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

[Amith2006]

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.

 

[Hootenanny]

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);
T = \frac{\lambda x}{L}
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);
E_p = \int_{0}^{e} \frac{\lambda x}{L} \;\; dx = \frac{\lambda e^2}{2L}

 

[Doc Al]

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);
T = \frac{\lambda x}{L}

You're missing the area. If \lambda is Young's modulus, then:
T = A \frac{\lambda x}{L}

 

[Hootenanny]

You're missing the area. If \lambda is Young's modulus, then:
T = A \frac{\lambda x}{L}

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 \lambda 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 Could you enlighten my Doc Al?

 

[Doc Al]

Not sure about bringing enlightenment before having coffee, but the standard definition of Young's modulus is Stress (F/A) over Strain (\Delta L / L). 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:

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.

 

[Hootenanny]

Not sure about bringing enlightenment before having coffee

I've just had a big cup

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

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 \lambda are Newtons

However, I know from my physics that youngs modulus is defined as;
\lambda = \frac{FL}{Ax}
which should leave units as N\cdot m^{-3}. Perhaps the applied mathematics textbook is using a different constant and incorrectly naming it the 'modulus of elasticity'?

 

[Doc Al]

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.

 

[Hootenanny]

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.