91Ó°ÊÓ

A particle can move along only an \(x\) axis, where conservative forces act on it (Fig. 8-66 and the following table). The particle is released at \(x=5.00 \mathrm{~m}\) with a kinetic energy of \(K=14.0 \mathrm{~J}\) and \(\mathrm{a}\) potential energy of \(U=0\). If its motion is in the negative direction of the \(x\) axis, what are its (a) \(K\) and (b) \(U\) at \(x=2.00 \mathrm{~m}\) and its (c) \(K\) and \((\) d \() U\) at \(x=0\) ? If its motion is in the positive direction of the \(x\) axis, what are its (e) \(K\) and (f) \(U\) at \(x=11.0 \mathrm{~m}\), its \((\mathrm{g}) \mathrm{K}\) and (h) \(U\) at \(x=12.0 \mathrm{~m}\), and its (i) \(K\) and \((\mathrm{j}) U\) at \(x=13.0 \mathrm{~m} ?(\mathrm{k})\) Plot \(U(x)\) versus \(x\) for the range \(x=0\) to \(x=13.0 \mathrm{~m}\). Next, the particle is released from rest at \(x=0\). What are (l) its kinetic energy at \(x=5.0 \mathrm{~m}\) and \((\mathrm{m})\) the maximum positive position \(x_{\max }\) it reaches? (n) What does the particle do after it reaches \(x_{\max } ?\) $$ \begin{array}{lc} \hline {\text { Range }} & {\text { Force }} \\ \hline 0 \text { to } 2.00 \mathrm{~m} & \vec{F}_{1}=+(3.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 2.00 \mathrm{~m} \text { to } 3.00 \mathrm{~m} & \vec{F}_{2}=+(5.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 3.00 \mathrm{~m} \text { to } 8.00 \mathrm{~m} & F=0 \\ 8.00 \mathrm{~m} \text { to } 11.0 \mathrm{~m} & \vec{F}_{3}=-(4.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 11.0 \mathrm{~m} \text { to } 12.0 \mathrm{~m} & \vec{F}_{4}=-(1.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 12.0 \mathrm{~m} \text { to } 15.0 \mathrm{~m} & F=0 \\ \hline \end{array} $$

Step by step solution

01

Understanding the Problem

The problem provides information about a particle that moves along the x-axis subject to different conservative forces over specified regions. The task is to calculate the kinetic energy (K) and potential energy (U) of the particle at different positions along the x-axis, based on the changes in these energies due to the work done by these forces. We'll solve for the energies at each specified position, taking into account the direction of the particle's motion.

02

Calculate Potential Energy Function U(x)

We can determine the potential energy as a function of position x by integrating the force F over each region. The potential energy U is obtained from the work done by the force: \Forces are given for different intervals: - From 0 to 2 m, \( F_1 = +(3.00 \ \text{N}) \)- From 2 to 3 m, \( F_2 = +(5.00 \ \text{N}) \)- From 3 to 8 m, \( F = 0 \)- From 8 to 11 m, \( F_3 = -(4.00 \ \text{N}) \)- From 11 to 12 m, \( F_4 = -(1.00 \ \text{N}) \)The potential energy change is \( \Delta U = - \int F \, dx \). We need to determine \( U(x) \) for each region.

03

Calculate U(x) in each Region

1. For 0 to 2 m: The work done by \( F_1 \) is \( W_1 = -\int_0^x 3 \, dx = -3x \). Therefore, \( U(x) = 3x \).2. For 2 to 3 m: The work done is \( W_2 = -\int_2^x 5 \, dx = -5(x - 2) \). Therefore, \( U(x) = 6 + 5(x - 2) \).3. For 3 to 8 m: Since \( F = 0 \), \( U(x) = \text{constant} = 11 \).4. For 8 to 11 m: \( W_3 = -\int_8^x (-4) \, dx = 4(x - 8) \). Thus, \( U(x) = 11 - 4(x - 8) \).5. For 11 to 12 m: \( W_4 = -\int_{11}^x (-1) \, dx = 1(x - 11) \). Thus, \( U(x) = -1 + x - 11 \).

