Answer
(a) 0.407 m
(b) 397 N
Work Step by Step
The kinetic frictional force on the bottom block,
$f_{k1}=\mu _{k}(m_{1}+m_{2})g-(1)$
The maximum static frictional force on the top block,
$f_{s2}^{max}=\mu_{s}m_{2}g-(2)$
Let's apply Newton's second law to the bottom block,
$F-f_{k1}-kx=0-(3)$
Let's apply Newton's second law to the top block,
$f_{s2}^{max}-kx=0$
$x=\frac{f_{s2}^{max}}{k}$
(2)=>
$x=\frac{\mu_{s}m_{2}g}{k}$ ; Let's plug known values into this equation.
$x=\frac{(0.9)(15\space kg)(9.8\space m/s^{2})}{325\space N/m}=0.407\space m$
(b) (3)=>
$F=kx+f_{k1}$
(1)=>
$F=kx+\mu _{k}(m_{1}+m_{2})g=(325\space N/m)(0.407\space m)+(0.6)(45\space kg)(9.8\space m/s^{2})=397\space N$
