A 3.0-kg and a 5.0-kg box rest side-by-side on a perfectly smooth,level floor.A horizontal force of 32 N is exerted on the 3.0-kg box pushing it against the 5.0-kg box,and,as a result,both boxes slide along the floor.How hard do the two boxes push against each other?
A)12 N
B)20 N
C)24 N
D)32 N
E)0 N

Question

Quizplus · Accepted Answer

The force exerted on the 3.0-kg box is distributed across both boxes, resulting in a combined mass of 8.0 kg. The acceleration of the system can be calculated using Newton's second law, $F = ma$, where $F = 32 \, \text{N}$ and $m = 8.0 \, \text{kg}$. The acceleration, $a$, is the same for both boxes since they move together. The force that the 3.0-kg box exerts on the 5.0-kg box (or vice versa) can be found by applying $F = ma$ to either box individually. Using the 3.0-kg box, for instance, we find that the force it exerts is equal to its mass times the acceleration of the system. Since the acceleration is $a = F/m = 32 \, \text{N} / 8.0 \, \text{kg} = 4.0 \, \text{m/s}^2$, the force the 3.0-kg box exerts on the 5.0-kg box is $3.0 \, \text{kg} \times 4.0 \, \text{m/s}^2 = 12 \, \text{N}$. However, this explanation overlooks the correct approach to solving the problem, which involves understanding that the force between the boxes is actually determined by the acceleration of the system and the effective force acting on the combined mass. The correct force of interaction is found by considering the system's acceleration and how it applies to the internal force between the two masses, leading to the realization that the initial setup of the explanation might not directly lead to the correct answer without considering the dynamics of the system as a whole.

A 30-Kg and a 5