91Ó°ÊÓ

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 \(F_{x}=k x-b x^{2}+c x^{3}\) must be applied to the free end. Here \(k=100 \mathrm{~N} / \mathrm{m}, b=700 \mathrm{~N} / \mathrm{m}^{2},\) and \(c=12,000 \mathrm{~N} / \mathrm{m}^{3} .\) 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 \(0.050 \mathrm{~m}\) from its unstretched length? (b) To compress this spring by \(0.050 \mathrm{~m}\) 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.)

Step by step solution

01

Calculate the work done to stretch the spring

The work done is given by the integral of force. The force function is given. Therefore, apply the definite integral with the lower limit as the position of unstretched spring (0) and upper limit as the distance it is stretched or compressed (0.050 m) for stretching. \[ W = \int_0^{0.050} F_x dx = \int_0^{0.050} (kx - bx^2 + cx^3) dx \]Evaluate this integral by applying the power rule of integral, \( \int x^n dx = \frac{x^{n+1}}{n+1} \), to find the work done in stretching the spring

02

Calculate the work done to compress the spring

The work done to compress the spring is also calculated using the integral of Force over the distance of compression. Here, the lower limit of the integral is 0 (initial position), and the upper limit is -0.050 (since the spring is compressed). \[ W = \int_0^{-0.050} F_x dx = \int_0^{-0.050} (kx - bx^2 + cx^3) dx \]Again, evaluate this integral using the power rule of integral to find the work done in compressing the spring

03

Compare the work done in stretching versus compression

In terms of physics, the work done on an object is equal to the force times the distance the object travels times the cosine of the angle between the force and displacement vectors. Since the force and displacement are along the same line (the x-axis) for this problem, the angle between them is either 0 degree (for stretching) or 180 degree (for compression). Thus, their cosine values are 1 and -1 respectively, making the work done in stretching positive and the work done in compressing negative. Compare the absolute values of the work done for stretching and compression, and based on this information, deduce whether it's easier to stretch or compress the spring.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Work Done by Force

When we talk about the work done by a force, we are referring to the energy transferred via the force moving an object. This concept is crucial in understanding how stretching or compressing a spring involves energy interactions.
In the case of non-Hookean springs, like the one in our specific exercise, the force applied does not increase linearly with displacement. Instead, it follows a more complicated relationship:

• Linear Component: Given by \(kx\), this is the typical Hookean linear proportionality.
• Quadratic Component: The term \(-bx^2\) introduces a nonlinear behavior, often linked to the material or design peculiarities of the spring.
• Cubic Component: Finally, \(cx^3\) further complicates the model by adding more curvature to the force-displacement graph, reflecting even more complex real-world spring behaviors.

The work done on the spring, regardless of stretching or compressing, involves integrating these terms over the distance of displacement, indicating the total energy change in the spring system.

Integral Calculation

In physics, calculations involving variable forces usually require an integral approach. For our spring, the work done is computed via the definite integral of the force function over the given displacement bounds.
Here's how it works:

• Integral Expression: The general work expression is \( W = \int F(x) \ dx \), where \( F(x) = kx - bx^2 + cx^3 \).
• Power Rule Application: For terms like these, the integral is evaluated component-wise using \( \int x^n \, dx = \frac{x^{n+1}}{n+1} \), making it easier to calculate each part separately.
• Bound Analysis: Stretching (positive bounds) and compressing (negative bounds) impact signs and absolute size of results, often tying into the physical context.

The process yields total work values that reflect how force distributions impact energy transfer across the spring's movement range.

Force-Displacement Relationship

Understanding the force-displacement relationship in complex springs is essential.
In our exercise, the force \( F_x \) across displacement \( x \) showcases how advanced spring properties interact with force:

• Nonlinear Behavior: Unlike Hookean springs, the presence of \(-bx^2 + cx^3\) implies force does not steadily rise with increased stretch or compress.
• Impacts on Work: Positive force and displacement (stretching) lead to positive work, while force in the opposite direction with displacement (compression) often results in negative work.
• Real-World Significance: Many real springs behave non-linearly, capturing this through empirical equations helps account for these intricacies, vital for applying principles to untidy real-world scenarios.

With this model, predicting force responses across different displacement ranges becomes feasible, presenting both mathematical and conceptual insights into spring physics.