- #1
novamatt
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Homework Statement
The question posed - Calculate the work done in a steel wire (Esteel = 200GPa) 2.0m long and 0.010cm² in cross-sectional area extends by 2.5mm when loaded.
so
Length (L) = 2m
Area (A) = 1(x10 to the -6)m
Extension (x) = 0.0025m
Ep? = 2(x10 to the 11) Nm?
Homework Equations
I know work done = Force x distance and EP = 1/2F[itex]\chi[/itex]
The Attempt at a Solution
my attemp at solving this as follows:
1/2 2(x10 to the eleven)Nm x 0.0025m
= 1(x10 to the eleven)Nm x 0.0025m
= 2.5(x10 to the seven) Nm²
this just does not seem right to me... I haven't used half of the information give in the problem and I'm pretty sure potential energy is not neccissarly the same as work done. Also if the was the case wouldn't the spring constant reach it's ultimate tensile stress or fracture point.
Please help I was away when we covered this in college and the notes provided do not seem to help at all.