Answer
See the detailed answer below.
Work Step by Step
a) Since the two forces of 9.8 N and 7.0 N are action-reaction pairs, so it seems that object 1 is above object 2.
We can see that the weight of the first object is a downward-reaction force on object 2. And the force of 7.0 N is a friction force.
We also can see that there is friction between two objects.
We can assume that the first object is a block of weight 9.8 N (a mass of 1 kg-block) and the other object is a toy truck of weight 19.6 N (a mass of 2 kg-block) that is pulled horizontally by a kid at a constant force of 21.0 N and rolls on the floor without friction. Find the acceleration of the toy truck that makes the block is on the verge of sliding backward (but it does not).
b)
$$\sum F_{x,truck}=F_{boy}-f_s=m_{truck}a_x$$
Thus,
$$a_x=\dfrac{F_{boy}-f_s}{m_{truck}}\tag 1$$
$$\sum F_{x,block}=f_s=m_{block}a_x$$
Plugging into (1);
$$a_x=\dfrac{F_{boy}-m_{block}a_x}{m_{truck}} $$
Thus,
$$a_xm_{truck}+m_{block}a_x= F_{boy} $$
$$a_x(m_{truck}+m_{block})= F_{boy} $$
$$a_x=\dfrac{ F_{boy} }{ m_{truck}+m_{block}}=\dfrac{21}{1+2} $$
$$a_x=\color{red}{\bf 7}\;\rm m/s^2 $$
