Answer
$a=-2.2\frac{m}{s^2}$
Work Step by Step
$F_{TA}=\mu_kN_A=(0.3)(m_Ag\cos(51^o))$
$m_Aa=T-m_Ag\sin(51^o)-F_{TA}$
$F_{TB}=\mu_kN_B=(0.3)(m_Bg\cos(21^o))$
$m_Ba=-T+m_Bg\sin(21^o)-F_{TB}$
$m_Aa+m_Ba=m_Bg\sin(21^o)-m_Ag\sin(51^o)-F_{TA}-F_{TB}$
$(7.0kg)a=(5.0kg)(-9.8\frac{m}{s^2})\sin(21^o)-(2.0kg)(-9.8\frac{m}{s^2})\sin(51^o)-(0.3)(-9.8\frac{m}{s^2})\Big((2.0kg)\cos(51^o)+(5.0kg)\cos(21^o)\Big)$
$(7.0kg)a=(-1.54)(-9.8\frac{m}{s^2})$
$a=-2.2\frac{m}{s^2}$
