Answer
(a) $\mu_s = 0.41$
(b) $F = 40~N$
Work Step by Step
(a) The applied force is equal in magnitude to the force of static friction. We can find the coefficient of static friction:
$mg~\mu_s = 12.0~N$
$\mu_s = \frac{12.0~N}{mg}$
$\mu_s = \frac{12.0~N}{(3.0~kg)(9.80~m/s^2)}$
$\mu_s = 0.41$
(b) We can assume that the required force $F$ is equal in magnitude to the force of static friction. We can find $F$:
$F = F_N~\mu_s$
$F = (3.0~kg+7.0~kg)~g~\mu_s$
$F = (3.0~kg+7.0~kg)(9.80~m/s^2)(0.41)$
$F = 40~N$
