Answer
(a) $\mu_k = 0.556$
(b) The acceleration of block B is $a = 2.13~m/s^2$ directed up.
Work Step by Step
(a) Since the system moves at a constant speed, the force of kinetic friction must be equal in magnitude to the weight of block B.
$F_f = (m_B)~g$
$(m_A)~g~\mu_k = (m_B)~g$
$\mu_k = \frac{m_B~g}{m_A~g} = \frac{25.0~N}{45.0~N}$
$\mu_k = 0.556$
(b) Since the total weight of the two blocks and the cat is 115.0 N, the total mass $M$ of the two blocks and the cat is $\frac{115.0~N}{g}$. We can use a force equation to find the acceleration of the system.
$\sum F = Ma$
$F_f - (m_B)~g = Ma$
$F_N~\mu_k - (m_B)~g = Ma$
$a = \frac{F_N~\mu_k - (m_B)~g}{M}$
$a = \frac{(90.0~N)(0.556)-(25.0~N)}{(115.0~N/g)}$
$a = 2.13~m/s^2$
If the system is in motion, the acceleration of block B is $a = 2.13~m/s^2$ directed up. This means that the system will slow down and come to rest.
