91Ó°ÊÓ

CALC 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. (a) How much work must be done to stretch this spring by 0.050 m from its unstretched length? (b) How much work must be done 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.)

Step by step solution

01

Analyze the Work Formula

The work done in stretching or compressing the spring can be found using the formula for work done by a variable force, given as \[ W = \int_{x_0}^{x_f} F_x \, dx \]where \( F_x = kx - bx^2 + cx^3 \), and \( x_0 \) and \( x_f \) are the initial and final positions respectively.

02

Define Initial and Final Positions

For part (a), to stretch the spring by 0.050 m, the initial position is \( x_0 = 0 \) and the final position is \( x_f = 0.050 \, \text{m} \). For part (b), to compress the spring by 0.050 m, the initial position is \( x_0 = 0 \) and the final position is \( x_f = -0.050 \, \text{m} \).

03

Integrate for Stretching

For stretching, substitute into the work formula:\[ W_{ ext{stretch}} = \int_{0}^{0.050} (100x - 700x^2 + 12000x^3) \, dx \]Calculate the definite integral to find:\[ W_{ ext{stretch}} = \left[ 50x^2 - \frac{700}{3}x^3 + 3000x^4 \right]_0^{0.050} = 50(0.050)^2 - \frac{700}{3}(0.050)^3 + 3000(0.050)^4 \]

04

Calculate Result for Stretching

Substitute the values into the expression:\[ W_{ ext{stretch}} = 50(0.0025) - \frac{700}{3}(0.000125) + 3000(0.00000625) \]Calculate each term:- First term: \( 50 \times 0.0025 = 0.125 \)- Second term: \( \frac{700}{3} \times 0.000125 \approx 0.029167 \)- Third term: \( 3000 \times 0.00000625 = 0.01875 \)Finally calculate the work:\[ W_{ ext{stretch}} = 0.125 - 0.029167 + 0.01875 = 0.114583 \, \text{J} \]

05

Integrate for Compression

Similarly, for compressing, use:\[ W_{ ext{compress}} = \int_{0}^{-0.050} (100x - 700x^2 + 12000x^3) \, dx \]Notice that because of symmetry and the signs, the calculated terms will be opposite.

06

Calculate Result for Compression

Substituting the limits for compression:\[ W_{ ext{compress}} = \left[ 50x^2 - \frac{700}{3}x^3 + 3000x^4 \right]_0^{-0.050} = -\left(50(0.050)^2 - \frac{700}{3}(0.050)^3 + 3000(0.050)^4 \right) = -0.114583 \, \text{J} \]

07

Compare Stretching vs Compression

The magnitudes of the work done in stretching and compressing are equal (\( 0.114583 \) J), but in opposite directions because of the force required. Since we need negative applied work to compress further than the natural length compared to stretching, it indicates that it is qualitatively easier to compress than stretch, especially as these terms grow at different rates for large x.

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Key Concepts

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

Variable Force in Nonlinear Springs

In this exercise, we're dealing with a nonlinear spring. Unlike linear springs that obey Hooke's Law (\( F = kx \)), nonlinear springs have force expressions that depend on higher powers of displacement. Here, the spring force is given by \( F_x = kx - bx^2 + cx^3 \), where the parameters represent nonlinearity:

• \( k \) is the linear stiffness.
• \( b \) and \( c \) introduce quadratic and cubic terms, making the force variable as the spring is compressed or stretched.

To translate this into how a spring behaves, imagine that as you compress or stretch it, the exerted force isn't constant. Instead, it varies non-linearly with these additional terms. The influence of these variable forces is crucial for understanding how much work is needed to move the spring from one length to another.

Understanding the Work-Energy Principle

In terms of physics, doing work on an object is about transferring energy to it. For springs, the work-energy principle helps us calculate how much energy it requires to change its length. The amount of work done is related to the area under the force versus displacement curve, represented mathematically by an integral.
To determine the work done (\( W \)) while stretching or compressing this nonlinear spring, we use the formula:\[ W = \int_{x_0}^{x_f} F_x \, dx \]Here:

• \( x_0 \) is the initial position (unstretched or uncompressed).
• \( x_f \) is the final position (after stretching or compressing).
• \( F_x \) is the variable force as derived from the given expression.

When the spring is stretched or compressed, work is positive if it supplements the system's energy, and negative if it takes energy away. This concept shows us why more energy is sometimes needed to move in one direction due to the non-linearity.

Evaluating Work with Definite Integrals

The calculation of work in this nonlinear spring problem is performed using definite integrals. Integrals sum up tiny strips of area under the curve, giving us the total work done between two points. Definite integrals are handy because they precisely calculate the work over specific limits of stretch or compression.
For a nonlinear spring, evaluate the definite integral:\[ W = \int_{x_0}^{x_f} (kx - bx^2 + cx^3) \, dx \]This technique involves:

• Substituting the force equation into the integral.
• Calculating the definite integral to find exact work values (e.g., 0.114583 J for both stretching and compressing this spring by 0.050 m).

Notice that when working with definite integrals, the direction (sign) of the work is important, as it reflects whether work is being done on or by the system. This explains why energy requirements differ between stretching and compressing nonlinear springs.