Answer
The wire stretches a distance of $2.2~cm$
Work Step by Step
(a) $Y = \frac{F/A}{\Delta L/L}$
$Y$ is Young's modulus
$F$ is the force
$A$ is the cross-sectional area
$\Delta L$ is the change in length
$L$ is the original length
We can find the change in length:
$\Delta L = \frac{F~L}{Y~A}$
$\Delta L = \frac{(5000~N)(2.0~m)}{(9.2\times 10^{10}~N/m^2)(5.0\times 10^{-6}~m^2)}$
$\Delta L = 2.2\times 10^{-2}~m = 2.2~cm$
The wire stretches a distance of $2.2~cm$.
