Young's Modulus question!?

Question

Consider a steel guitar string of initial length L=1m and cross-sectional area A=.500mm^2 . The Young's modulus of the steel is Y= 2.0*10^ 11Pa . How far (Delta L) would such a string stretch under a tension of 1500 N?Express your answer in millimeters using two significant figures. and in mm.

I 've tried lots of times this problem and I keep getting it wrong could someone help me out I've tried lots of times and I dont get it right.

so far none of the answers given are right anyone out there with another answer? : (

I got it the answer is 15 mm I just used the equation F= k delta x for springs and I found k for the guitar string by deriving k=(Y*A)/L

Answer 1

Your problem may be in units, Young's modulus is the ratio of stress to strain, not the ratio of force to elongation. It has units of KiloNewtons per square meter or KPa. The "linear" stress-strain relationship, which holds well for most metals, is
σ = E*ε
The cross sectional area is 5.0*10^-7 m^2. Then the average normal stress is σ = F / A = 3.0*10^6 KPa. Dividing by Young's modulus E, gives an average strain of 1.5*10^-5, a dimensionless number. Since the unstretched length is exactly 1m, the total elongation is ΔL = L*ε, or 1.5*10^-3 mm.

I hope that is correct. If it doesn't check out, please disregard this message!

Answer 2

L = 1m
A = 500mm^2 = 500 * 10^-6 m
Y = 2 * 10^11
T = 1500N
deltaL = ?

T = F/L
F = T*L
=1500 * 1
=1500Nm

Y = FL/A delta L
deltaL = FL/AY
=1500*1 / 5*10^-4*2*10^11
=1500/10*10^7
=1500/10^8
=1500*10^-8
=1.5*10^-11m
1.5*10^-13mm