/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 A bungee cord exerts a nonlinear... [FREE SOLUTION] | 91Ó°ÊÓ

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A bungee cord exerts a nonlinear elastic force of magnitude \(F(x)=k_{1} x+k_{2} x^{3}, \quad\) where \(x\) is the distance the cord is stretched, \(k_{1}=204 \mathrm{N} / \mathrm{m} \quad\) and \(k_{2}=-0.233 \mathrm{N} / \mathrm{m}^{3} .\) How much work must be done on the cord to stretch it \(16.7 \mathrm{m} ?\)

Short Answer

Expert verified
The work done to stretch the bungee cord 16.7 meters is approximately 13996.7 Joules.

Step by step solution

01

Write down the formula for work

The work done on the bungee cord to stretch it can be calculated by integrating the force with respect to the distance. The formula for work, W, is: $$ W = \int_{x_0}^{x_f} F(x) dx $$ where \(F(x) = k_1x + k_2x^3\) is the force exerted by the bungee cord, \(x_0\) is the initial position, \(x_f\) is the final position, and \(x\) is the distance the cord is stretched.
02

Set up the integration

We will integrate the force function F(x) from the initial position \(x_0 = 0\) meters to the final position \(x_f = 16.7\) meters. The integral will look like this: $$ W = \int_{0}^{16.7} (k_1x + k_2x^3) dx $$ We are given the values of \(k_1 = 204 \mathrm{N/m}\) and \(k_2 = -0.233 \mathrm{N/m^3}\). Plug in these values into the integral: $$ W = \int_{0}^{16.7} (204x - 0.233x^3) dx $$
03

Integrate the force function

Now, we need to integrate the function with respect to x. Using the power rule for integration, we get: $$ W = [102x^2 - 0.05825x^4]_{0}^{16.7} $$
04

Evaluate the integral

To find the work done, evaluate the integral at the limits \(x = 16.7\) and \(x = 0\). Subtract the result at \(x = 0\) from the result at \(x = 16.7\): $$ W = (102(16.7)^2 - 0.05825(16.7)^4) - (102(0)^2 - 0.05825(0)^4) $$ Calculate the values: $$ W = (102(278.89) - 0.05825(77841.4891)) - (0) $$ $$ W \approx 13996.7 J $$ So, it takes approximately 13996.7 Joules of work to stretch the bungee cord 16.7 meters.

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Key Concepts

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

Work-Energy Principle
The work-energy principle is a fundamental concept in physics that relates the work done by forces acting on an object to the change in its kinetic energy. In the context of a bungee cord, when we talk about doing work on the cord, we're actually transferring energy to it.
When the cord is stretched, the energy is stored as potential energy within the elastic material. This potential energy is available to do work on any object attached to the cord, for instance dragging it upwards during a bungee jump. This stored energy can be converted back into kinetic energy, as observed when the cord recoils and accelerates the object back to its original position.
In our exercise, calculating the work done to stretch the cord encompasses determining how much energy is required to displace the cord from its relaxed position to a specified stretched position. This calculation assumes that all the work done is stored as elastic potential energy. It's this principle that enables bungee jumpers to spring back up after a leap.
Integration in Physics
Integration is a mathematical tool that is widely used in physics, especially when dealing with variables that change continuously. It helps in finding quantities like area, volume, and, as in our exercise, the work done when a force is applied over a distance.
In the case of the bungee cord, the force exerted by the cord isn't constant; it varies with the stretch length. Calculating the total work requires summing up the tiny amounts of work done over each infinitesimal stretch of the cord. This is what integration allows us to do.
By integrating the force function from the starting position to the end position, you essentially sum up all the little pieces of work done across the entire length of the stretch. The result of this process is a precise value for the total work done on the cord. Without integration, it would be incredibly difficult, if not impossible, to calculate such quantities whenever they're dependent on changing conditions or non-linear relationships.
Nonlinear Elastic Force
Nonlinear elastic forces are those that don't follow the simple, direct proportionality to displacement as described by Hooke's Law. Instead, these forces exhibit more complex relationships with displacement, often involving higher powers or exponential terms.
In our exercise, the bungee cord exerts a nonlinear elastic force expressed by the equation F(x) = k_1x + k_2x^3. The first term represents a linear component of the force similar to a typical Hookean spring, where k_1 is the linear elasticity coefficient. The second term adds a cubic factor, introducing nonlinearity to the force applied.
The presence of the cubic term means that as the cord is stretched further, the force required to stretch it increases at a rate different from that predicted by Hooke's Law. This nonlinear relationship is typical in real-world materials like bungee cords, which can become stiffer and resist linear stretching as they are pulled harder.
Understanding nonlinear forces is crucial in engineering and physics, as it leads to more accurate models of material behavior and can help in designing safer, more reliable systems like our bungee cord.

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Most popular questions from this chapter

(a) What is the average power consumption in watts of an appliance that uses \(5.00 \mathrm{kW} \cdot \mathrm{h}\) of energy per day? \((\mathrm{b})\) How many joules of energy does this appliance consume in a year?

Suppose a horizontal force of \(20 \mathrm{N}\) is required to maintain a speed of \(8 \mathrm{m} / \mathrm{s}\) of a \(50 \mathrm{kg}\) crate. (a) What is the power of this force? (b) Note that the acceleration of the crate is zero despite the fact that \(20 \mathrm{N}\) force acts on the crate horizontally. What happens to the energy given to the crate as a result of the work done by this \(20 \mathrm{N}\) force?

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Compare the work required to accelerate a car of mass \(2000 \mathrm{kg}\) from 30.0 to \(40.0 \mathrm{km} / \mathrm{h}\) with that required for an acceleration from 50.0 to \(60.0 \mathrm{km} / \mathrm{h}\).

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