/*! 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 84 A certain spring is found \(n o ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A certain spring is found \(n o t\) to conform to Hooke's law. The force (in newtons) it exerts when stretched a distance \(x\) (in meters) is found to have magnitude \(52.8 x+38.4 x^{2}\) in the direction opposing the stretch. (a) Compute the work required to stretch the spring from \(x=0.500 \mathrm{~m}\) to \(x=1.00 \mathrm{~m} .\) (b) With one end of the spring fixed, a particle of mass \(2.17 \mathrm{~kg}\) is attached to the other end of the spring when it is stretched by an amount \(x=1.00 \mathrm{~m} .\) If the particle is then released from rest, what is its speed at the instant the stretch in the spring is \(x=0.500 \mathrm{~m} ?\) (c) Is the force exerted by the spring conservative or nonconservative? Explain.

Short Answer

Expert verified
(a) 31 J; (b) 5.34 m/s; (c) Conservative force.

Step by step solution

01

Understanding Work Done on a Spring

When computing the work required to stretch a spring, we utilize the integral of the force function over the displacement. In this case, the force the spring exerts is given by \( F(x) = 52.8x + 38.4x^2 \). To find the work done, the formula is \( W = \int_{x_i}^{x_f} F(x) \, dx \), where \( x_i \) and \( x_f \) are the initial and final positions, respectively.
02

Compute the Work from x = 0.500 m to x = 1.00 m

Substitute the given force into the work integral: \[ W = \int_{0.5}^{1.0} (52.8x + 38.4x^2) \, dx \]. Evaluate this integral:1. \( \int 52.8x \, dx = 26.4x^2 \), evaluated from 0.5 to 1.2. \( \int 38.4x^2 \, dx = 12.8x^3 \), evaluated from 0.5 to 1.Now calculate each:\( 26.4(1^2) - 26.4(0.5^2) = 26.4 - 6.6 = 19.8 \) J,\( 12.8(1^3) - 12.8(0.5^3) = 12.8 - 1.6 = 11.2 \) J.So, the total work is \( 19.8 + 11.2 = 31 \) J.
03

Setup for Kinetic Energy Calculation

The change in potential energy of the spring equals the change in kinetic energy of the particle as it moves from \( x = 1.00 \text{ m} \) to \( x = 0.500 \text{ m} \). The work done in this process is given by the previous calculation of the negative work, due to potential energy decrease. Use the formula \( \Delta KE = \int_{x_f}^{x_i} F(x) \), where the initial speed is zero.
04

Compute the Speed of the Particle

Calculate the change in potential energy, equivalent to the work done when returning from \( x = 1 \text{ m} \) to \( x = 0.5 \text{ m} \), i.e., \(-31 \text{ J}\). Since the initial kinetic energy is zero, the kinetic energy at 0.5 m is \( \frac{1}{2} m v^2 = 31 \text{ J}\). Solving for \( v \) gives \( v = \sqrt{\frac{2 \times 31}{2.17}} \approx 5.34 \text{ m/s}\).
05

Nature of the Spring Force

The force exerted by the spring is considered conservative because the work done depends only on the starting and ending points, not the path taken. In conservative forces, energy is conserved within the system.

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.

Work Done on a Spring
When considering the work done on a spring, we need to look at the force law governing the spring's behavior. Typically, according to Hooke's law, this force is proportional to the displacement, expressed as \( F(x) = -kx \), where \( k \) is the spring constant. However, the spring in our exercise deviates from Hooke's law with a force described by \( F(x) = 52.8x + 38.4x^2 \). This indicates a more complex relationship.

To find the work done when stretching the spring from one position to another, integrate the force over the displacement. Specifically, compute \( W = \int_{x_i}^{x_f} F(x) \, dx \) where \( x_i \) and \( x_f \) are the initial and final stretch amounts. This integral accounts for the variable nature of the force. In this example, compute \( W = \int_{0.5}^{1.0} (52.8x + 38.4x^2) \, dx \), leading to a total work of 31 Joules.
  • Key point: The work done is the energy required to move the spring over a particular displacement.
Kinetic Energy Calculation
Kinetic energy is a measure of an object's energy due to its motion. Calculating it involves understanding the conversion of potential energy into kinetic energy as forces act upon an object, like a particle attached to a spring. Initially at rest and subject to conservative forces, the particle's kinetic energy when the spring retracts from \( x = 1.00 \) m to \( x = 0.500 \) m comes from the stored potential energy.

