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Chapter 7: Problem 17

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?

Short Answer

Expert verified
(a)(i) 4U_0, (ii) \(\frac{1}{4}U_0\); (b)(i) \(\sqrt{2}x_0\), (ii) \(\frac{1}{\sqrt{2}}x_0\)

Step by step solution

01

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.
02

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 \( x_0 \).
03

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 \( 4U_0 \).
04

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 \( \frac{1}{4}U_0 \).
05

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.
06

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.

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Key Concepts

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

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:
  • \( U = \frac{1}{2}kx^2 \)
Here, \( U \) is the potential energy, \( k \) is the spring constant, and \( x \) is the displacement from equilibrium. This formulation shows that the potential energy stored in a spring is directly proportional to the square of the displacement.
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:
  • The formula \( U = \frac{1}{2}kx^2 \) incorporates this constant.
  • The potential energy is proportional to the spring constant.
So, if a spring's constant is high, even small displacements will result in greater potential energy storage. In practical terms, choosing a spring with the right spring constant is crucial depending on the task. Whether cushioning impacts in cars or determining vibrations in electronic devices, the spring constant is a key player.
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.
  • 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.
For instance, if you increase the displacement from the equilibrium, the spring stores more potential energy. This stored energy can be calculated by substituting the new displacement into the potential energy formula.
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.

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Most popular questions from this chapter

Bungee Jump. A bungee cord is 30.0 \(\mathrm{m}\) long and, when stretched a distance \(x,\) it exerts a restoring force of magnitude \(k x\) Your father-in- law (mass 95.0 \(\mathrm{kg}\) ) stands on a platform 45.0 \(\mathrm{m}\) above the ground, and one end of the cord is tied securely to his ankle and the other end to the platform. You have promised him that when he steps off the platform he will fall a maximum distance of only 41.0 \(\mathrm{m}\) before the cord stops him. You had several bungee cords to select from, and you tested them by stretching them out, tying one end to a tree, and pulling on the other end with a force of 380.0 \(\mathrm{N}\) . When you do this, what distance will the bungee cord that you should select have stretched?

Up and Down the Hill. A \(28-\mathrm{kg}\) rock approaches the foot of a hill with a speed of 15 \(\mathrm{m} / \mathrm{s}\) . This hill slopes upward at a constant angle of \(40.0^{\circ}\) above the horizontal. The coefficients of static and kinetic friction between the hill and the rock are 0.75 and 0.20 , respectively. (a) Use energy conservation to find the maximum height above the foot of the hill reached by the rock. (b) Will the rock remain at rest at its highest point, or will it slide back down the hill? (c) If the rock does slide back down, find its speed when it returns to the bottom of the hill.

Bouncing Ball. A 650 -gram rubber ball is dropped from an initial height of \(250 \mathrm{m},\) and on each bounce it returns to 75\(\%\) of its previous height. (a) What is the initial mechanical energy of the ball, just after it is released from its initial height? (b) How much mechanical energy does the ball lose during its first bounce? What happens to this energy? (c) How much mechanical energy is lost during the second bounce?

A \(120-\mathrm{kg}\) mail bag hangs by a vertical rope 3.5 \(\mathrm{m}\) long. \(\mathrm{A}\) postal worker then displaces the bag to a position 2.0 \(\mathrm{m}\) sideways from its original position, always keeping the rope taut. (a) What horizontal force is necessary to hold the bag in the new position? (b) As the bag is moved to this position, how much work is done (i) by the rope and (ii) by the worker?

When it is burned, 1 gallon of gasoline produces \(1.3 \times 10^{8} \mathrm{J}\) of energy. A \(1500-\mathrm{kg}\) car accelerates from rest to 37 \(\mathrm{m} / \mathrm{s}\) in 10 \(\mathrm{s}\) . The engine of this car is only 15\(\%\) efficient (which is typical), meaning that only 15\(\%\) of the energy from the combustion of the gasoline is used to accelerate the car. The rest goes into things like the internal kinetic energy of the engine parts as well as heating of the exhaust air and engine. (a) How many gallons of gasoline does this car use during the acceleration? (b) How many such accelerations will it take to burn up 1 gallon of gas?
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