Answer
(a) $1.7\times10^{8}N/m^{2}$
(b) $0.018$
Work Step by Step
(a) Maximum stress = $\frac{F}{A}=\frac{6.8\times10^{4}N}{4\times10^{-4}m^{2}}=1.7\times10^{8}N/m^{2}$
(b) Let's apply equation 10.17 $F=Y(\frac{\Delta L}{L_{0}})A$ to find the strain.
$\frac{\Delta L}{L_{0}}=\frac{1}{Y}(\frac{F}{A})$ ; Let's plug known values into this equation.
$\frac{\Delta L}{L_{0}}=(\frac{1}{9.4\times10^{9}N/m^{2}})(1.7\times10^{8}N/m^{2})\approx0.018$
Here the value of Y is taken from table 10.1
