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A mass resting on a horizontal, frictionless surface is attached to one end of a spring; the other end is fixed to a wall. It takes \(3.6 \mathrm{~J}\) of work to compress the spring by \(0.13 \mathrm{~m}\). If the spring is compressed, and the mass is released from rest, it experiences a maximum acceleration of \(15 \mathrm{~m} / \mathrm{s}^{2} .\) Find the value of \((a)\) the spring constant and \((b)\) the mass.

Short Answer

Expert verified
The spring constant is approximately 426.03 N/m, and the mass is about 3.69 kg.

Step by step solution

01

Determine Spring Constant

To find the spring constant \( k \), we use the work-energy principle for springs, given as \( W = \frac{1}{2}k x^2 \), where \( W = 3.6 \text{ J} \) is the work done and \( x = 0.13 \text{ m} \) is the compression. Solving for \( k \), we rearrange and substitute:\[ k = \frac{2W}{x^2} = \frac{2 \times 3.6}{(0.13)^2} \].Calculating this gives:\[ k \approx 426.03 \text{ N/m} \].
02

Calculate the Mass

Given that the maximum acceleration \( a = 15 \text{ m/s}^2 \) occurs when the spring force is maximum, use Newton's second law \( F = ma \) and Hooke's law \( F = kx \). Hence, \( ma = kx \).Solving for the mass \( m \), we have:\[ m = \frac{kx}{a} = \frac{426.03 \times 0.13}{15} \].Upon calculation:\[ m \approx 3.69 \text{ kg} \].

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Key Concepts

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

Hooke's Law
Hooke's Law describes the relationship between the force exerted by a spring and the amount it is stretched or compressed. This law states that the force needed to extend or compress a spring by a distance \( x \) is proportional to that distance. Mathematically, it is expressed as:
  • \( F = kx \)
where \( F \) is the force applied to the spring (in newtons), \( k \) is the spring constant (a measure of the spring's stiffness), and \( x \) is the displacement of the spring from its equilibrium position (in meters).
The spring constant \( k \) tells us how stiff the spring is: larger \( k \) values indicate stiffer springs. In the exercise, the spring constant was calculated using Hooke's Law in combination with the work-energy principle. This provides insight into how much force is required to compress or extend the spring by a specific distance.
work-energy principle
The work-energy principle connects the work done on an object with its energy changes. For springs, this principle is often applied to calculate how much work is done during compression or extension, linking it to the potential energy stored in the spring. The formula used here is:
  • \( W = \frac{1}{2}kx^2 \)
where \( W \) is the work done on the spring (in joules), \( k \) is the spring constant, and \( x \) is the displacement.
In practical terms, when a spring is compressed or extended, it stores energy. This stored energy can be calculated, and in our exercise, it allowed us to determine the spring constant by rearranging the formula to solve for \( k \). This demonstrates how the work-energy principle can be a powerful tool in both problem-solving and understanding the transfer and storage of energy in physical systems.
Newton's second law
Newton's second law is fundamental in understanding mechanical motion and can be expressed as:
  • \( F = ma \)
where \( F \) is the force applied to an object (in newtons), \( m \) is the object's mass (in kilograms), and \( a \) is the acceleration produced (in meters per second squared).
In the context of our exercise, Newton's second law works hand-in-hand with Hooke's Law to find the mass attached to the spring. Once we know the force exerted by the spring at maximum compression (using Hooke's Law), the mass can be derived using the relationship between force and acceleration provided by Newton's second law.
Understanding this law is crucial because it not only helps calculate mass but also provides insight into how various forces and mass interact to affect an object's motion. It forms the basis for analyzing mechanical systems in equilibrium and motion, as shown in the exercise when determining the mass with a given acceleration.

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Most popular questions from this chapter

(II) At \(t=0,\) a \(785-\mathrm{g}\) mass at rest on the end of a horizontal spring \((k=184 \mathrm{N} / \mathrm{m})\) is struck by a hammer which gives it an initial speed of 2.26 \(\mathrm{m} / \mathrm{s} .\) Determine \((a)\) the period and frequency of the motion, \((b)\) the amplitude, (c) the maximum acceleration, (d) the position as a function of time, \((e)\) the total energy, and \((f)\) the kinetic energy when \(x=0.40 A\) where \(A\) is the amplitude.

(II) A vertical spring of spring constant 115 \(\mathrm{N} / \mathrm{m}\) supports a mass of 75 \(\mathrm{g}\) . The mass oscillates in a tube of liquid. If the mass is initially given an amplitude of 5.0 \(\mathrm{cm}\) , the mass is observed to have an amplitude of 2.0 \(\mathrm{cm}\) after 3.5 \(\mathrm{s}\) . Estimate the damping constant b. Neglect buoyant forces.

An elastic cord is \(65 \mathrm{~cm}\) long when a weight of \(75 \mathrm{~N}\) hangs from it but is \(85 \mathrm{~cm}\) long when a weight of \(180 \mathrm{~N}\) hangs from it. What is the "spring" constant \(k\) of this elastic cord?

(II) Agent Arlene devised the following method of measuring the muzzle velocity of a rifle (Fig. \(33 ) .\) She fires a bullet into a 4.648 -kg wooden block resting on a smooth surface, and attached to a spring of spring constant \(k=142.7 \mathrm{N} / \mathrm{m} .\) The bullet, whose mass is \(7.870 \mathrm{g},\) remains embedded in the wooden block. She measures the maximum distance that the block compresses the spring to be 9.460 \(\mathrm{cm} .\) What is the speed \(v\) of the bullet?

A small fly of mass \(0.25 \mathrm{~g}\) is caught in a spider's web. The web oscillates predominately with a frequency of \(4.0 \mathrm{~Hz}\). (a) What is the value of the effective spring stiffness constant \(k\) for the web? ( \(b\) ) At what frequency would you expect the web to oscillate if an insect of mass \(0.50 \mathrm{~g}\) were trapped?

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