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A force of \(4.1 \mathrm{N}\) is required to stretch a certain spring by \(1.4 \mathrm{cm} .\) (a) How far must this spring be stretched for its potential energy to be \(0.020 \mathrm{J} ?\) (b) How much stretch is required for the spring potential energy to be \(0.080 \mathrm{J} ?\)

Short Answer

Expert verified
(a) Stretch the spring by 0.044 m; (b) Stretch it by 0.088 m.

Step by step solution

01

Understand Spring Potential Energy

The potential energy (PE) stored in a spring is given by the formula: \(PE = \frac{1}{2} k x^2\), where \(k\) is the spring constant and \(x\) is the displacement from the equilibrium position.
02

Calculate Spring Constant

Use Hooke's Law \(F = kx\) to find \(k\). Given \(F = 4.1 \ \text{N}\) and \(x = 1.4 \ \text{cm} = 0.014 \ \text{m}\), we calculate \(k = \frac{F}{x} = \frac{4.1}{0.014} \ \text{N/m}\).
03

Calculate Displacement for 0.020 J

Set \(PE = 0.020 \ \text{J}\) and use the formula \(PE = \frac{1}{2} k x^2\). Solve for \(x\):\[0.020 = \frac{1}{2} \times \frac{4.1}{0.014} \times x^2\]. Calculate \(x\).
04

Calculate Displacement for 0.080 J

Set \(PE = 0.080 \ \text{J}\) and use the same formula. Solve: \[0.080 = \frac{1}{2} \times \frac{4.1}{0.014} \times x^2\]. Calculate \(x\) for this potential energy.

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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 mechanics that describes the linear relationship between the force applied to a spring and the displacement caused by that force. It is expressed mathematically by the formula \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement from the spring's equilibrium position. This law is crucial for understanding how springs and elastic materials behave under force. It essentially tells us that the further a spring is stretched or compressed, the more force is required to do so, provided the material remains within its elastic limit. This concept lays the groundwork for calculating other properties like potential energy in springs.
Spring Constant
The spring constant, denoted as \( k \), is a measure of a spring's stiffness. It is a key part of Hooke's Law. When you know the force applied to a spring and the displacement it caused, you can calculate the spring constant using the formula \( k = \frac{F}{x} \). This calculation is crucial because the spring constant allows us to predict how a spring will respond to different forces. A higher spring constant means a stiffer spring that requires more force for a given displacement. To find \( k \) for our spring, we took the force of \( 4.1 \ \text{N} \) and the displacement of \( 0.014 \ \text{m} \), resulting in a spring constant \( k = \frac{4.1}{0.014} \ \text{N/m} \). Understanding \( k \) helps us further explore the behavior of the spring under various conditions.
Displacement Calculation
Calculating the displacement is a central part of understanding how energy is stored in a spring. We use the potential energy formula for a spring: \( PE = \frac{1}{2} k x^2 \). To find how far a spring must be stretched for a certain potential energy, we rearrange this formula to solve for \( x \). For example, if the desired potential energy is \( 0.020 \ \text{J} \), we set up the equation: \( 0.020 = \frac{1}{2} \times \frac{4.1}{0.014} \times x^2 \). Solving this equation will yield the displacement \( x \) needed to achieve the specified energy. Similarly, for a potential energy of \( 0.080 \ \text{J} \), we use the same approach to calculate \( x \). This helps us understand the precise adjustments needed to attain different energy states in the spring.
Force and Energy in Springs
The concepts of force and energy are intricately linked when it comes to springs. The force exerted by, or required to move, a spring is a function of its displacement, dictated by Hooke's Law. This force is what allows a spring to store energy. The energy stored in a spring is called elastic or potential energy, captured by the formula \( PE = \frac{1}{2} k x^2 \). This formula indicates that the energy is proportional to the square of the displacement. So, doubling the displacement increases the energy by a factor of four, showcasing the non-linear relationship between displacement and energy. Understanding this relationship helps us calculate either one, knowing the values of other parameters, which is key in applications involving mechanical systems like shock absorbers or load stabilizers.

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Most popular questions from this chapter

A mass is attached to the bottom of a vertical spring. This causes the spring to stretch and the mass to move downward. (a) Does the potential energy of the spring increase, decrease, or stay the same during this process? Explain. (b) Does the gravitational potential energy of the Earth-mass system increase, decrease, or stay the same during this process? Explain.

A 0.21 apple falls from a tree to the ground, \(4.0 \mathrm{m}\) below. Ignoring air resistance, determine the apple's kinetic energy, \(K\) the gravitational potential energy of the system, \(U\), and the total mechanical energy of the system, \(E\), when the apple's height above the ground is (a) \(4.0 \mathrm{m},\) (b) \(3.0 \mathrm{m},\) (c) \(2.0 \mathrm{m}\), (d) \(1.0 \mathrm{m},\) and (e) \(0 \mathrm{m}\). Take ground level to be \(y=0\). (e) \(0 \mathrm{m}\). Take ground level to be \(y=0\).

At an amusement park, a swimmer uses a water slide to enter the main pool. If the swimmer starts at rest, slides without friction, and descends through a vertical height of \(2.31 \mathrm{m},\) what is her speed at the bottom of the slide?

A block of mass \(m=0.95 \mathrm{kg}\) is connected to a spring of force constant \(k=775 \mathrm{N} / \mathrm{m}\) on a smooth, horizontal surface. (a) Plot the potential energy of the spring from \(x=-5.00 \mathrm{cm}\) to \(x=5.00 \mathrm{cm}\) (b) Determine the turning points of the block if its speed at \(x=0\) is \(1.3 \mathrm{m} / \mathrm{s}\)

Predict/Explain When a ball of mass \(m\) is dropped from rest from a height \(h,\) its kinetic energy just before landing is \(K\). Now, suppose a second ball of mass \(4 m\) is dropped from rest from a height \(h / 4 .\) (a) Just before ball 2 lands, is its kinetic energy \(4 K, 2 K, K, K / 2,\) or \(K / 4 ?\) (b) Choose the best explanation from among the following: I. The two balls have the same initial energy. II. The more massive ball will have the greater kinetic energy. III. The reduced drop height results in a reduced kinetic energy.

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