Problem 17
Question
A spring stores potential energy \(U_{0}\) when it is compressed a distance \(x_{0}\) from its uncompressed length. (a) In terms of \(U_{0},\) how much energy does it store when it is compressed (i) twice as much and (ii) half as much? (b) In terms of \(x_{0},\) how much must it be compressed from its uncompressed length to store (i) twice as much energy and (ii) half as much energy?
Step-by-Step Solution
Verified Answer
(a)(i) 4U_0, (ii) \(\frac{1}{4}U_0\); (b)(i) \(\sqrt{2}x_0\), (ii) \(\frac{1}{\sqrt{2}}x_0\)
1Step 1: Understand Hooke's Law
A spring's potential energy when compressed or stretched is given by the equation \( U = \frac{1}{2}kx^2 \), where \( U \) is the potential energy, \( k \) is the spring constant, and \( x \) is the displacement from the equilibrium position.
2Step 2: Express the Given Potential Energy
From the problem, \( U_0 = \frac{1}{2}k x_0^2 \). This is the energy stored when the spring is compressed by \( x_0 \).
3Step 3: Calculate Energy for Double Compression
To find the energy stored when the spring is compressed twice as much, substitute \( 2x_0 \) into the energy equation: \( U = \frac{1}{2}k(2x_0)^2 = 4 \cdot \frac{1}{2}k x_0^2 = 4U_0 \). So, the energy stored is \( 4U_0 \).
4Step 4: Calculate Energy for Half Compression
To find the energy stored when the spring is compressed half as much, substitute \( \frac{1}{2}x_0 \) into the energy equation: \( U = \frac{1}{2}k(\frac{1}{2}x_0)^2 = \frac{1}{4} \cdot \frac{1}{2}k x_0^2 = \frac{1}{4}U_0 \). So, the energy stored is \( \frac{1}{4}U_0 \).
5Step 5: Find Compression for Double Energy
To have double the potential energy, \( 2U_0 = \frac{1}{2}kx^2 \). Setting this equal to the energy equation \( 2 \cdot \frac{1}{2}k x_0^2 \), we solve for \( x \) to get \[ x = \sqrt{2}x_0 \]. The spring must be compressed \( \sqrt{2}x_0 \) to store twice as much energy.
6Step 6: Find Compression for Half Energy
To have half the potential energy, \( \frac{1}{2}U_0 = \frac{1}{2}kx^2 \). Setting \( \frac{1}{2} \cdot \frac{1}{2}k x_0^2 \), solve for \( x \) to get \[ x = \frac{1}{\sqrt{2}}x_0 \]. The spring must be compressed \( \frac{1}{\sqrt{2}}x_0 \) to store half as much energy.
Key Concepts
Potential EnergySpring ConstantDisplacement from Equilibrium
Potential Energy
Springs are incredible devices when it comes to storing energy, and this stored energy is known as potential energy. In the context of springs, potential energy is the energy stored when a spring is compressed or stretched. This type of energy depends on how much you compress or extend the spring away from its resting position, known as equilibrium.
Understanding the equation for potential energy of a spring is key. It is given by the formula:
So, if you double the displacement, the energy doesn't just double; it grows by a factor of four! Similarly, halving the displacement reduces the energy to a quarter of its original value. This gives us insight into how sensitive the potential energy is to changes in displacement.
Understanding the equation for potential energy of a spring is key. It is given by the formula:
- \( U = \frac{1}{2}kx^2 \)
So, if you double the displacement, the energy doesn't just double; it grows by a factor of four! Similarly, halving the displacement reduces the energy to a quarter of its original value. This gives us insight into how sensitive the potential energy is to changes in displacement.
Spring Constant
The spring constant, often denoted as \( k \), is a fundamental concept when discussing springs within physics and engineering. It determines how stiff or flexible a spring is.
Think of the spring constant as a measure of toughness or rigidity. A higher value indicates a stiffer spring, which requires more force to achieve the same displacement compared to a spring with a lower \( k \).
The spring constant relates directly to how a spring stores potential energy:
Think of the spring constant as a measure of toughness or rigidity. A higher value indicates a stiffer spring, which requires more force to achieve the same displacement compared to a spring with a lower \( k \).
The spring constant relates directly to how a spring stores potential energy:
- The formula \( U = \frac{1}{2}kx^2 \) incorporates this constant.
- The potential energy is proportional to the spring constant.
Displacement from Equilibrium
Displacement from equilibrium is a simple yet powerful concept in understanding spring mechanics. It refers to how much the spring is either compressed or stretched from its natural, resting position.
This displacement, denoted by \( x \), plays a crucial role in determining the potential energy stored in a spring.
Moreover, this concept is vital in the module above, as varying the displacement directly affects how much energy is stored. Understanding this teaches us how mechanical systems behave and guides us in designing systems where springs are integral.
This displacement, denoted by \( x \), plays a crucial role in determining the potential energy stored in a spring.
- The relationship is expressed in the energy formula: \( U = \frac{1}{2}kx^2 \).
- This shows that potential energy increases with the square of the displacement.
Moreover, this concept is vital in the module above, as varying the displacement directly affects how much energy is stored. Understanding this teaches us how mechanical systems behave and guides us in designing systems where springs are integral.
Other exercises in this chapter
Problem 15
A force of 800 \(\mathrm{N}\) stretches a certain spring a distance of 0.200 \(\mathrm{m} .\) (a) What is the potential energy of the spring when it is stretche
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A spring of negligible mass has force constant \(k=\) 1600 \(\mathrm{N} / \mathrm{m}\) . (a) How far must the spring be compressed for 3.20 \(\mathrm{J}\) of po
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