This potential energy change (equal to the work already computed but negative when considering energy perspective \(-31\) J) converts entirely into the particle's kinetic energy. Using the kinetic energy formula \( \Delta KE = \frac{1}{2} mv^2 \), where \( m \) is mass and \( v \) is velocity, solve for \( v \) to find the speed. For the given mass of 2.17 kg, \( v = \sqrt{\frac{2 \times 31}{2.17}} \approx 5.34 \text{ m/s} \).
  • Remember: Kinetic energy changes when work is performed, influencing an object's speed.
Conservative Forces
Forces like the spring force can be categorized as conservative if the work they perform only depends on initial and final positions and not on the actual path taken. This concept is essential because it implies that the energy within a system can be perfectly transferred or conserved, allowing reversible processes.

In our spring scenario, the work computed in both directions (stretching and retracting) reflects the unchanging nature of energy despite the deviation from Hooke's law. The force's conservative nature tells us the system's total mechanical energy remains constant.
  • Core concept: The conservation of energy principle underpins conservative force contexts.
Hooke's Law Deviation
Hooke's Law is a foundational principle in spring mechanics that states that force varies linearly with displacement (\( F = -kx \)). However, not all springs strictly follow this law. When deviations occur, as in the exercise with a non-linear force \( F(x) = 52.8x + 38.4x^2 \), other physical characteristics, like material properties of the spring, affect stiffness and force exertion.

This deviation means calculating force and work requires more complex mathematical handling, i.e., integrating over a polynomial expression instead of a simple linear one. Deviations from Hooke's law are common in real-world materials when subjected to large deformations or under special temperature conditions.
  • Practical insight: Deviation from Hooke's law highlights complex interactions in real-world materials and informs us to adapt calculations.

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

Figure 8 -31 shows a ball with mass \(m=0.341 \mathrm{~kg}\) attached to the end of a thin rod with length \(L=0.452 \mathrm{~m}\) and negligible mass. The other end of the rod is pivoted so that the ball can move in a vertical circle. The rod is held horizontally as shown and then given enough of a downward push to cause the ball to swing down and around and just reach the vertically up position, with zero speed there. How much work is done on the ball by the gravitational force from the initial point to (a) the lowest point, (b) the highest point, and (c) the point on the right level with the initial point? If the gravitational potential energy of the ball-Earth system is taken to be zero at the initial point, what is it when the ball reaches (d) the lowest point, (e) the highest point, and (f) the point on the right level with the initial point? (g) Suppose the rod were pushed harder so that the ball passed through the highest point with a nonzero speed. Would \(\Delta U_{g}\) from the lowest point to the highest point then be greater than, less than, or the same as it was when the ball stopped at the highest point?

The potential energy of a diatomic molecule (a two-atom system like \(\mathrm{H}_{2}\) or \(\mathrm{O}_{2}\) ) is given by $$ U=\frac{A}{r^{12}}-\frac{B}{r^{6}} $$ where \(r\) is the separation of the two atoms of the molecule and \(A\) and \(B\) are positive constants. This potential energy is associated with the force that binds the two atoms together. (a) Find the \(e q u i-\) librium separation \(-\) that is, the distance between the atoms at which the force on each atom is zero. Is the force repulsive (the atoms are pushed apart) or attractive (they are pulled together) if their separation is (b) smaller and (c) larger than the equilibrium separation?

From the edge of a cliff, a \(0.55 \mathrm{~kg}\) projectile is launched with an initial kinetic energy of \(1550 \mathrm{~J}\). The projectile's maximum upward displacement from the launch point is \(+140 \mathrm{~m}\). What are the (a) horizontal and (b) vertical components of its launch velocity? (c) At the instant the vertical component of its velocity is \(65 \mathrm{~m} / \mathrm{s},\) what is its vertical displacement from the launch point?

An outfielder throws a baseball with an initial speed of \(81.8 \mathrm{mi} / \mathrm{h} .\) Just before an infielder catches the ball at the same level, the ball's speed is \(110 \mathrm{ft} / \mathrm{s}\). In foot-pounds, by how much is the mechanical energy of the ball-Earth system reduced because of air drag? (The weight of a baseball is 9.0 oz.)

A spring \((k=200 \mathrm{~N} / \mathrm{m})\) is fixed at the top of a frictionless plane inclined at angle \(\theta=40^{\circ}(\) Fig. \(8-59) .\) A \(1.0 \mathrm{~kg}\) block is projected up the plane, from an initial position that is distance \(d=0.60 \mathrm{~m}\) from the end of the relaxed spring, with an initial kinetic energy of 16 J. (a) What is the kinetic energy of the block at the instant it has compressed the spring \(0.20 \mathrm{~m} ?\) (b) With what kinetic energy must the block be projected up the plane if it is to stop momentarily when it has compressed the spring by \(0.40 \mathrm{~m} ?\)

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.