Answer
$4.4\times10^{-6}m$
Work Step by Step
Please find the attached image first.
Here we use equation 10.18 $F=S(\frac{\Delta X}{L_{0}})A$ to find the relative displacement of the upper surface of the block.
$F=S(\frac{\Delta X}{L_{0}})A=\gt \Delta X=\frac{FL_{0}}{SA}$
$ \Delta X=\frac{(770\space N)(0.04\space m)}{(3.5\times10^{10}N/m^{2})(0.01\space m)(0.02\space m)}=4.4\times10^{-6}m$
So, the relative displacement of the upper surface of the block = $4.4\times10^{-6}m$
