Answer
(a) $v = 0.218~m/s$
(b) $T = 11.7~N$
Work Step by Step
(a) Let's consider the system of both blocks. We can find the acceleration of the system. The total mass $M$ of both blocks is 3.55 kg.
$\sum F = Ma$
$m_B~g - F_f = Ma$
$a = \frac{m_B~g - m_A~g~\mu_k}{M}$
$a = \frac{(1.30~kg)(9.80~m/s^2) - (2.25~kg)(9.80~m/s^2)(0.450)}{3.55~kg}$
$a = 0.794~m/s^2$
We can use the acceleration to find the speed $v$.
$v^2 = v_0^2+2ad = 0+2ad$
$v = \sqrt{2ad} = \sqrt{(2)(0.794~m/s^2)(0.0300~m)}$
$v = 0.218~m/s$
(b) We can use block B to find the tension $T$ in the cord.
$\sum F = ma$
$m_B~g - T = m_B ~a$
$T = (m_B)(g-a) = (1.30~kg)(9.80~m/s^2 - 0.794~m/s^2)$
$T = 11.7~N$
