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An ideal 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 the spring store when it is compressed (i) twice as much and (ii) half as much? (b) In terms of \(x_{0}\). how much must the spring 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) The potential energy is 4 times of \( U_{0} \) when it is compressed twice as much. (ii) The potential energy is a quarter of \( U_{0} \) when it is compressed half as much. (b) (i) The compression distance required to store twice of \( U_{0} \) energy is twice of \( x_{0} \). (ii) The compression distance required to store half of \( U_{0} \) energy is the same as \( x_{0} \).

Step by step solution

01

Calculate Energy for Double Compression

For the spring that is compressed twice as much, the distance is \( 2x_{0} \). Thus, the energy is calculated as \( U = 0.5k(2x_{0})^2 = 0.5k4x_{0}^2 = 2kx_{0}^2 = 4U_{0} \). The potential energy is 4 times of \( U_{0} \) when it is compressed twice as much.
02

Calculate Energy for Half Compression

For the spring compressed half as much, the distance is \( 0.5x_{0} \). Thus, the energy is calculated as \( U = 0.5k(0.5x_{0})^2 = 0.5k(0.25x_{0}^2) = 0.125kx_{0}^2 = 0.25U_{0} \). The potential energy is a quarter of \( U_{0} \) when it is compressed half as much.
03

Calculate Compression for Double Energy

For the spring to store twice as much energy, the compression distance is calculated by equating \( 2U_{0} \) to the potential energy formula. Thus, \( 2U_{0} = 0.5kx^2 \). From which, \( x = sqrt(4U_{0}/k) = sqrt(4)x_{0} = 2x_{0} \). The compression distance required to store twice of \( U_{0} \) energy is twice of \( x_{0} \).
04

Calculate Compression for Half Energy

For the spring to store half as much energy, the compression distance is calculated by equating \( 0.5U_{0} \) to the potential energy formula. Thus, \( 0.5U_{0} = 0.5kx^2 \). From which, \( x = sqrt(U_{0}/k) = x_{0} \). The compression distance required to store half of \( U_{0} \) energy is the same as \( x_{0} \).

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Key Concepts

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

Hooke's Law
When we talk about springs and their potential energy, one fundamental principle comes to mind: Hooke's Law. This law is named after 17th-century British physicist Robert Hooke, who discovered that the force needed to stretch or compress a spring is directly proportional to the distance it's stretched or compressed.

Mathematically, we express this as:
\[ F = -kx \],
where \( F \) is the force applied to the spring, \( k \) is the spring constant that measures the stiffness of the spring, and \( x \) is the displacement from the spring's original length. The negative sign indicates that the direction of the force is opposite to the displacement - meaning the spring resists the stretch or compression.

In our original exercise, we've applied Hooke's Law to determine the relationship between the spring's compression and the potential energy stored. Understanding this correlation allows us to answer more complex questions, such as how much energy a spring will store under varying degrees of compression.
Elastic Potential Energy
Elastic potential energy is the kind of energy stored in elastic materials as a result of their stretching or compressing. Springs are a perfect example of an object that can store this type of energy, which is directly linked to Hooke's Law.

The equation for elastic potential energy in a spring is given by:
\[ U = \frac{1}{2}kx^2 \],
where \( U \) represents the elastic potential energy, \( k \) is the spring constant, and \( x \) is the displacement of the spring from its equilibrium position.

The step-by-step solution provided in the exercise clearly demonstrates this relationship. For example, when the spring is compressed to half its original displacement (\( x_{0} \)), the potential energy becomes a quarter of what it was at full displacement, because the equation includes the square of the displacement (\( x^2 \)). On the contrary, compressing it twice as much results in four times the energy, showing the quadratic relationship between displacement and stored energy.
Mechanical Energy Conservation
The principle of conservation of mechanical energy tells us that in a closed system, where only conservative forces (like gravity or spring force) act, the total mechanical energy remains constant. This total includes both the potential energy, like what's stored in a spring, and the kinetic energy of moving objects.

For a spring system, as the spring goes from compressed to released, the stored elastic potential energy converts into kinetic energy of whatever mass is attached to the spring, and vice versa. This energy transformation continues perpetually in an ideal system with no energy losses.

