Answer
(a) F = 1.44 N
(b) F = 1.80 N
Work Step by Step
(a) We can consider the system of both block A and block B. Since the system is moving at a constant speed, the force $F$ must be equal in magnitude to the force of kinetic friction exerted on block B by the ground.
$F = F_N~\mu_k = (4.80~N)(0.300)$
$F = 1.44~N$
(b) We can consider the system of block B. Since the system is moving at a constant speed, the force $F$ must be equal in magnitude to the total force of kinetic friction exerted on block B by the ground and by block A.
$F = F_N~\mu_k + (m_A~g)~\mu_k$
$F = (4.80~N)(0.300) + (1.20~N)(0.300)$
$F = 1.80~N$
