Answer
$5.7\times10^{-2}m$
Work Step by Step
We can write,
The force on the rock $(F_{2})=mg=40\times9.8\space N=392\space N$
by the plunger
Let's apply equation 11.3 $F_{1}=F_{2}(\frac{A_{1}}{A_{2}}) $ to find the force on the piston $(F_{1})$
$F_{1}=392\space N(\frac{15\space cm^{2}}{65\space cm^{2}})=90.46\space N$
We know that, this value is equal to the force which the spring exerts on the piston.
For the spring, we can write,
$F=kx=>x=\frac{F}{k}$ ; Let's plug known values into this equation.
$x=\frac{90.46\space N}{1600\space N/m}=5.7\times10^{-2}m$
