Answer
The maximum amplitude such that the small block will not slip off the large block is $0.16~meters$
Work Step by Step
We can find the maximum possible acceleration that could be provided by the force of static force on the block of mass $m$ as:
$ma = mg~\mu_s$
$a = g~\mu_s$
The maximum force exerted by the spring occurs when the spring is stretched or compressed to the amplitude $A$. Let $a = g~\mu_s$. Therefore;
$F = (M+m)~a$
$kA = (M+m)~g~\mu_s$
$A = \frac{(M+m)~g~\mu_s}{k}$
$A = \frac{(6.0~kg+1.25~kg)(9.80~m/s^2)(0.30)}{130~N/m}$
$A = 0.16~meters$
The maximum amplitude such that the small block will not slip off the large block is $0.16~meters$.
