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Chapter 6: Problem 6

For a certain type of nonlinear spring, the force required to keep the spring stretched a distance \(s\) is given by the formula \(F=k s^{4 / 3}\). If the force required to keep it stretched 8 inches is 2 pounds, how much work is done in stretching this spring 27 inches?

Short Answer

Expert verified
The work done to stretch the spring 27 inches is approximately 117.16 inch-pounds.

Step by step solution

01

Understand the formula for force

The force required to stretch the spring is given by the formula \(F = k s^{4/3}\), where \(F\) is the force, \(k\) is a constant, and \(s\) is the distance stretched.
02

Determine the constant \(k\)

We know \(F = 2\) pounds when \(s = 8\) inches. Substitute these values into the force equation: \[ 2 = k \cdot 8^{4/3} \]Solve for \(k\):\[ k = \frac{2}{8^{4/3}} \] Calculate \(8^{4/3}\), which is \( (2^3)^{4/3} = 2^4 = 16 \), hence:\[ k = \frac{2}{16} = \frac{1}{8} \].
03

Write the force function with \(k\)

Now that we have \(k = \frac{1}{8}\), substitute \(k\) back into the force equation:\[ F(s) = \frac{1}{8} s^{4/3} \].
04

Set up the integral for work done

Work done by a force is the integral of the force function with respect to distance. We need to calculate work done to stretch the spring from 0 to 27 inches:\[ W = \int_{0}^{27} \frac{1}{8} s^{4/3} \, ds \].
05

Compute the integral

To find the integral, compute:\[ W = \frac{1}{8} \int_{0}^{27} s^{4/3} \, ds \].The antiderivative of \(s^{4/3}\) is \(\frac{s^{7/3}}{7/3} = \frac{3}{7} s^{7/3}\). Therefore,\[ W = \frac{1}{8} \left[ \frac{3}{7} s^{7/3} \right]_{0}^{27} \].Evaluate this from 0 to 27:\[ W = \frac{1}{8} \times \frac{3}{7} \left[ 27^{7/3} - 0^{7/3} \right] \].
06

Evaluate \( 27^{7/3} \)

Compute \( 27^{7/3} \):\( 27^{7/3} = (3^3)^{7/3} = 3^7 \). Thus, \( 3^7 = 2187 \).
07

Calculate the work done

Substitute back \( 2187 \) for \(27^{7/3}\) in the equation for work:\[ W = \frac{1}{8} \times \frac{3}{7} \times 2187 \].Simplify:\[ W = \frac{6561}{56} = 117.1607 \] approximately.The work done is approximately 117.16 inch-pounds.

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Key Concepts

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

Nonlinear Springs
Nonlinear springs differ from ordinary linear springs by having a force that does not increase proportionally with the distance they are stretched. Instead, the force they exert follows a nonlinear pattern. In the given exercise, the force is expressed by the formula:
  • \( F = k s^{4/3} \)
  • Where \( F \) is the force, \( s \) is the distance the spring is stretched, and \( k \) is a constant unique to that particular spring.
Unlike linear springs, where Hooke's Law \( (F = kx) \) states a direct proportionality to stretch distance, nonlinear springs increase in complexity. This complexity requires integration methods to determine work done over a variable stretch.For a nonlinear spring, knowing the specific form of the equation is vital for solving problems and understanding how much work is needed for a given stretch.
Work and Energy in Physics
In physics, work refers to the energy transferred when an object is moved over a distance by an external force. The formula to calculate work is the integral of force over the distance it acts upon:
  • \( W = \int F(s) \, ds \)
  • Where \( W \) is the work done, \( F(s) \) is the force as a function of distance \( s \), and \( s \) itself can vary over the path defined by the integral limits.
For this exercise, the moving of the spring from 0 to 27 inches required integrating the force function from the unstressed position to the stretched position. The calculated value of work represents the total energy transferred to the spring to stretch it the specified distance. This approach emphasizes how integration techniques can be used to account for changing forces.
Integration Techniques
Integration is a fundamental calculus tool used to find the work done by nonlinear springs, as seen with the function given \( F(s) = \frac{1}{8} s^{4/3} \). Calculating work requires performing definite integrals:
  • The integral \( \int_{0}^{27} s^{4/3} \, ds \) calculates the net work done for stretching from 0 to 27 inches.
  • The antiderivative or the result of integrating \( s^{4/3} \) is \( \frac{3}{7} s^{7/3} \), emphasizing the need to have a strong grasp of integral calculus.
When solving such problems, it is important to accurately compute integral results including constants and bounds. Always remember to evaluate the function at the upper and lower limits of the integral to find the net work done correctly.In calculus, understanding these integral solutions enhances problem-solving capabilities, especially in physics-related applications involving energy and forces.

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Most popular questions from this chapter

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