Answer
(a) $0.048m$
(b) Yes
Work Step by Step
(a) We can find the required compression as follows:
$\Sigma F_x=F-N=0$
$\implies Kx-\frac{mg}{\mu_s}=0$
This can be rearranged as:
$x=\frac{mg}{\mu_s K}$
We plug in the known values to obtain:
$x=\frac{(0.27)(9.81)}{(0.46)(120)}$
$x=0.048m$
(b) We know that $x=\frac{mg}{\mu_s K}$. This equation shows us that the spring displacement is directly proportional to the mass of the block. Thus, if we double the mass, the spring displacement will also be doubled.
