Problem 36 - University Physics with Modern Physics

Question

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.

Step-by-Step Solution

Answer

The 2 kg box accelerates at \(7.5 \text{ m/s}^2\) and the 3 kg box at \(5 \text{ m/s}^2\).

Step 1: Identify Given Values

We have two boxes on a smooth horizontal floor. The boxes have masses of \(2.0 \text{ kg}\) and \(3.0 \text{ kg}\). The spring between them has a force constant of \(k = 250 \text{ N/m}\) and is initially compressed by \(6.0 \text{ cm} = 0.06 \text{ m}\). We need to find the accelerations of each box right after they are released.

Step 2: Use Hooke's Law to Find Force

According to Hooke's Law, the force exerted by a spring is given by \( F = kx \), where \( k \) is the spring constant and \( x \) is the displacement. Substitute the values: \( F = 250 \text{ N/m} \times 0.06 \text{ m} = 15 \text{ N} \). This force is exerted equally on both boxes in opposite directions.

Step 3: Apply Newton's Second Law to Each Box

For the 2.0 kg box, apply Newton's second law: \( F = ma \). The force acting on it is 15 N, so \( 15 \text{ N} = 2.0 \text{ kg} \times a_1 \). Solving for \( a_1 \), we have \( a_1 = \frac{15 \text{ N}}{2.0 \text{ kg}} = 7.5 \text{ m/s}^2 \).For the 3.0 kg box, similarly, \( F = ma \). The force is the same (15 N but in the opposite direction), so \( 15 \text{ N} = 3.0 \text{ kg} \times a_2 \). Solving for \( a_2 \), we have \( a_2 = \frac{15 \text{ N}}{3.0 \text{ kg}} = 5 \text{ m/s}^2 \).

Step 4: Draw Free-Body Diagrams

For each box, draw a free-body diagram. For the 2 kg box, draw a force arrow pointing to the right labeled \(15 \text{ N}\). For the 3 kg box, draw a force arrow pointing to the left also labeled \(15 \text{ N}\). These represent the forces exerted by the spring immediately after release.

Key Concepts

Hooke's Law

Hooke's Law is a principle of physics that states the force required to compress or extend a spring is proportional to the distance the spring is stretched or compressed.
This relationship is expressed in the formula:

• \( F = kx \)

Here, \( F \) is the force exerted by the spring, \( k \) is the spring's constant (measured in Newtons per meter), and \( x \) is the displacement from the spring's original length.
If you have a spring constant of \( 250 \text{ N/m} \) and compress the spring by \( 0.06 \text{ m} \), the spring exerts a force of \( 15 \text{ N} \).
In the context of our problem, this force is what propels the boxes apart once released.

spring force

The spring force is the push or pull exerted by a spring.
It acts in the opposite direction of the displacement that caused it. In our case, with two boxes compressed by a spring, each box experiences a force in the opposite direction to the compression.
The spring, with a force constant of \( 250 \text{ N/m} \), applies a force of \( 15 \text{ N} \) to each box upon release.

• The spring force on the 2 kg box pushes it to the right.
• The spring force on the 3 kg box pushes it to the left.

Understanding this directional force is crucial for determining the motion of each box.

acceleration calculations

Acceleration is a measure of how quickly an object changes its velocity. According to Newton’s Second Law, acceleration\( (a) \) can be calculated using the formula:

• \( a = \frac{F}{m} \)

where \( F \) is the force applied and \( m \) is the mass of the object.
In our exercise:

• The 2.0 kg box experiences an acceleration of \( 7.5 \text{ m/s}^2 \) due to a 15 N force.
• The 3.0 kg box experiences an acceleration of \( 5 \text{ m/s}^2 \) with the same force, but applied in the opposite direction.

This difference in acceleration occurs because Newton's Second Law considers both force and mass. With smaller mass on the 2 kg box, it accelerates more.

free-body diagrams

Free-body diagrams are essential tools used to represent the forces acting on an object.
They help visualize the magnitude and direction of all forces, aiding in understanding the motion that results.

• For the 2 kg box, the free-body diagram shows a single arrow pointing to the right, labeled 15 N.
• For the 3 kg box, the arrow points to the left with the same 15 N label.

These diagrams are a simplified way to model the physical scenario, focusing on the forces at play at a snapshot in time, right when the spring is released.
Using these diagrams aids in applying Newton's laws effectively to solve problems related to forces and motion.