.php?t=113574); wonder if that fellow is using the same text.
 
  • #8
Looks like he is, it's a pretty standard text for A-Level Mathematics, funny I haven't noticed it before. I'll just have to remember that my physics exam uses the 'proper' youngs modulus! Thanks for you help.
 

1. How do you calculate the energy stored in a wire?

In order to calculate the energy stored in a wire, you will need to use the formula: Energy = (1/2) x (Young's Modulus) x (Cross-sectional Area) x (Change in Length)^2.

2. What is Young's Modulus?

Young's Modulus is a measure of the stiffness or elasticity of a material. It is defined as the ratio of stress (force per unit area) to strain (change in length per unit length) in a material under tension or compression.

3. How does the length of a wire affect the energy stored?

The longer the wire, the greater the energy stored. This is because the change in length is squared in the energy formula, so a longer wire will have a larger change in length and therefore, a larger amount of energy stored.

4. What is the relationship between the cross-sectional area and the energy stored in a wire?

The cross-sectional area of a wire does not have a direct relationship with the energy stored. However, a larger cross-sectional area can result in a higher Young's Modulus, which will increase the energy stored in the wire.

5. Can the energy stored in a wire be negative?

No, the energy stored in a wire cannot be negative. This is because the energy formula includes the squared value of the change in length, which will always result in a positive value.

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