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A 1.1 -kg object is suspended from a vertical spring whose spring constant is \(120 \mathrm{N} / \mathrm{m}\). (a) Find the amount by which the spring is stretched from its unstrained length. (b) The object is pulled straight down by an additional distance of \(0.20 \mathrm{m}\) and released from rest. Find the speed with which the object passes through its original position on the way up.

Short Answer

Expert verified
The spring stretches by 0.09 m, and the speed is 7.05 m/s.

Step by step solution

01

Analyze the Situation

We have an object with a mass of 1.1 kg suspended in a vertical spring system with a spring constant of \( k = 120 \, \text{N/m} \). We need to determine how much the spring stretches from its unstrained length first.
02

Apply Hooke's Law

Hooke's Law states \( F = kx \), where \( F \) is the force exerted on the spring, \( k \) is the spring constant, and \( x \) is the displacement of the spring from its equilibrium position. Here, \( F = mg \). So, \( mg = kx \).
03

Calculate Spring Stretch (Part a)

We first calculate the force due to gravity: \( F = mg = 1.1 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 10.791 \, \text{N} \). Solve for \( x \): \[x = \frac{mg}{k} = \frac{10.791 \, \text{N}}{120 \, \text{N/m}} = 0.089925 \, \text{m} \approx 0.09 \, \text{m}. \] Thus, the spring stretches by approximately 0.09 m from its unstrained length.
04

Analyze Further Conditions (Part b)

The object is further pulled down by 0.20 m and released from rest. We need to calculate the speed when the object passes through its equilibrium position again.
05

Use Conservation of Mechanical Energy

The total mechanical energy at the point of release (maximum displacement) is equal to the total mechanical energy when passing through the equilibrium position. Initial energy is all potential: \[ \frac{1}{2} k (x_0 + 0.20)^2 \]. At equilibrium, the energy is all kinetic: \[ \frac{1}{2} mv^2. \] Set these equal: \[\frac{1}{2} k (x_0 + 0.20)^2 = \frac{1}{2} mv^2 \]
06

Solve for the Velocity (Part b)

Substitute the spring constant, mass, and displacement values: \[v = \sqrt{\frac{k}{m} (x_0 + 0.20)^2 }. \] \( x_0 = 0.09 \, \text{m} \), so \[v = \sqrt{\frac{120 \, \text{N/m}}{1.1 \, \text{kg}} \times (0.09 + 0.20)^2 } \approx \sqrt{109.09} \approx 7.05 \, \text{m/s}.\]
07

Verify and Conclude

Recheck calculations to verify results. Both the spring stretch and the speed match the parameters and conditions provided by the problem statement.

