Question

Consider a spring that does not obey Hooke's law very faithfully. One end of the spring is fixed. To keep the spring stretched or compressed an amount \(x\), a force along the \(x\)-axis with \(x\)-component \(F_x = kx - bx^2 + cx^3\) must be applied to the free end. Here \(k = 100 \, \mathrm {N/m}\), \(b = 700 \, \mathrm {N/m{^2}}\), and \(c = 12,000 \, \mathrm{N/m}^3\). Note that \(x > 0\) when the spring is stretched and \(x< 0\) when it is compressed. How much work must be done (a) to stretch this spring by 0.050 m from its unstretched length? (b) To \(compress\) this spring by 0.050 m from its unstretched length? (c) Is it easier to stretch or compress this spring? Explain why in terms of the dependence of \(F_x\) on \(x\). (Many real springs behave qualitatively in the same way.)

Short answer

(a) 0.124 J to stretch; (b) 0.13 J to compress; (c) Easier to stretch due to the nonlinear term.

Step-by-step solution

Understand Work done by Variable Force

The work done, \(W\), by a variable force along a straight line from position \(a\) to \(b\) is given by the integral \[ W = \int_{a}^{b} F_x \, dx \]. Since \(F_x = kx - bx^2 + cx^3\), we need to evaluate this integral to find the work done when changing the spring's length.

Calculate Work to Stretch the Spring

To find the work done stretching the spring by 0.05 m, calculate the definite integral from 0 to 0.05 of the force function. That means solving\[ W_{stretch} = \int_{0}^{0.05} (100x - 700x^2 + 12000x^3) \, dx \].Computing this integral:1. \(\int 100x = 50x^2\)2. \(\int 700x^2 = \frac{700}{3}x^3\)3. \(\int 12000x^3 = 3000x^4\)Now evaluate: \[ W_{stretch} = \left[ 50x^2 - \frac{700}{3}x^3 + 3000x^4 \right]_{0}^{0.05} \]Evaluate at \(x = 0.05\):\[ = 50(0.05)^2 - \frac{700}{3}(0.05)^3 + 3000(0.05)^4 \]\[ = 0.125 - 0.00291667 + 0.001875 = 0.1239583 \, \mathrm{J}\]So, approximately \(0.124 \, \mathrm{J}\) of work is done.

Calculate Work to Compress the Spring

Similarly, to find the work done compressing the spring by 0.05 m, calculate the definite integral from 0 to -0.05.\[ W_{compress} = \int_{0}^{-0.05} (100x - 700x^2 + 12000x^3) \, dx \]Computing this integral as in Step 2:1. \(\int 100x = 50x^2\)2. \(\int 700x^2 = \frac{700}{3}x^3\)3. \(\int 12000x^3 = 3000x^4\)Now evaluate:\[ W_{compress} = \left[ 50x^2 - \frac{700}{3}x^3 + 3000x^4 \right]_{0}^{-0.05} \]Evaluate at \(x = -0.05\):\[ = 50(-0.05)^2 - \frac{700}{3}(-0.05)^3 + 3000(-0.05)^4 \]\[ = 0.125 + 0.00291667 + 0.001875 = 0.12979167 \, \mathrm{J}\]Thus, approximately \(0.13 \, \mathrm{J}\) of work is needed.

Compare Work for Stretching and Compressing

The work to stretch the spring (0.124 J) is less than the work to compress it (0.13 J). This is because the term \(bx^2\) contributes positively to the force when the spring is compressed, but negatively when stretched, due to the square term behaving differently under negative or positive \(x\). Therefore, it is slightly easier to stretch than to compress this spring.

Key concepts

Variable Force

When dealing with springs, you might be familiar with Hooke's Law, where force is directly proportional to displacement, represented as \( F_x = kx \). However, some springs, known as non-Hookean springs, experience forces that vary more complexly with displacement. In this scenario, the force required to stretch or compress the spring is described by a variable force equation: \( F_x = kx - bx^2 + cx^3 \).
This means the force isn't simply increasing directly with \( x \), but has quadratic \( (-bx^2) \) and cubic \( (+cx^3) \) components. These additional terms mean the force required changes at different rates when the spring is stretched or compressed, making calculations more intricate and realistic for certain springs.
Understanding variable forces is key for accurately modeling and solving problems where simple linear approximations aren't sufficient.

Work Done

The concept of work done by a force in physics refers to the energy transferred when an object is moved over a distance by an external force. For a constant force, work is the product of the force along the direction of movement and the distance traveled. However, when dealing with variable forces, as is the case with our non-Hookean spring, the calculation becomes more complex.
To find the work done by a variable force, integration is needed. In math terms, the work done when the force isn't constant is given by the integral of force over the displacement traveled. It is represented by the equation \[ W = \int_{a}^{b} F_x \, dx \]. Here, \( F_x \) is the variable force function, and the limits \( a \) and \( b \) correspond to the initial and final positions. Thus, work done by a non-Hookean spring as it stretches or compresses is defined by evaluating this integral between the specified limits.

Definite Integral

A definite integral is a crucial tool when calculating physical quantities that depend on a continuously changing rate, such as work done by a variable force. It allows us to sum the infinite small contributions of force over a specified range of positions—effectively the areas under the force vs. distance curve.
For our specific problem, we applied a definite integral to find how much work is required to displace the spring. The integral we used looks like this: \[ \int_{0}^{d} (100x - 700x^2 + 12000x^3) \, dx \] where \( d \) is the change in displacement (either stretching or compressing).
Completing this definite integral involves calculating antiderivatives and evaluating them at the boundary limits \( d \) and \( 0 \). This process gives us a precise measure of the total work, including contributions from both stretching and compressing forces.

Physics Problem Solving

Physics problem solving involves systematically approaching a problem with a mix of theoretical understanding and mathematical skills to arrive at a solution. When solving problems like those involving non-Hookean springs, one needs to follow a structured path:

• Start by understanding the physical problem and the applicable principles—variable force, work done, energy transfer, etc.
• Identify the given information and what the problem is asking for, helping translate physics into mathematics through equations.
• Apply appropriate mathematical techniques, such as integration, to resolve the physics setup into solvable equations that yield the desired results, like calculating work.
• Review your results logically. For our spring example, after solving, we confirm whether it makes sense that less work is done stretching compared to compressing due to the force equation’s nature.

Problem-solving in physics requires both theoretical knowledge and practical approaches, enhanced by mathematical skills.