/*! 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 13 A vertical spring stores 0.962 J... [FREE SOLUTION] | 91Ó°ÊÓ

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A vertical spring stores 0.962 J in spring potential energy when a \(3.5-\mathrm{kg}\) mass is suspended from it. (a) By what multiplicative factor does the spring potential energy change if the mass attached to the spring is doubled? (b) Verify your answer to part (a) by calculating the spring potential energy when a a \(7.0-\mathrm{kg}\) mass is a ttached to the spring.

Short Answer

Expert verified
The spring potential energy quadruples, increasing to 3.848 J with the 7.0 kg mass.

Step by step solution

01

Understand the Relationship

The potential energy stored in a spring is given by the formula \( U = \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position. This displacement \( x \) is proportional to the weight \( mg \) exerted by the mass on the spring.
02

Calculate Initial Displacement

For a mass \( m = 3.5 \) kg, calculate the displacement \( x \) using the relationship \( mg = kx \). Rearrange it to find \( x = \frac{mg}{k} \), but for energy calculation we start with energy \( U = \frac{1}{2} k x^2 = 0.962 \) J.
03

Set Up Equation for Double Mass

If the mass is doubled to \( 7.0 \) kg, the displacement \( x' \) becomes \( x' = \frac{2mg}{k} \) because the weight is now \( 2mg \). Substituting \( x' \) back into the potential energy formula gives \( U' = \frac{1}{2} k (2x)^2 \).
04

Calculate New Potential Energy

By simplifying \( U' = \frac{1}{2} k (2x)^2 = \frac{1}{2} k \, 4x^2 = 4 \times \frac{1}{2} k x^2 = 4U \). Hence, the energy quadruples.
05

Verify with Specific Calculation

The initial spring potential energy with \( 3.5 \) kg is \( 0.962 \) J. When the mass is doubled, substitute the potential energy formula for the doubling effect, we use \( U' = 4 \times 0.962 = 3.848 \) J.

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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 fundamental principle in physics that describes how objects deform in response to an applied force. It explains that the force required to extend or compress a spring by a distance is directly proportional to that distance. This relationship can be mathematically expressed as:\[ F = kx \]Where:
  • \( F \) is the force exerted by or on the spring (in Newtons).
  • \( k \) is the spring constant, a measure of the spring's stiffness or resistance to deformation.
  • \( x \) is the displacement of the spring from its equilibrium position (in meters).

In context, when a mass is attached to a vertical spring, the spring stretches until the force applied by the mass (due to gravity) is balanced by the force of the spring, reaching equilibrium. This means \( kx = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity (approximately \( 9.8 \text{ m/s}^2 \)).
Understanding Hooke’s Law is essential because it lays the groundwork for analyzing the potential energy stored in the spring, which will be discussed next.
Spring Constant
The spring constant, denoted as \( k \), is a crucial parameter in Hooke's Law. It determines the stiffness of a spring and how much force is needed to change its length by a certain amount.
A high spring constant means the spring is stiffer and more resistant to deformation, whereas a low spring constant indicates a more easily deformed spring.
In our exercise, calculating the spring's potential energy involved using the spring constant within the formula:\[ U = \frac{1}{2} k x^2 \]Here, the spring constant \( k \) helps compute the potential energy \( U \) by determining how the displacement \( x \) contributes to this stored energy.
Once a displacement is known, determining the spring constant helps quantify the elasticity and behavior of the spring under various loads. For instance, doubling the mass more than doubled the displacement, impacting the potential energy stored.
Mechanical Energy
Mechanical energy encompasses both the potential and kinetic energy within a mechanical system. In the case of springs, we focus primarily on potential energy, which is energy stored as a result of its position or configuration.
Spring potential energy is specifically given by the formula:\[ U = \frac{1}{2} k x^2 \]This expression quantifies the energy stored in a spring when it is compressed or stretched from its equilibrium position. Importantly, this potential energy changes based on the displacement \( x \).
In our context, the energy increased significantly when the mass doubled because the displacement increased, thus increasing the stored mechanical energy fourfold.
By understanding mechanical energy, especially through the lens of spring systems, students can grasp how forces and movements translate into energy changes, providing insights into various physical scenarios. Mechanical energy allows us to explore the conversion and conservation of energy in diverse settings.

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Most popular questions from this chapter

A pendulum bob with a mass of 0.13 kg is attached to a string with a length of \(0.95 \mathrm{m}\). We choose the potential energy to be zero when the string makes an angle of \(90^{\circ}\) with the vertical. (a) Find the potential energy of this system when the string makes an angle of \(45^{\circ}\) with the vertical. (b) Is the magnitude of the change in potential energy from an angle of \(90^{\circ}\) to \(45^{\circ}\) greater than, less than, or the same as the magnitude of the change from \(45^{\circ}\) to \(0^{\circ}\) ? Explain. (c) Calculate the potential energy of the system when the string is vertical.

An \(81.0-k g\) in-line skater does +3420 J of nonconservative work by pushing against the ground with his skates. In addition, friction does -715 J of nonconservative work on the skater. The skater's initial and final speeds are \(2.50 \mathrm{m} / \mathrm{s}\) and \(1.22 \mathrm{m} / \mathrm{s},\) respectively. (a) Has the skater gone uphill, downhill, or remained at the same level? Explain. (b) Calculate the change in height of the skater.

A trapeze artist of mass \(m\) swings on a rope of length \(L\) Initially, the trapeze artist is at rest and the rope makes an angle \(\theta\) with the vertical. (a) Find the tension in the rope when it is vertical. (b) Explain why your result for part (a) depends on \(L\) in the way it does.

The work required to stretch a certain spring from an elongation of \(4.00 \mathrm{cm}\) to an elongation of \(5.00 \mathrm{cm}\) is \(30.5 \mathrm{J}\). (a) Is the work required to increase the elongation of the spring from 5.00 \(\mathrm{cm}\) to \(6.00 \mathrm{cm}\) greater than, less than, or equal to \(30.5 \mathrm{J}\) ? Explain. (b) Verify your answer to part (a) by calculating the required work.

A player passes a \(0.600-\mathrm{kg}\) basketball downcourt for a fast break. The ball leaves the player's hands with a speed of \(8.30 \mathrm{m} / \mathrm{s}\) and slows down to \(7.10 \mathrm{m} / \mathrm{s}\) at its highest point. (a) lgnoring air resistance, how high above the release point is the ball when it is at its maximum height? (b) How would doubling the ball's mass affect the result in part (a)? Explain.

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