A 2.0 -kg box and a 3.0 -kg box on a perfectly smooth...

A 2.0 -kg box and a 3.0 -kg box on a perfectly smooth horizontal floor have a spring of force constant 250 $\mathrm{N} / \mathrm{m}$ compressed between them. If the initial compression of the spring is $6.0 \mathrm{cm},$ find the acceleration of each box the instant after they are released. Be sure to include free-body diagrams of each box as part of your solution.

Key Concepts

Newton's Second Law

Newton's Second Law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. This principle is used to determine the acceleration of each box by dividing the force applied to each box (derived from the spring) by its respective mass.

Hooke's Law

Hooke's Law describes the behavior of springs, stating that the force exerted by a spring is proportional to its displacement from the equilibrium position (F = k?x). In this problem, the compressed spring provides an instantaneous force on each box, and this relationship is used to calculate that force.

Free-Body Diagrams

Free-body diagrams are visual representations that isolate an object and depict all external forces acting upon it. These diagrams are crucial in analyzing the forces on each box, including the spring force and any other forces, to correctly apply Newton's Second Law and determine the resulting acceleration.

Action-Reaction Principle

The Action-Reaction Principle, as stated in Newton's Third Law, indicates that forces between two interacting bodies are equal in magnitude and opposite in direction. This concept explains how the spring exerts equal and opposite forces on the two boxes, leading to differing accelerations based on their masses.

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