/*! 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 72 A 6.8 -kg bowling ball is attach... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 6.8 -kg bowling ball is attached to the end of a nylon cord with a cross- sectional area of \(3.4 \times 10^{-5} \mathrm{m}^{2}\) . The other end of the cord is fixed to the ceiling. When the bowling ball is pulled to one side and released from rest, it swings downward in a circular arc. At the instant it reaches its lowest point, the bowling ball is 1.4 \(\mathrm{m}\) lower than the point from which it was released, and the cord is stretched \(2.7 \times 10^{-3} \mathrm{m}\) from its unstrained length. What is the unstrained length of the cord? Hint: When calculating any quantity other than the strain, ignore the increase in the length of the cord.

Short Answer

Expert verified
The unstrained length of the cord is calculated using energy conservation and Hooke's Law, finding the force and strain relationship.

Step by step solution

01

Identify Energy Conversion

The potential energy lost by the bowling ball when it swings downward will be converted into kinetic energy at the lowest point of its path. The potential energy (PE) lost can be expressed as \( PE = mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h \) is the height change.
02

Calculate Potential Energy

Given the mass \( m = 6.8 \text{ kg} \), gravitational acceleration \( g = 9.8 \text{ m/s}^2 \), and height \( h = 1.4 \text{ m} \), the lost potential energy is \( PE = 6.8 \times 9.8 \times 1.4 \).
03

Use Kinetic Energy to Find Speed at Lowest Point

The kinetic energy at the lowest point (KE) is equal to the lost potential energy: \( KE = PE = \frac{1}{2}mv^2 \). Solve for \( v \) (velocity) to find the speed of the bowling ball at the lowest point.
04

Relate Force and Strain

Hooke's Law describes the relationship between the force on the cord and the strain: \( F = k \Delta L \), where \( k \) is the spring constant and \( \Delta L = 2.7 \times 10^{-3} \text{ m} \) is the stretch. Young's Modulus \( Y \) is used to find \( k \) by the formula \( k = \frac{Y A}{L} \).
05

