Problem 62 - University Physics with Modern Physics

Question

A 3.00-kg block is connected to two ideal horizontal springs having force constants \(k_1 = 25.0\) N/cm and \(k_2 = 20.0\) N/cm (\(\textbf{Fig. P7.62}\)). The system is initially in equilibrium on a horizontal, frictionless surface. The block is now pushed 15.0 cm to the right and released from rest. (a) What is the maximum speed of the block? Where in the motion does the maximum speed occur? (b) What is the maximum compression of spring 1?

Step-by-Step Solution

Answer

(a) Maximum speed is approximately 5.81 m/s at equilibrium. (b) Maximum compression of spring 1 is 20.1 cm.

Step 1: Convert Units

The force constants are given in N/cm, which need to be converted to N/m.For spring 1: \[k_1 = 25.0 \text{ N/cm} = 25.0 \times 100 \text{ N/m} = 2500 \text{ N/m}\]For spring 2: \[k_2 = 20.0 \text{ N/cm} = 20.0 \times 100 \text{ N/m} = 2000 \text{ N/m}\]Similarly, convert the displacement from cm to meters:\[ x = 15.0 \text{ cm} = 0.15 \text{ m}\]

Step 2: Calculate Total Effective Spring Constant

Since the block is connected to two springs in parallel, the combined spring constant is the sum of the individual spring constants:\[k_{total} = k_1 + k_2 = 2500 \text{ N/m} + 2000 \text{ N/m} = 4500 \text{ N/m}\]

Step 3: Apply Conservation of Energy

At maximum displacement, all the energy is stored in the springs as potential energy. As the block passes through equilibrium, all energy is kinetic.Initial total potential energy:\[ U = \frac{1}{2} k_{total} x^2 = \frac{1}{2} \times 4500 \text{ N/m} \times (0.15 \text{ m})^2\]Calculate to find:\[ U = \frac{1}{2} \times 4500 \times 0.0225 = 50.625 \text{ J}\]At maximum speed, the kinetic energy equals this potential energy. Thus,\[ \frac{1}{2} mv^2 = 50.625\]

Step 4: Solve for Maximum Speed

Rearrange the kinetic energy equation to solve for speed \(v\):\[v = \sqrt{\frac{2E_k}{m}} = \sqrt{\frac{2 \times 50.625}{3.00}}\]Calculate the answer:\[v = \sqrt{\frac{101.25}{3.00}} = \sqrt{33.75} \approx 5.81 \text{ m/s}\]

Step 5: Determine Where Maximum Speed Occurs

The maximum speed occurs when all the potential energy has been converted to kinetic energy. This happens when the block passes through the equilibrium position.

Step 6: Calculate Maximum Compression of Spring 1

At maximum compression of spring 1, all energy is stored again as potential energy in that spring only.If spring 2 contributes no force, then:\[U = \frac{1}{2} k_1 x_1^2 = 50.625 \text{ J} \]Solve for \(x_1\):\[x_1 = \sqrt{\frac{2 \times 50.625}{2500}}\]Calculate the answer:\[x_1 = \sqrt{0.0405} \approx 0.201 \text{ m} = 20.1 \text{ cm}\]

Key Concepts

Spring Constant

The spring constant, often denoted as \( k \), is a fundamental concept in understanding harmonic motion. It essentially measures a spring's stiffness. The larger the spring constant, the stiffer the spring. In our context, two springs are involved, each having different constants \( k_1 = 2500 \text{ N/m} \) and \( k_2 = 2000 \text{ N/m} \). This exercise demonstrates a scenario where both springs are in parallel, meaning their combined effect must be considered when determining the system's behavior.
The total effective spring constant in such a parallel setup is the sum of the individual constants. Thus,

• \( k_{total} = k_1 + k_2 = 4500 \text{ N/m} \)

This combined spring constant provides a comprehensive measure of the system's overall stiffness. Understanding this is crucial for analyzing the system's potential energy and how it'll behave when displaced.

Conservation of Energy

Conservation of energy is a pivotal principle in physics, asserting that energy cannot be created or destroyed, only transformed from one form to another. In the setup with our block and springs, this principle helps us explore how energy transitions between potential and kinetic states.
The system initially shows all energy stored as potential due to the displacement of the springs. As the block is released and moves towards equilibrium, this potential energy (\( U \)) is converted to kinetic energy (\( E_k \))

• Initial potential energy: \( U = \frac{1}{2}k_{total}x^2 \)
• When the block reaches its maximum speed, \( E_k = U \)

This conversion indicates the maximum kinetic energy is equal to the potential energy when the block is at its farthest from the equilibrium position.

Kinetic Energy

Kinetic energy is the energy of motion. When the block moves after being released, it transitions from potential to kinetic energy as it passes through equilibrium.
At this point, all the stored energy in the springs has converted to kinetic energy, which is calculated as:

• \( E_k = \frac{1}{2}mv^2 \)

For the block, kinetic energy becomes maximum at equilibrium because it possesses the highest velocity at that point. Using the derived kinetic energy formula:

• \( v = \sqrt{\frac{2E_k}{m}} \)

Inserting our values reveals that at maximum speed, \( v \approx 5.81 \text{ m/s} \). This reflects the speed transitions from potential energy at maximum displacement.

Potential Energy

Potential energy in springs is stored due to the deformation, either compression or stretching from equilibrium. It's described by the formula \( U = \frac{1}{2}kx^2 \), where \( x \) is the displacement from equilibrium.
This exercise highlights how potential energy is initially stored when the block is pushed 0.15 meters. It converts entirely to kinetic energy as the block rushes back to equilibrium.
During parts of motion where the springs compress or extend, this stored potential energy reaches its peak. Specifically, the maximum compression of spring 1 is found using:

• \( x_1 = \sqrt{\frac{2U}{k_1}} \)

Calculating gives \( x_1 \approx 20.1 \text{ cm} \). This figure represents the point where spring 1 is fully storing the system's mechanical energy—ensuring a full grasp of potential energy's role.