04

Energy Conservation and Calculating Kinetic Energies

Using the conservation of mechanical energy \( K + U = \text{constant} = 14 \). We'll use this to find K and U at various points:- At \( x = 2.00 \mathrm{~m} \): \( U(2.00) = 6 \), thus \( K = 14 - 6 = 8 \).- At \( x = 0 \): \( U(0) = 0 \), so \( K = 14 - 0 = 14 \).- In the positive x-direction, at \( x=11.0 \mathrm{~m} \): Since \( U(11.0) = 1 \), \( K = 14 - 1 = 13 \).- At \( x=12.0 \mathrm{~m} \): \( U(12.0) = 0 \), \( K = 14 \).- At \( x=13.0 \mathrm{~m} \): \( U(13.0) = 1 \), \( K = 13 \).

05

Plot Potential Energy Function U(x)

Plot the potential energy function \( U(x) \) versus \( x \) from \( x = 0 \) to \( x = 13.0 \mathrm{~m} \). The plot should show the changes in potential energy at the specified regions as derived from the integration of force expressions in Step 3.

06

Determine Kinetic Energy at x=5.0 m (Released from Rest at x=0)

When the particle is released from rest at \( x=0 \), the kinetic energy at \( x = 5.0 \mathrm{~m} \) can be found using initial conditions: \( K(0) = 0 \) and \( U(5) = \text{from plot or calculation} \). Since \( U(5) \) is known and \( U(0) = 0 \), we calculate \( K(5) = U(0) - U(5) \).

07

Determine Maximum Position x_max and Behavior Beyond x_max

To find \( x_{\max} \), identify the farthest position where the total mechanical energy equals the potential energy \( U(x) \). Beyond this point, the particle will reverse its motion or stop depending on the total energy being equal to the maximum potential energy available.

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Key Concepts

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

Kinetic Energy

Kinetic energy is the energy that a particle possesses due to its motion. It is given by the formula \( K = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity of the particle. In the context of the exercise, kinetic energy varies as the particle moves along the x-axis because of the work done by conservative forces.
- At the starting point \( x = 5.00 \text{ m} \), the kinetic energy \( K \) is given as \( 14.0 \text{ J} \). This establishes a baseline for calculating changes in kinetic energy.
- As the particle progresses through different sections governed by varying forces, its velocity and thus kinetic energy change.
- When the particle's kinetic energy at different positions is needed, it is recalculated using the relation mentioned in the exercise: \( K(x) = \text{constant} - U(x) \). This constant represents the total mechanical energy, and \( U(x) \) represents the potential energy at position \( x \).
Understanding how kinetic energy interacts with potential energy and conservative forces is key to analyzing this system.

Potential Energy

Potential energy refers to the stored energy of a system due to its position or configuration. For conservative forces, the change in potential energy is directly related to the work done by or against those forces.
- The potential energy function \( U(x) \) indicates how energy is stored in the system as the particle moves. Integrating the force over distance gives this function, showing how potential energy changes.
- For example, between \( x = 0 \) and \( x = 2.00 \text{ m} \), potential energy changes are calculated using \( U(x) = 3x \) due to a constant force of \( +3 \text{ N} \).
- In sections where the force is zero, such as from \( x = 3.00 \text{ m} \) to \( x = 8.00 \text{ m} \), the potential energy remains constant, demonstrating that no work is done by or against conservative forces.
By understanding this concept, one can determine the energy storage capabilities of the system depending on its state.

Mechanical Energy Conservation

Mechanical energy conservation is the principle that in a closed system subject to conservative forces, total mechanical energy remains constant. This is an integral part of analyzing particle motion along the x-axis in this exercise.
- The sum of kinetic and potential energy at any point along the particle's path should equal the initial total mechanical energy: \( K + U = \text{constant} \). In this problem, the constant is initially given as \( 14 \text{ J} \).
- As the particle moves, its kinetic and potential energies transform into each other while their sum remains fixed, assuming no non-conservative forces like friction are present.
- For practical calculations, when a particle transitions from one energy state to another, its kinetic energy increases if potential energy decreases and vice versa.
Understanding mechanical energy conservation helps in predicting and calculating the motion and behavior of particles in conservative fields. It assists in determining points like maximum displacement and analyzing how particles behave over time.