Calculate Young's Modulus and Solve for Unstrained Length

Given the cord's cross-sectional area \( A = 3.4 \times 10^{-5} \text{ m}^2 \), use Young’s Modulus to rearrange and solve for the unstrained length \( L \). The Young's Modulus is constant for the material, and after finding \( F \) using \( F = ma \) at the lowest point, use \( \Delta L = \frac{F L}{Y A} \) to solve for \( L \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Energy Conversion
In this exercise, we witness an excellent example of energy conversion. When the 6.8 kg bowling ball is released from rest, it transforms the potential energy it holds due to its elevated position into kinetic energy as it swings downward.
This transformation underscores one of the core principles of physics: the conservation of energy. As the bowling ball descends, the gravitational potential energy it initially possesses decreases, and correspondingly, its kinetic energy increases.
This energy conversion can be mathematically represented as:
  • Potential Energy (PE) decreases by: \[ PE = mgh \] where \( m \) is mass, \( g \) is gravitational acceleration, and \( h \) is the height from which it falls.
  • At the lowest point in its swing, the kinetic energy (KE) is at its maximum and equal to the lost potential energy: \[ KE = \frac{1}{2}mv^2 = PE \]
This illustrates how these forms of energy are interchangeable in a closed system, permitting us to track one type of energy to determine the characteristics of another.
Potential Energy
Potential energy is a form of stored energy that depends on an object's position relative to a zero position. In this case, the bowling ball has gravitational potential energy because it is suspended at a certain height above the ground.
Gravitational potential energy is determined by the formula:
\[ PE = mgh \] where:
  • \( m \) is the mass of the object (6.8 kg).
  • \( g \) is the acceleration due to gravity (approximately 9.8 m/s² on Earth).
  • \( h \) is the height change (1.4 meters in this scenario).
As the bowling ball drops, its height decreases, leading to a reduction in potential energy. This lost potential energy is precisely what gets transformed into kinetic energy, causing the ball to accelerate as it moves downward. Understanding this concept helps in appreciating how objects store and convert energy based on their position and motion.
Hooke's Law
Hooke's Law is fundamental to understanding how materials respond to force, and it describes the behavior of elastic materials like the nylon cord in this problem. The law is expressed as:
\[ F = k \Delta L \] where:
  • \( F \) is the force applied to the material.
  • \( k \) is the spring constant, reflecting the stiffness of the material.
  • \( \Delta L \) is the change in length of the material due to the applied force, given here as 2.7 x 10^{-3} m.
In this context, Hooke's Law helps us understand how the cord stretches when the bowling ball exerts a force on it during its motion. The spring constant \( k \) can be determined using the material's properties and is key to calculating other quantities such as the unstrained length of the cord by connecting it with concepts like Young's Modulus.
Young's Modulus
Young's Modulus is a measure of a material's ability to withstand changes in length when under lengthwise tension or compression. It characterizes the stiffness of a given material and is crucial in understanding how materials like the nylon cord behave when forces are applied.
Young's Modulus \( Y \) is used in conjunction with other constants and variables to calculate the spring constant \( k \), essential for applying Hooke's Law. The relationship is given by:
\[ k = \frac{Y A}{L} \] where:
  • \( Y \) is Young's Modulus.
  • A is the cross-sectional area of the material (3.4 x 10^{-5} m²).
  • \( L \) is the unstrained length of the material.
Understanding and applying Young's Modulus allows us to solve for the unstrained length of the cord by rearranging and solving the equation for \( L \). This insight into material behavior under stress reveals the interconnectedness of theoretical physics principles and real-world observations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that increases from 0 to \(3.0 \times 10^{-2} \mathrm{N}\) The length and radius of the collagen are, respectively, 2.5 and \(0.091 \mathrm{cm},\) and Young's modulus is \(3.1 \times 10^{6} \mathrm{N} / \mathrm{m}^{2} .\) (a) If the stretching obeys Hooke's law, what is the spring constant \(k\) for collagen? (b) How much work is done by the variable force that stretches the collagen? (See Section 6.9 for a discussion of the work done by a variable force.)

A 0.60 -kg metal sphere oscillates at the end of a vertical spring.As the spring stretches from 0.12 to 0.23 \(\mathrm{m}\) (relative to its unstrained length), the speed of the sphere decreases from 5.70 to 4.80 \(\mathrm{m} / \mathrm{s}\) . What is the spring constant of the spring?

A person bounces up and down on a trampoline, while always staying in contact with it. The motion is simple harmonic motion, and it takes 1.90 s to complete one cycle. The height of each bounce above the equilibrium position is 45.0 \(\mathrm{cm}\) . Determine (a) the amplitude and (b) the angular frequency of the motion. (c) What is the maximum speed attained by the person?

The drawing shows a top view of a frictionless horizontal surface, where there are two springs with particles of mass \(m_{1}\) and \(m_{2}\) attached to them. Each spring has a spring constant of 120 \(\mathrm{N} / \mathrm{m}\) . The particles are pulled to the right and then released from the positions shown in the drawing. How much time passes before the particles are side by side for the first time at \(x=0 \mathrm{m}\) if \(\quad\) (a) \(m_{1}=m_{2}=\) 3.0 \(\mathrm{kg}\) and \(\quad(\mathrm{b}) m_{1}=3.0 \mathrm{kg}\) and \(m_{2}=27 \mathrm{kg} ?\)

Two physical pendulums (not simple pendulums) are made from meter sticks that are suspended from the ceiling at one end. The sticks are uniform and are identical in all respects, except that one is made of wood (mass \(=0.17 \mathrm{kg}\) and the other of metal (mass \(=0.85 \mathrm{kg}\) . They are set into oscillation and execute simple harmonic motion. Determine the period of \((\mathrm{a})\) the wood pendulum and \((\mathrm{b})\) the metal pendulum.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.