## College Physics (4th Edition)

The wire stretches a distance of $2.2~cm$
(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 $2.2~cm$.