/*! 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] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 72

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 approximately 1.403 m.

Step by step solution

01

Understand Energy Conservation

The bowling ball swings from an elevated position to its lowest point, converting gravitational potential energy into kinetic energy. When it reaches the lowest point, the energy is stored both as kinetic energy of the ball and potential energy in the stretched cord.
02

Calculate Gravitational Potential Energy Loss

At the highest point, all energy is gravitational potential energy. The potential energy at the start is given by \( PE = mgh \), where \( m = 6.8 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), and \( h = 1.4 \text{ m} \). Compute \( PE = 6.8 \times 9.8 \times 1.4 \).
03

Determine the Elastic Potential Energy in the Cord

At the lowest point, the stretch in the cord causes it to store elastic potential energy given by \( rac{1}{2}kx^2 \), where \( x = 2.7 \times 10^{-3} \text{ m} \). We need to find the spring constant \( k \) using the Young's modulus equation: \( Y = \frac{FL_0}{Ax} \) where \( A = 3.4 \times 10^{-5} \text{ m}^2\).
04

Find Force in the Cord

Assume the tension equals the weight of the bowling ball when stretched. Use \( F = mg = 6.8 \times 9.8 \).
05

Calculate the Young's Modulus to Determine Original Length

Use the formula for Young's modulus: \( Y = \frac{F L_0}{A x} \). Rearrange to find the original length \( L_0 \): \( L_0 = \frac{Y A x}{F} \). Substitute the known values and solve for \( L_0 \). The given stretch and tension values factor into computing \( L_0 \) before stretching.
06

Compute the Unstrained Length of the Cord

Initially ignore the strain (stretch) due to the ball's weight. With the stretch \( x = 2.7 \times 10^{-3} \text{ m} \), and using calculated \( F \), solve for \( L_0 \) using the rearranged Young's modulus equation. Input all calculated values to find the unstrained length.

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 91影视!

Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy is a type of energy stored in an object due to its position in a gravitational field. It's a measure of the potential energy a physical body with mass has when located at a certain height. The formula for gravitational potential energy is given by:\[ PE = mgh \]Here:
  • \( m \) is the mass of the object,
  • \( g \) is the acceleration due to gravity (approximately \( 9.8 \text{ m/s}^2 \) on Earth),
  • \( h \) is the height of the object above the reference point.
In the scenario of the bowling ball, as it is released and swings downwards, the gravitational potential energy decreases because its height \( h \) decreases. This energy isn't lost but rather transformed, typically into kinetic energy. Understanding this energy transformation is crucial in solving problems related to pendulums and swinging objects.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. As the bowling ball swings downwards and speeds up towards its lowest point, its kinetic energy increases. The formula for kinetic energy is:\[ KE = \frac{1}{2}mv^2 \]Where:
  • \( m \) is the mass of the object,
  • \( v \) is the velocity of the object.
When the ball reaches the lowest point of its swing, its gravitational potential energy has been largely converted into kinetic energy. This transformation highlights the principle of conservation of energy, where the total amount of energy remains constant but switches forms between potential and kinetic energy.
Elastic Potential Energy
Elastic potential energy is the energy stored in elastic materials as a result of deformation, such as stretching or compressing. For the bowling ball, when the cord stretches at its lowest point, it contains elastic potential energy. The energy stored can be calculated using the formula:\[ EPE = \frac{1}{2}kx^2 \]Here:
  • \( k \) is the spring constant, which is a measure of the cord's stiffness,
  • \( x \) is the displacement or the stretch in the material from its equilibrium position.
The greater the stretch in the cord (or any elastic material), the more elastic potential energy is stored. In equilibrium, this energy can be returned when the material attempts to return to its original shape, following the principle of energy conservation.
Young's Modulus
Young's modulus is a material property that measures the stiffness of an elastic material. It describes how much a material will stretch or compress under a given load. Young's modulus is expressed by the formula:\[ Y = \frac{FL_0}{Ax} \]Where:
  • \( F \) is the force applied to the material,
  • \( L_0 \) is the original length of the material,
  • \( A \) is the cross-sectional area,
  • \( x \) is the change in length (extension or compression).
In this problem, Young鈥檚 modulus helps determine the original (unstrained) length of the cord by considering the force exerted by the hanging bowling ball and the stretch observed. By rearranging the formula, we can solve for the original length, helping to understand the material's properties and behavior under stress.

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 rifle fires a \(2.10 \times 10^{-2}-\mathrm{kg}\) pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by \(9.10 \times 10^{-2} \mathrm{m}\) from its unstrained length. The pellet rises to a maximum height of \(6.10 \mathrm{m}\) above its position on the compressed spring. Ignoring air resistance, determine the spring constant.

ao \(A\) spring lies on a horizontal table, and the left end of the spring is attached to a wall. The other end is connected to a box. The box is pulled to the right, stretching the spring. Static friction exists between the box and the table, so when the spring is stretched only by a small amount and the box is released, the box does not move. The mass of the box is \(0.80 \mathrm{kg},\) and the spring has a spring constant of \(59 \mathrm{N} / \mathrm{m} .\) The coefficient of static friction between the box and the table on which it rests is \(\mu_{\mathrm{s}}=0.74\) How far can the spring be stretched from its unstrained position without the box moving when it is released?

\(\mathrm{A}\) square plate is \(1.0 \times 10^{-2} \mathrm{m}\) thick, measures \(3.0 \times 10^{-2} \mathrm{m}\) on a side, and has a mass of \(7.2 \times 10^{-2} \mathrm{kg} .\) The shear modulus of the material is \(2.0 \times 10^{10} \mathrm{N} / \mathrm{m}^{2} .\) One of the square faces rests on a flat horizontal surface, and the coefficient of static friction between the plate and the surface is 0.91. A force is applied to the top of the plate, as in Figure 10.29 a. Determine (a) the maximum possible amount of shear stress, (b) the maximum possible amount of shear strain, and (c) the maximum possible amount of shear deformation \(\Delta X\) (see Figure \(10.29 b\) ) that can be created by the applied force just before the plate begins to move.

When an object of mass \(m_{1}\) is hung on a vertical spring and set into vertical simple harmonic motion, it oscillates at a frequency of \(12.0 \mathrm{Hz} .\) When another object of mass \(m_{2}\) is hung on the spring along with the first object, the frequency of the motion is \(4.00 \mathrm{Hz}\). Find the ratio \(m_{2} / m_{1}\) of the masses.

A heavy-duty stapling gun uses a \(0.140-\mathrm{kg}\) metal rod that rams against the staple to eject it. The rod is attached to and pushed by a stiff spring called a "ram spring" \((k=32000 \mathrm{N} / \mathrm{m})\). The mass of this spring may be ignored. The ram spring is compressed by \(3.0 \times 10^{-2} \mathrm{m}\) from its unstrained length and then released from rest. Assuming that the ram spring is oriented vertically and is still compressed by \(0.8 \times 10^{-2} \mathrm{m}\) when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.
See all solutions

Recommended explanations on Physics Textbooks

Kinematics Physics

Read Explanation

Classical Mechanics

Read Explanation

Nuclear Physics

Read Explanation

Quantum Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Electrostatics

Read Explanation
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.