Answer
(a) $20N$
(b) same
Work Step by Step
(a) We can find the required force as follows:
$F=\frac{mg}{\mu_s}$
We plug in the known values to obtain:
$F=\frac{(1.6)(9.81)}{0.79}$
$F=20N$
(b) If we increase the applied force greater than that found in part(a), even then the force of static friction will remain the same because it balances the weight of the board ($f_s=mg$).
