Consider the system shown in Fig.E5.26. Block A weighs $50N$ and block B weighs $25.0N.$ Once block B is set into downward motion. it descends at a constant speed

(b) A cat, also of weight $50N ,$ falls asleep on top of Block A. If Block B is now set into downwards motion, what is its acceleration (magnitude and direction)?

Mass of block A:-

$m_A=\frac{50}{10}=5kg$

mass of block B:

$m_B=\frac{25}{10}=2.5kg$ ($g=10m/se{c}^2$)

(b) Since 

$$\begin{gathered} W_A=50N \\ {W}_B=25N \end{gathered}$$

and $W_C=50N$ ( $W_C=$ weight of cat)

The block B is set into Downward motion with acceleration $\stackrel{\rightharpoonup }{a}.$ So, we get from Newton Seconds law or motion:

$$\begin{gathered} m_{B}g-T=m_B.a\to \left(1\right) \\ \because F_{Net}=m\stackrel{\rightharpoonup }{a} \\ T-\left(M_a+M_c\right)g\times {\mu }_k=\left(M_a+M_c\right).\stackrel{\rightharpoonup }{a}\to \left(2\right) \end{gathered}$$

Hence frictional force:

$F_f={\mu }_k\left(m_A+m_C\right).g$

and $m_C=\frac{W_C}{10}=5kg \left(m_C=mass of cat\right)$

Using equation ($1$) and ($2$)

$$\begin{gathered} m_{B}g-{\mu }_k\left(m_A+m_C\right)g=\left(m_B+m_A+m_C\right).a \\ a=\frac{m_{B}g-{\mu }_k\left(m_A+m_C\right).g}{\left(m_B+m_A+m_C\right)} \left({\mu }_k=0.5\right) \end{gathered}$$

Putting all values:

$a=-2m/se{c}^2$

So, magnitude of acceleration is $2m/se{c}^2$ and direction will be upward($\uparrow$).