Understanding the conservation of mechanical energy aids in analyzing systems like roller coasters, pendulums, or indeed, springs. Our exercise suggests analyzing how much a spring must be compressed to store certain amounts of energy, and this involves applying the concept of mechanical energy conservation to predict the system's behavior without actually observing it in action.
Energy Transformation
Energy transformation is all about the conversion of energy from one form to another. In the context of spring systems, we often see the transformation between elastic potential energy and kinetic energy.

When a spring is compressed, as in our textbook example, we're doing work on the spring and thus inputting energy into the system. This energy is stored as elastic potential energy. Once the spring is released, that stored energy is transformed into kinetic energy as it pushes back to return to its original shape, potentially causing an attached mass to move.

In a real-world scenario, these transformations are not entirely efficient—some energy is always lost to the environment as heat due to friction. However, in ideal textbook problems like ours, we assume perfect energy conversions to simplify the learning and problem-solving process for students. These examples lay the groundwork for understanding more complex energy interactions in the physical world.

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Most popular questions from this chapter

\(\mathrm{A} 2.50 \mathrm{~kg}\) mass is pushed against a horizontal spring of force constant \(25.0 \mathrm{~N} / \mathrm{cm}\) on a frictionless air table. The spring is attached to the tabletop, and the mass is not attached to the spring in any way. When the spring has been compressed enough to store \(11.5 \mathrm{~J}\) of potential energy in it, the mass is suddenly released from rest. (a) Find the greatest speed the mass reaches. When does this occur? (b) What is the greatest acceleration of the mass, and when does it occur?

A 1.20 kg piece of cheese is placed on a vertical spring of negligible mass and force constant \(k=1800 \mathrm{~N} / \mathrm{m}\) that is compressed \(15.0 \mathrm{~cm} .\) When the spring is released, how high does the cheese rise from this initial position? (The cheese and the spring are not attached.)

A small rock with mass \(m\) is released from rest at the inside rim of a large, hemispherical bowl (point \(A\) ) that has radius \(R\), as shown in Fig. E7.9. If the normal force exerted on the rock as it slides through its lowest point (point \(B\) ) is twice the weight of the rock, how much work did friction do on the rock as it moved from \(A\) to \(B ?\) Express your answer in terms of \(m, R,\) and \(g\)

Two blocks are attached to either end of a light rope that passes over a light, frictionless pulley suspended from the ceiling. One block has mass \(8.00 \mathrm{~kg},\) and the other has mass \(6.00 \mathrm{~kg}\). The blocks are released from rest. (a) For a \(0.200 \mathrm{~m}\) downward displacement of the \(8.00 \mathrm{~kg}\) block, what is the change in the gravitational potential energy associated with each block? (b) If the tension in the rope is \(T\), how much work is done on each block by the rope? (c) Apply conservation of energy to the system that includes both blocks. During the \(0.200 \mathrm{~m}\) downward displacement, what is the total work done on the system by the tension in the rope? What is the change in gravitational potcntial energy associated with the system? Use energy conservation to find the speed of the \(8.00 \mathrm{~kg}\) block after it has descended \(0.200 \mathrm{~m} .\)

CALC The potential energy of two atoms in a diatomic molecule is approximated by \(U(r)=\left(a / r^{12}\right)-\left(b / r^{6}\right),\) where \(r\) is the spacing between atoms and \(a\) and \(b\) are positive constants. (a) Find the force \(F(r)\) on one atom as a function of \(r .\) Draw two graphs: one of \(U(r)\) versus \(r\) and one of \(F(r)\) versus \(r\). (b) Find the equilibrium distance between the two atoms. Is this equilibrium stable? (c) Suppose the distance between the two atoms is equal to the equilibrium distance found in part (b). What minimum energy must be added to the molecule to dissociate it - that is, to separate the two atoms to an infinite distance apart? This is called the dissociation energy of the molecule. (d) For the molecule CO, the cquilibrium distance between the carbon and oxygen atoms is \(1.13 \times 10^{-10} \mathrm{~m}\) and the dissociation energy is \(1.54 \times 10^{-18} \mathrm{~J}\) per molecule. Find the values of the constants \(a\) and \(b\).

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