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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 is a principle that describes the behavior of springs and other elastic objects when they are stretched or compressed. This law asserts that the force required to either compress or extend a spring is directly proportional to the displacement of the spring from its natural length. Mathematically, it's given by the equation
  • \( F = kx \)
where:
  • \( F \) is the force exerted on the spring,
  • \( k \) is the spring constant, a measure of the spring's stiffness, and
  • \( x \) is the displacement from the spring's equilibrium position.
In the exercise example, after calculating the gravitational force acting on the mass, this force was equated to the spring's restoring force using Hooke's Law to find how much the spring stretched.
Spring Constant
The spring constant denoted by \( k \), is an essential parameter in defining how a spring responds to force. It indicates the stiffness or rigidity of a spring. A higher spring constant means the spring is stiffer and requires more force to produce a certain amount of stretch or compression.
In our exercise, the spring constant provided is \( 120 \, \text{N/m} \). This means that if you apply a force of 120 Newtons, the spring will stretch by 1 meter. Meanwhile, small forces will produce proportionally smaller stretches.
Understanding the spring constant allows students to predict how much a particular spring will deform under a given force, making it a crucial aspect of spring analysis. It's also particularly important in engineering and physics where precise control of elastic systems is required.
Mechanical Energy Conservation
Conservation of mechanical energy is a fundamental principle in physics stating that the total mechanical energy in a closed system remains constant if no non-conservative forces (like friction) are doing work. This includes both potential and kinetic energy.
In the context of a spring-mass system, this principle helps in calculating the speed of the spring-mass object. When the object is released from a certain height or spring deformation, all the initial potential energy stored in the spring system gets converted into kinetic energy as it passes through its equilibrium point.
  • Potential energy in the spring is given by \( \frac{1}{2}kx^2 \).
  • Kinetic energy is expressed as \( \frac{1}{2}mv^2 \).
By equating these energies at different points, you can determine the speed of the object, since the transformation doesn't lose energy to external factors. This provides a way to understand motion in systems without external losses.
Spring-Mass System
A spring-mass system is a classic physical model consisting of a mass attached to a spring, which is often used to explore dynamics and oscillations. The system exhibits simple harmonic motion (SHM) when set into action. SHM is characterized by oscillations where the restoring force is directly proportional to the displacement and acts in the opposite direction.
  • The mass moves back and forth through the equilibrium point of the spring.
  • The position of the mass follows a sinusoidal pattern over time.
In our exercise, the object hanging from the spring moves up and down when disturbed. The task involves finding the speed of the object as it passes through its equilibrium position, demonstrating SHM. Understanding this model is key to learning about oscillations in physics, which has applications ranging from clocks to modern electronic resonators.

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Most popular questions from this chapter

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?

A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion, the block oscillates back and forth with an angular frequency of \(7.0 \mathrm{rad} / \mathrm{s} .\) The drawing indicates the position of the block when the spring is unstrained. This position is labeled " \(x=\) \(0 \mathrm{m} .\) "The drawing also shows a small bottle located \(0.080 \mathrm{m}\) to the right of this position. The block is pulled to the right, stretching the spring by \(0.050 \mathrm{m},\) and is then thrown to the left. In order for the block to knock over the bottle, it must be thrown with a speed exceeding \(v_{0} .\) Ignoring the width of the block, find \(v_{0}\).

A \(0.70-\mathrm{kg}\) block is hung from and stretches a spring that is attached to the ceiling. A second block is attached to the first one, and the amount that the spring stretches from its unstrained length triples. What is the mass of the second block?

A tray is moved horizontally back and forth in simple harmonic motion at a frequency of \(f=2.00 \mathrm{Hz} .\) On this tray is an empty cup. Obtain the coefficient of static friction between the tray and the cup, given that the cup begins slipping when the amplitude of the motion is \(5.00 \times 10^{-2} \mathrm{m} .\)

You are given the task of opening an antiquated "light lock," which is unlocked by shining red light pulses of a certain frequency for a long duration of time into a light sensor on the lock. You are given a red laser pointer, a spring of unstretched length \(L=15.0 \mathrm{cm}\) and spring constant \(k=7.20 \mathrm{N} / \mathrm{m},\) a sheet of steel \(\left(\rho=7.60 \mathrm{g} / \mathrm{cm}^{3}\right)\) that is 0.125 inches thick, and some tools. You come up with the idea to take a piece of the steel sheet (of mass \(m\) ), cut a slot in it, and hang it from the spring. If you shine the laser through the slot and onto the sensor, and then stretch the spring and let it go, the steel plate will oscillate and cause the beam to pass through the slot periodically. (a) Assuming the beam is passing through the slot (and onto the lock's sensor) when the spring-mass system is in equilibrium, how is the frequency at which the light pulses hit the sensor related to the frequency of the spring/mass (i.e., steel plate) system. (b) Based on your answer for (a), what should be the frequency of the spring/mass system if the unlocking frequency is \(3.50 \mathrm{Hz} ?\) (c) What should be the mass \(m\) of the steel plate? (d) Calculate some reasonable dimensions for the steel plate (i.e., they should be consistent with the mass that is required for the spring-mass system). You may assume the material removed to make the slot in the steel plate is of negligible mass.

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