Question

an elephant pushes two heavy boxes, box a and box b, to the right across a rough horizontal surface, as shown in the figure. consider two situations: 1. the boxes are moving with a constant velocity toward the right. 2. the boxes are accelerating toward the right. in each situation, compare the magnitude of the force (f_{ab}) that box a exerts on box b with the magnitude of the force (f_{ba}) that box b exerts on box a. moving with constant velocity accelerating a (f_{ab}=f_{ba}) (f_{ab}=f_{ba}) b (f_{ab}=f_{ba}) (f_{ab}>f_{ba}) c (f_{ab}>f_{ba}) (f_{ab}=f_{ba}) d (f_{ab}>f_{ba}) (f_{ab}>f_{ba})

Explanation:

Step1: Recall Newton's third - law

Newton's third law states that for every action, there is an equal and opposite reaction. When two objects interact, the force exerted by object A on object B ($F_{AB}$) and the force exerted by object B on object A ($F_{BA}$) are a pair of action - reaction forces.

Step2: Apply to both situations

Whether the boxes are moving with a constant velocity (equilibrium situation with net force zero) or accelerating (non - equilibrium situation with a non - zero net force), Newton's third law always holds. So, $F_{AB}=F_{BA}$ in both cases.

Answer:

A. $F_{AB} = F_{BA}$ (Moving with constant velocity), $F_{AB} = F_{BA}$ (Accelerating)