Problem 67 - University Physics with Modern Physics

Question

A certain spring is found not to obey Hooke's law; it exerts a restoring force \(F_{x}(x)=-\alpha x-\beta x^{2}\) if it is stretched or compressed, where \(\alpha=60.0 \mathrm{N} / \mathrm{m}\) and \(\beta=18.0 \mathrm{N} / \mathrm{m}^{2} .\) The mass of the spring is negligible. (a) Calculate the potentinl-energy function \(U(x)\) for this spring. Let \(U=0\) when \(x=0\) (b) An object with mass 0.900 \(\mathrm{kg}\) on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00 \(\mathrm{m}\) to the right (the \(+x\) -direction) to stretch the spring, and released. What is the speed of the object when it is 0.50 \(\mathrm{m}\) to the right of the \(x=0\) equilibrium position?

Step-by-Step Solution

Answer

(a) \( U(x) = 30.0 x^2 + 6.0 x^3 \). (b) The speed at 0.50 m is approximately 7.85 m/s.

Step 1: Understand the Restoring Force Function

The restoring force for this spring is given by the equation \( F_x(x) = -\alpha x - \beta x^2 \), where \( \alpha = 60.0 \ \text{N/m} \) and \( \beta = 18.0 \ \text{N/m}^2 \). To begin, note that the force is not a simple Hookean force because it includes a non-linear term \( -\beta x^2 \).

Step 2: Potential Energy Calculation

The potential energy \( U(x) \) can be determined by integrating the force function. Since force is the negative derivative of potential energy, we find:\[ U(x) = -\int F_x(x) \, dx = \int (\alpha x + \beta x^2) \, dx \]

Step 3: Integrate to Find the Potential Energy

Perform the integration:\[ U(x) = \int (\alpha x + \beta x^2) \, dx = \frac{1}{2} \alpha x^2 + \frac{1}{3} \beta x^3 + C \]We set \( U(0) = 0 \) for the potential energy function, implying the constant \( C = 0 \). Thus,\[ U(x) = \frac{1}{2} \alpha x^2 + \frac{1}{3} \beta x^3 \].

Step 4: Potential Energy Function at a Specific Position

Substitute \( \alpha \) and \( \beta \) into the potential energy function to check values:\[ U(x) = \frac{1}{2} (60.0) x^2 + \frac{1}{3} (18.0) x^3 \]Simplifying yields the specific form:\[ U(x) = 30.0 x^2 + 6.0 x^3 \].

Step 5: Analyze the Energy Conservation (Part b)

To find the speed at \( x = 0.50 \ \text{m} \), we use energy conservation:The total mechanical energy (E) at any position is the sum of kinetic (K) and potential energy (U) and remains constant:\[ E = K + U(x) \].Initially, when the object is pulled to \( x = 1.00 \ \text{m} \) and released, its kinetic energy is zero, and all the energy is potential:\[ E = U(1.0) \].

Step 6: Initial Potential Energy

Calculate the initial potential energy when \( x = 1.0 \ \text{m} \):\[ U(1.0) = 30.0(1.0)^2 + 6.0(1.0)^3 = 36.0 \ \text{J} \].Hence, \( E = 36.0 \ \text{J} \).

Step 7: Total Energy at x = 0.50 m

When the object moves to \( x = 0.50 \ \text{m} \):\[ U(0.50) = 30.0(0.50)^2 + 6.0(0.50)^3 = 7.5 + 0.75 = 8.25 \ \text{J} \].

Step 8: Determine Kinetic Energy at x = 0.50 m

Use the conservation of energy:\[ K + U(0.50) = E \]\[ K = E - U(0.50) = 36.0 - 8.25 = 27.75 \ \text{J} \].

Step 9: Solve for Speed

The kinetic energy is given by \( K = \frac{1}{2}mv^2 \). Thus,\[ 27.75 = \frac{1}{2}(0.900)v^2 \]Solve for \( v \):\[ v^2 = \frac{2 \times 27.75}{0.900} \approx 61.667 \]\[ v \approx \sqrt{61.667} \approx 7.85 \ \text{m/s} \].

Key Concepts

Potential Energy

Potential energy is a fascinating concept in physics that represents the stored energy of an object based on its position or configuration. For springs, this energy can change depending on how much the spring is stretched or compressed. In this case, the non-Hookean spring does not follow the regular linear rule of Hooke's law.
An exciting aspect of potential energy in our case deals with the specific function for this spring, which was determined through the process of integration. As the force exerted by the spring is non-linear, the potential energy function integrates the force equation given by \( F(x) = -\alpha x - \beta x^2 \). Through integration, the derived potential energy function is \( U(x) = \frac{1}{2} \alpha x^2 + \frac{1}{3} \beta x^3 \). The potential energy is zero at the equilibrium position \( x = 0 \).
This form of potential energy highlights how an object's stored energy can vary with both linear and quadratic terms, which reflects the unique behavior of a non-Hookean spring.

Conservation of Energy

The conservation of energy principle is a cornerstone of physics, as it asserts that the total energy in an isolated system remains constant. In the case of the spring problem at hand, both potential energy and kinetic energy play key roles. When the spring is stretched to \( x = 1.00 \, \text{m} \), the system holds potential energy, calculated to be 36.0 Joules. As we release the object, it begins to move, converting potential energy into kinetic energy.
At any point in time, the sum of potential and kinetic energy equals the total energy initially in the system, ensuring energy is conserved. As the object moves to \( x = 0.50 \, \text{m} \), its potential energy decreases to 8.25 Joules, meaning the remaining energy becomes kinetic. This energy transfer and balance confirm that the total energy remains at 36.0 Joules, highlighting the importance and application of the conservation of energy.

Kinetic Energy

Kinetic energy refers to the energy an object possesses due to its motion. In our spring problem, after the initial release from the stretched position, potential energy is converted into kinetic energy as the object accelerates. Kinetic energy is expressed as \( K = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity.
From the conservation of energy, we determined the kinetic energy at \( x = 0.50 \, \text{m} \) to be approximately 27.75 Joules. Using the kinetic energy formula, we can calculate the object's velocity. Substituting the kinetic energy value into the equation allows us to find the speed of the object, which is critical for understanding motion dynamics.
This conversion between potential and kinetic energy in non-Hookean springs showcases the complex interactions in systems where forces are non-linear.

Integration in Physics

Integration in physics is a powerful mathematical tool used to calculate quantities like potential energy when dealing with varying forces. In this problem, the force exerted by the spring involves both linear \( -\alpha x \) and nonlinear \( -\beta x^2 \) terms. To find the potential energy from the force, we perform integration on the force function \( F(x) \).
The integration process results in \( U(x) = \frac{1}{2} \alpha x^2 + \frac{1}{3} \beta x^3 \), capturing how potential energy is stored as a function of position. This explicit function is crucial for assessing the behavior of the spring as it stretches and compresses.
Understanding integration in this context helps elucidate how energy varies with position, especially in systems where forces are not described by simple, linear equations.