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A spring with a spring constant of \(3200 \mathrm{~N} / \mathrm{m}\) is initially stretched until the elastic potential energy of the spring is \(1.44 \mathrm{~J}\). \((U=0\) for the relaxed spring.) What is \(\Delta U\) if the initial stretch is changed to (a) a stretch of \(2.0 \mathrm{~cm},(\mathrm{~b})\) a compression of \(2.0 \mathrm{~cm}\), and (c) a compression of \(4.0 \mathrm{~cm}\) ?

Short Answer

Expert verified
(a) -0.8 J, (b) -0.8 J, (c) 1.12 J.

Step by step solution

01

Given Values and Formula for Elastic Potential Energy

We are given that the spring constant, \( k \), is \( 3200 \mathrm{~N/m} \). The elastic potential energy \( U \) for the initially stretched spring is \( 1.44 \mathrm{~J} \). The formula for elastic potential energy is given by \( U = \frac{1}{2} k x^2 \), where \( x \) is the displacement from the relaxed position.
02

Find Initial Stretch Displacement

First, we need to find the initial stretch \( x_0 \) that results in \( 1.44 \mathrm{~J} \) of energy. Using the formula, we have:\[ 1.44 = \frac{1}{2} \times 3200 \times x_0^2 \]Solving for \( x_0^2 \):\[ x_0^2 = \frac{1.44 \times 2}{3200} \]\[ x_0^2 = 0.0009 \]Taking the square root gives:\[ x_0 = 0.03 \text{ m} \]
03

Calculate Change in Elastic Potential Energy for New Stretch

Now, calculate the elastic potential energy for the new stretch of \( 2.0 \mathrm{~cm} \) and find the change \( \Delta U \):Convert \( 2.0 \mathrm{~cm} \) to meters: \( 0.02 \text{ m} \). Use the formula for potential energy:\[ U_1 = \frac{1}{2} \times 3200 \times (0.02)^2 \]\[ U_1 = \frac{1}{2} \times 3200 \times 0.0004 \]\[ U_1 = 0.64 \text{ J} \]\[ \Delta U = U_1 - 1.44 = 0.64 - 1.44 = -0.8 \text{ J} \]
04

Calculate Change in Elastic Potential Energy for Compression

For compression with \( 2.0 \mathrm{~cm} \): it is the same calculation as the stretch as the sign doesn't affect the squared term in the energy formula. Thus, \( \Delta U = -0.8 \text{ J} \).
05

Calculate Change in Elastic Potential Energy for 4.0 cm Compression

Convert \( 4.0 \mathrm{~cm} \) to meters: \( 0.04 \text{ m} \). Compute potential energy:\[ U_2 = \frac{1}{2} \times 3200 \times (0.04)^2 \]\[ U_2 = \frac{1}{2} \times 3200 \times 0.0016 \]\[ U_2 = 2.56 \text{ J} \]\[ \Delta U = U_2 - 1.44 = 2.56 - 1.44 = 1.12 \text{ J} \]

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Key Concepts

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

Spring Constant
The spring constant, denoted by the symbol \( k \), is a fundamental property of a spring that indicates its stiffness. It shows the amount of force needed to stretch or compress the spring by a unit distance. For example, if a spring has a spring constant of \( 3200 \, \text{N/m} \), it means that you need to exert 3200 newtons of force to stretch or compress the spring by 1 meter. In this exercise, the spring constant is a given value, which simplifies the calculations of elastic potential energy when the spring is either stretched or compressed. Knowing the spring constant helps determine how much energy is stored in the spring for a given displacement.
Displacement
Displacement in the context of elastic potential energy refers to the amount a spring is stretched or compressed from its relaxed, unstrained position. It is usually denoted by \( x \) and is crucial in calculating how much energy a spring stores. Displacement can be either positive or negative depending on whether the spring is stretched or compressed, respectively.

In this exercise, the displacement from the relaxed position varies for different scenarios: a stretch of 2.0 cm, a compression of 2.0 cm, and a compression of 4.0 cm. However, when using the energy formula, the displacement is used in its squared form, which means both stretching and compressing over the same distance result in the same magnitude of potential energy.
Compression
A compression in a spring refers to shortening its length compared to its natural, relaxed length. It is just as easily calculated as stretching because both involve a change in displacement from the spring's equilibrium position.

Despite the direction (inward for compression, outward for stretching), the mathematical treatment of compression in energy terms remains consistent with stretching. This is shown by the squared term \((x^2)\) in the energy formula. Thus, when a spring is compressed by a certain amount, the physical laws governing the system ensure that the potential energy stored is the same as for an equivalent stretch, as was demonstrated by cases (a) and (b) in this exercise.
Energy Formula
The energy formula for calculating elastic potential energy is:
  • \( U = \frac{1}{2} k x^2 \)
This formula allows you to compute the energy stored in a spring when it is displaced from its relaxed state. Here:
  • \( U \) is the elastic potential energy.
  • \( k \) is the spring constant.
  • \( x \) is the displacement from the relaxed position of the spring.

The formula highlights that the energy stored is proportional to the square of the displacement \((x^2)\) and directly proportional to the spring constant \( k \). It shows that doubling the displacement increases the stored energy fourfold due to the square term. This mathematical insight is key to solving problems and understanding concepts related to springs in physics.

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