Answer
(a) The mass of block C is 12.9 kg.
(b) The tension in the cord attached to block A is 47.2 N.
The tension in the cord attached to block C is 101 N.
Work Step by Step
(a) Let's consider the system of all three blocks.
$\sum F = (m_A + m_B + m_C)a$
$m_C~g - m_A~g - m_B~g~\mu_k = (m_A + m_B + m_C)a$
$m_C(g-a)= (m_A + m_B)a + m_A~g + m_B~g~\mu_k$
$m_C= \frac{(m_A + m_B)a + m_A~g + m_B~g~\mu_k}{g-a}$
$m_C= \frac{(4.00~kg + 12.0~kg)(2.00~m/s^2) + (4.00~kg)(9.80~m/s^2) + (12.0~kg)(9.80~m/s^2)(0.25)}{9.80~m/s^2-2.00~m/s^2}$
$m_C = 12.9~kg$
The mass of block C is 12.9 kg
(b) We can find the tension $T_A$ in the cord connected with block A.
$T_A - m_A~g = m_A~a$
$T_A = m_A(a+g) = (4.00~kg)(2.00~m/s^2+9.80~m/s^2)$
$T_A = 47.2~N$
We can find the tension $T_C$ in the cord connected with block C.
$m_C~g - T_C = m_C~a$
$T_C = m_C(g-a) = (12.9~kg)(9.80~m/s^2 - 2.00~m/s^2)$
$T_C = 101~N$
