Problem 239 For the following exercises, fin... [FREE SOLUTION]

Chapter 2: Problem 239

For the following exercises, find the mass of the twodimensional object that is centered at the origin. A force of \(F=20 x-x^{3} \mathrm{~N}\) stretches a nonlinear spring by \(x\) meters. What work is required to stretch the spring from \(x=0\) to \(x=2 \mathrm{m?}\)

Short Answer

Expert verified

36 Joules

Step by step solution

Understand the Problem

We need to find the work done in stretching a spring from 0 meters to 2 meters with a given force function. The force function is \( F(x) = 20x - x^3 \). The work done when stretching or compressing a spring is the integral of the force over the distance through which it acts.

Set Up the Integral for Work

The work done \( W \) is given by the integral of the force function over the interval from \( x = 0 \) to \( x = 2 \). Thus, the expression for work is: \[ W = \int_{0}^{2} (20x - x^3) \, dx \]

Evaluate the Integral

To find the work, evaluate the integral: \[ \begin{align*} W &= \int_{0}^{2} (20x - x^3) \, dx \ &= \left[ 10x^2 - \frac{x^4}{4} \right]_{0}^{2} \ &= \left( 10(2)^2 - \frac{(2)^4}{4} \right) - \left( 10(0)^2 - \frac{(0)^4}{4} \right) \ &= \left( 40 - 4 \right) = 36 \end{align*} \]

Conclusion

The calculated work done to stretch the spring from 0 to 2 meters is 36 Joules. Hence, it requires 36 Joules of work to stretch the spring over the given distance with the force function provided.

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Key Concepts

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

Work Integral

When calculating the work done by a force over a certain path, integrating the force function over that path is the key process. This is known as a work integral. When dealing with continuous forces, like the stretching of a spring, this approach enables us to calculate the work required precisely.
The formula to find the work done by a force described by a function is:

Identify the force function, in this case, given as a function of position.
Set the limits of integration, which correspond to the initial and final positions.
Evaluate the integral of the force function over these limits.

In our example, you would consider the force applied from the initial position 0 meters to the final position 2 meters. Therefore, the work is expressed as a definite integral: \[W = \int_{0}^{2} (20x - x^3) \, dx\]

Nonlinear Spring

Springs often follow Hooke's Law where force is proportional to displacement. However, a nonlinear spring does not adhere to this linear relationship. The force needed to stretch or compress a nonlinear spring varies in a non-linear manner.
The force function of a nonlinear spring is more complex than linear springs, typically depending on higher powers of displacement. In this exercise, the nonlinear spring has a force function:

\( F(x) = 20x - x^3 \)

This function indicates that as the spring stretches, the force initially increases nearly linearly but changes non-linearly as further displacement occurs.

Force Function

The force function tells us how much force is applied to displace a spring or object along a path. It varies depending on whether the spring is linear or nonlinear.
In our problem, the force applied to the nonlinear spring is given by the function:

\( F(x) = 20x - x^3 \)

This equation highlights that force is not constant but varies with the displacement \( x \). The first term, \( 20x \), suggests an increase in force proportional to displacement, typical of linear springs. However, the \( -x^3 \) term introduces a decrease in force for larger values of \( x \), showcasing nonlinearity.

Definite Integral

A definite integral helps compute the accumulated quantity, in this case, work, over a defined interval along a function鈥檚 domain. For work calculations with variable forces, the definite integral provides the total work done over a specific distance.
The steps involved are:

Set up the integral with the force function and limits of integration.
Evaluate the antiderivative of the function within the defined limits.
Subtract the fundamental theorem of calculus to find the total work.

For our spring problem, the definite integral is evaluated as:\[\left[ 10x^2 - \frac{x^4}{4} \right]_{0}^{2}\]By calculating this, we determine the exact work needed to stretch the spring from 0 to 2 meters and find it to be 36 Joules. This calculation integrates the variable force applied over the displacement path.

91影视

For the following exercises, find the mass of the twodimensional object that is centered at the origin. A force of \(F=20 x-x^{3} \mathrm{~N}\) stretches a nonlinear spring by \(x\) meters. What work is required to stretch the spring from \(x=0\) to \(x=2 \mathrm{m?}\)

Short Answer

Step by step solution

Understand the Problem

Set Up the Integral for Work

Evaluate the Integral

Conclusion

Key Concepts

Work Integral

Nonlinear Spring

Force Function

Definite Integral

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