/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 A single conservative force acti... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A single conservative force acting on a particle varies as \(\mathbf{F}=\left(-A x+B x^{2}\right) \hat{\mathbf{i}} N,\) where \(A\) and \(B\) are constants and \(x\) is in meters. (a) Calculate the potential-energy function \(U(x)\) associated with this force, taking \(U=0\) at \(x=0 .\) (b) Find the change in potential energy and the change in kinetic energy as the particle moves from \(x=2.00 \mathrm{m}\) to \(x=3.00 \mathrm{m}\).

Short Answer

Expert verified
The potential energy function associated with the force is \(U=\frac{A}{2}x^2 - \frac{B}{3}x^3\). The change in potential energy as the particle moves from \(x=2.00 m\) to \(x=3.00 m\) is \(\frac{A}{2} + 3B\). The change in kinetic energy for the same motion is \(-\frac{A}{2} -3B\).

Step by step solution

01

Calculation of Potential Energy Function

The potential energy function \(U(x)\) associated with this force can be calculated by taking the integral of the force with respect to displacement, the potential energy \(U(x)\) is given by \(U=-\int F dx\). After integrating, you have \(U=-\int (-Ax + Bx^2) dx\), and since given \(U=0\) at \(x=0\), the integration constant equals to zero which gives the potential energy \(U(x)\) function as , \(U=\frac{A}{2}x^2 - \frac{B}{3}x^3\)
02

Find the change in potential energy

The change in potential energy as the particle moves from \(x=2.00 m\) to \(x=3.00 m\) is given by \(ΔU = U(3)-U(2)\), substituting \(x=3m\) and \(x=2m\) into \(U(x)\), and finally get \(ΔU = U(3)-U(2) = \frac{A}{2}(3)^2 - \frac{B}{3}(3)^3 - \left(\frac{A}{2}(2)^2 - \frac{B}{3}(2)^3\right) = \frac{A}{2} + 3B\)
03

Find the change in kinetic energy

The change in kinetic energy is equal to the negative of the change in potential energy (since no external forces doing work), which means \(ΔK = - ΔU\). Substituting the value of \(ΔU\) from step 2, you get \(ΔK = - ΔU = -(\frac{A}{2} + 3B)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Conservative Force
In the realm of physics, a conservative force is one where the work done in moving a particle between two points is independent of the path taken. This concept is pivotal because it simplifies the analysis of many physical systems. Gravity and spring forces are classic examples of conservative forces.

When dealing with a conservative force, any work done by or against the force results in a change in the particle's potential energy, commonly denoted as U(x). In the exercise provided, the force \( \mathbf{F}=(-Ax + Bx^{2}) \hat{\mathbf{i}} N \) varies with the position x, indicating that it is derived from a potential energy function. To calculate this function, one integrates the negative of the force with respect to displacement.

The absence of a path dependency in a conservative force field also ensures the possibility of defining a scalar potential energy function. This energy is stored within the system and can be converted to other forms, such as kinetic energy, without any loss, adhering to the law of conservation of energy.
Kinetic Energy
While potential energy is associated with the position of an object within a force field, kinetic energy represents the energy an object possesses due to its motion. It is formally defined as \( \frac{1}{2}mv^{2} \), where m is the mass of the object and v its velocity.

In the exercise, as the particle moves from one point to another under the influence of the conservative force, its kinetic energy changes. This alteration in kinetic energy, denoted by \( ΔK \), is directly related to the work done by the conservative force. According to the work-energy theorem, the change in kinetic energy is equivalent to the work done on the particle.

In the context of the given problem, the negative change in potential energy equates to the change in kinetic energy due to the work done by the force as the particle moves, confirming that energy within the system is merely transformed from one form to another, not created or destroyed.
Energy Conservation
The principle of energy conservation is one of the fundamental concepts in physics. It states that the total energy in an isolated system remains constant—energy can neither be created nor destroyed, but only transformed from one form to another. This principle is crucial in understanding the relationship between potential energy and kinetic energy.

In the exercise's scenario, as the particle transitions between different positions in space, its potential energy changes. This change in potential energy results in an equivalent but opposite change in kinetic energy, which keeps the total mechanical energy of the particle constant if no external work is done on the system. The exercise explicitly demonstrates this principle by showing that the change in kinetic energy \( ΔK \) is the negative of the change in potential energy \( ΔU \), which implies that the overall energy of the system is conserved.

The concept is significant in academic exercises and real-world applications, providing a foundational understanding that underpins much of classical and modern physics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A \(5.00-\mathrm{kg}\) block free to move on a horizontal, frictionless surface is attached to one end of a light horizontal spring. The other end of the spring is held fixed. The spring is compressed \(0.100 \mathrm{m}\) from equilibrium and released. The speed of the block is \(1.20 \mathrm{m} / \mathrm{s}\) when it passes the equilibrium position of the spring. The same experiment is now repeated with the frictionless surface replaced by a surface for which the coefficient of kinetic friction is \(0.300 .\) Determine the speed of the block at the equilibrium position of the spring.

The potential energy of a system of two particles separated by a distance \(r\) is given by \(U(r)=A / r,\) where \(A\) is a constant. Find the radial force \(\mathbf{F}_{r}\) that each particle exerts on the other.

At 11: 00 A.M. on September \(7,2001,\) more than 1 million British school children jumped up and down for one minute. The curriculum focus of the "Giant Jump" was on earthquakes, but it was integrated with many other topics, such as exercise, geography, cooperation, testing hypotheses, and setting world records. Children built their own seismographs, which registered local effects. (a) Find the mechanical energy released in the experiment. Assume that 1050 000 children of average mass \(36.0 \mathrm{kg}\) jump twelve times each, raising their centers of mass by \(25.0 \mathrm{cm}\) each time and briefly resting between one jump and the next. The free-fall acceleration in Britain is \(9.81 \mathrm{m} / \mathrm{s}^{2},\) (b) Most of the energy is converted very rapidly into internal energy within the bodies of the children and the floors of the school buildings. Of the energy that propagates into the ground, most produces high-frequency "microtremor" vibrations that are rapidly damped and cannot travel far. Assume that \(0.01 \%\) of the energy is carried away by a long-range seismic wave. The magnitude of an earthquake on the Richter scale is given by $$ M=\frac{\log E-4.8}{1.5} $$ where \(E\) is the seismic wave energy in joules. According to this model, what is the magnitude of the demonstration quake? (It did not register above background noise overseas or on the seismograph of the Wolverton Seismic Vault, Hampshire. \()\)

Dave Johnson, the bronze medalist at the 1992 Olympic decathlon in Barcelona, leaves the ground at the high jump with vertical velocity component \(6.00 \mathrm{m} / \mathrm{s}\). How far does his center of mass move up as he makes the jump?

A single conservative force acts on a \(5.00-\mathrm{kg}\) particle. The equation \(F_{x}=(2 x+4)\) N describes the force, where \(x\) is in meters. As the particle moves along the \(x\) axis from \(x=1.00 \mathrm{m}\) to \(x=5.00 \mathrm{m},\) calculate (a) the work done by this force, (b) the change in the potential energy of the system, and \((\mathrm{c})\) the kinetic energy of the particle at \(x=5.00 \mathrm{m}\) if its speed is \(3.00 \mathrm{m} / \mathrm{s}\) at \(x=1.00 \mathrm{m}\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.