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Chapter 7: Problem 27

A climbing rope is designed to exert a force given by F = - kx + bx3 , where k = 244 N/m, b = 3.24 N/m3 , and x is the stretch in meters. Find the potential energy stored in the rope when it鈥檚 been stretched 4.68 m. Take U = 0 when the rope isn鈥檛 stretched鈥攖hat is, when x = 0. Is this more or less than if the rope were an ideal spring with the same spring constant k?

Short Answer

Expert verified
The potential energy stored in the rope when it鈥檚 been stretched 4.68 m can be calculated by integrating the given force function from 0 to x and then evaluating the result at x = 4.68 m. Then, this value can be compared to the potential energy of an ideal spring stretched the same distance by substituting into the potential energy equation for an ideal spring \( U = \frac{1}{2} kx^2 \). The comparison can be made based on which value is larger, the potential energy of the rope or that of the ideal spring.

Step by step solution

01

Calculate the potential energy of the rope

The potential energy \( U \) in the rope when it鈥檚 been stretched \( x = 4.68 m \) can be computed by integrating the given force equation from 0 to x. The integral will be: \( U = -\int_0^x F \, dx = -\int_0^x (-kx + bx^3) \, dx = \int_0^x (kx - bx^3) \, dx \). This integration gives the potential energy function which is then evaluated at 4.68 m.
02

Evaluate the potential energy at x = 4.68 m

Now, you can substitute the values \( k = 244 N/m \), \( b = 3.24 N/m^3 \) and \( x = 4.68 m \) into the resulting function from step 1. This will give the potential energy of the rope when it鈥檚 stretched 4.68 m.
03

Calculate the potential energy of an ideal spring

For a comparison purpose, calculate the potential energy for an ideal spring using the formula \( U = \frac{1}{2} kx^2 \). Substitute the values \( k = 244 N/m \) and \( x = 4.68 m \) into the equation.
04

Compare the potential energies

Now that you have the potential energy for both systems, the rope and the ideal spring, you can make a comparison. If the potential energy of the rope is greater than that of the ideal spring, then you can conclude that the rope stores more energy when stretched by the same distance. If it is less, then the ideal spring stores more energy.

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Key Concepts

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

Climbing Rope Physics
Climbing ropes are a vital component in rock climbing, offering safety and security to climbers. They are designed to withstand significant stresses and distribute forces effectively during a fall.
Unlike simple ropes, climbing ropes are dynamic. This means they can stretch under force, absorbing energy which reduces the shock load on a climber during a fall.
  • The stretching ability also helps prevent injuries and equipment damages.
  • Because of this, understanding the physics of how these ropes work involves more than just looking at them as linear springs.
  • The force involved when a climbing rope stretches isn鈥檛 linear. This is where complex physics comes into play.
Understanding the specific forces and energies in climbing ropes can save lives and improve climbing safety.
Hooke's Law
Hooke's Law is a fundamental principle explaining elasticity. It states that the force needed to extend or compress a spring is proportional to the distance it is stretched or compressed. It's often expressed as:
\( F = -kx \)
Here, \( F \) is the force applied, \( x \) is the distance the spring is stretched or compressed from its original position, and \( k \) is the spring constant.
  • Hooke鈥檚 Law implies a linear relationship between force and displacement, making it straightforward to calculate potential energy.
  • It provides a baseline understanding for analyzing elastic objects under small deformations, like an ideal spring.
In climbing ropes, however, deviations from Hooke鈥檚 Law occur, especially when more extensive stretching happens.
Spring Constant
The spring constant, denoted as \( k \), is a measure of a spring's stiffness in Hooke's Law. A higher spring constant means a stiffer spring that requires more force to stretch.
For the equation \( F = -kx \), the spring constant provides a quantifiable measure of elasticity:
  • With a spring constant of \( 244 \frac{N}{m} \), the climbing rope is relatively elastic, meaning it can stretch considerably with applied force.
  • The spring constant is critical in determining the amount of force needed to stretch any spring-based system.
In climbing, understanding the spring constant aids in predicting how a rope will behave under physical stress, ensuring proper dynamic responses during falls.
Nonlinear Elasticity
In nonlinear elasticity, the relationship between stress and strain doesn鈥檛 follow a simple linear path like Hooke's Law. This characteristic is especially important in materials subjected to large deformation, such as climbing ropes when stretched significantly.
Rather than behaving like an ideal spring, a climbing rope鈥檚 force response involves more complex terms. For example,
\( F = -kx + bx^3 \)
illustrates how additional terms represent the nonlinear behavior.
  • The cubic term \( bx^3 \) adds complexity beyond the linear behavior represented in Hooke's Law.
  • This nonlinear elasticity ensures the rope absorbs energy more effectively, providing better shock absorption during falls.
Understanding these nonlinear characteristics is crucial for designing ropes that are safe and effective in real climbing scenarios.

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Most popular questions from this chapter

A particle moves along the x-axis under the influence of a force F = ax2 + b, where a and b are constants. Find the potential energy as a function of position, taking U = 0 at x = 0.

Your roommate is writing a science fiction novel and asks your advice about a plot point. Her characters are mining ore on the Moon and launching it toward Earth. Bins with 1500 kg of ore will be launched by a large spring, to be compressed 17 m. It takes a speed of 2.4 km/s to escape the Moon鈥檚 gravity. What do you tell her is an appropriate spring constant?

In a railroad switchyard, a rail car of mass 28,600 kg starts from rest and rolls down an incline and onto a level stretch of track. It then hits a spring bumper at the end of the track. If the spring constant is 1.88 MN/m and if the spring compresses a maximum of 1.03 m, what鈥檚 the height at which the car started? Neglect friction.

The nuchal ligament is a cord-like structure that runs along the back of the neck and supports much of the head鈥檚 weight in animals like horses and cows. The ligament is extremely stiff for small stretches, but loosens as it stretches further, thus functioning as a biological shock absorber. Figure 7.17 shows the force鈥揹istance curve for a particular nuchal ligament; the curve can be modeled approximately by the expression F1x2 = 0.43x - 0.033x2 + 0.00086x3 , with F in kN and x in cm. Find the energy stored in the ligament when it鈥檚 been stretched (a) 8.0 cm and (b) 16 cm.

A mass \(m\) is dropped from height \(h\) above the top of a spring of constant \(k\) mounted vertically on the floor. Show that the spring's maximum compression is given by \((m g / k)(1+\sqrt{1+2 k h / m g})\).
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