Answer
(a) $T = 3.92~N$
(b) $a = 2.16~m/s^2$
Work Step by Step
(a) Since the 1.0-kg block does not move, the tension $T$ in the rope attached to the 1.0-kg block must be equal in magnitude to the force of friction exerted on the 1.0-kg block.
$T = F_f$
$T = (1.0~kg)(9.80~m/s^2)(0.40)$
$T = 3.92~N$
(b) Let's consider the system of the 2.0-kg block. We can use a force equation to find the acceleration of the 2.0-kg block.
$\sum F = (2.0~kg)~a$
$F - F_{f1}-F_{f2}= (2.0~kg)~a$
$a = \frac{F - F_{f1}-F_{f2}}{2.0~kg}$
$a = \frac{20~N - (1.0~kg)(9.80~m/s^2)(0.40)-(3.0~kg)(9.80~m/s^2)(0.40)}{2.0~kg}$
$a = 2.16~m/s^2$
