The system is released from rest with the cable taut. For the friction coefficients ${}_s=0\mathrm{.}05$ and ${}_k=0\mathrm{.}04,$ calculate the

acceleration of each body and the tension $T$ in the cable. Neglect the small mass and friction of the pulleys.

$aofA=$

Direction $=$

$aofB=$

Direction $=$ Downward

$T=mecse{c}^2$

$\frac{m}{se{c}^2}$