Answer
$$
T=9.4 \mathrm{N}
$$
Work Step by Step
For the $m_{2}=1.0$ kg block, application of Newton's laws result in
$$
\begin{array}{rl}{F \cos \theta-T-f_{k}=m_{2} a} & {x \text { axis }} \\ {F_{N}-F \sin \theta-m_{2} g=0} & {y \text { axis }}\end{array}
$$
since $f_{k}=\mu_{k} F_{N},$ these equations can be combined into an equation to solve for $a :$
$$
F\left(\cos \theta-\mu_{k} \sin \theta\right)-T-\mu_{k} m_{2} g=m_{2} a
$$
Similarly (but without the applied push) we analyze the $m_{1}=2.0$ kg block:
$$\begin{array}{rlrl}{T-f_{k}^{\prime}} & {=m_{1} a} & {} & {x \text { axis }} \\ {F_{N}^{\prime}-m_{1} g} & {=0} & {} & {y \text { axis }}\end{array}$$
Using $f_{k}=\mu_{k} F_{N}^{\prime},$ the equations can be combined:
$$
T-\mu_{k} m_{1} g=m_{1} a
$$
Subtracting the two equations for $a$ and solving for the tension, we obtain
$$
T=\frac{m_{1}\left(\cos \theta-\mu_{k} \sin \theta\right)}{m_{1}+m_{2}} F=\frac{(2.0 \mathrm{kg})\left[\cos 35^{\circ}-(0.20) \sin 35^{\circ}\right]}{2.0 \mathrm{kg}+1.0 \mathrm{kg}}(20 \mathrm{N})$$$$=9.4 \mathrm{N}
$$
