Electra metal wire wrapping poisons ratio on my phone
Hello,
PEstored = Work done on the wire during stretching= integral of (Fdx) [F=force, dx = small displacement]
But by hooke's law
Y = FL/Ax [F=force, L=length of wire, A= area, x = extension]
So, F= (AY/L)x
Thus PEstored = Integral of (AY/L)*(xdx)
= (AYl
2
)/2L [l = maximum extension]
= (1/2)*(AL)*(Y)*(l/L)
2
[(AL) = volume, (l/L) = a = strain]
Or PE = (volume)*(1/2)*Y*a
2
So PE/volume = (1/2)*Y*a
2
So substitute the value of young's modulus in Y and strain value in 'a'.
Hope this helps.....
Reply
Hello,
PEstored = Work done on the wire during stretching= integral of (Fdx) [F=force, dx = small displacement]
But by hooke's law
Y = FL/Ax [F=force, L=length of wire, A= area, x = extension]
So, F= (AY/L)x
Thus PEstored = Integral of (AY/L)*(xdx)
= (AYl
2
)/2L [l = maximum extension]
= (1/2)*(AL)*(Y)*(l/L)
2
[(AL) = volume, (l/L) = a = strain]
Or PE = (volume)*(1/2)*Y*a
2
So PE/volume = (1/2)*Y*a
2
So substitute the value of young's modulus in Y and strain value in 'a'.
Hope this helps.....
Hello Milan,
The Elastic potential energy is given by 1/2 *stress*strain
According to this we get,
=1/2*(Y*alpha)*(alpha)
=1/2*Y*(alpha)raised to 2
Substitute the values and you will get the answer.
Hope this helps and all the best!
