(a) The system of interest if the acceleration of the child in the wagon is to be calculated are the wagon and the children outside the wagon.
(b) The acceleration of the child-wagon system is 0.33 m/s².
(c) Acceleration of the child-wagon system is zero when the frictional force is 21 N.
Apply Newton's second law of motion;
∑F = ma
where;
∑F = 96 N - 75 N - 12 N = 9 N
Thus, the system of interest if the acceleration of the child in the wagon is to be calculated are;
→ → Ф ←
1st child friction wagon 2nd child
a = ∑F/m
a = 9 N / 27 kg
a = 0.33 m/s²
∑F = 96 N - 75 N - 21 N = 0 N
a = ∑F/m
a = 0/27 kg
a = 0 m/s²
Learn more about net force here: https://brainly.com/question/14361879
#SPJ1