/*! 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 41 A single conservative force acts... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 41

A single conservative force acts on a \(5.00-\mathrm{kg}\) particle. The equation \(F_{x}=(2 x+4) \mathrm{N}\) 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 \((c)\) the kinetic energy of the particle at \(x=5.00 \mathrm{m}\) if its speed is 3.00 \(\mathrm{m} / \mathrm{s}\) at \(x=1.00 \mathrm{m} .\)

Short Answer

Expert verified
The work done by the force is 72 Joules, the change in potential energy is -72 Joules, and the kinetic energy at x = 5.00 m is the sum of the initial kinetic energy (22.5 Joules) and the work done (72 Joules), resulting in 94.5 Joules.

Step by step solution

01

Calculate the Work Done by the Force

Work done by a variable force along a straight line is given by the integral of the force with respect to displacement. Compute the integral of the given force function with respect to x from the initial position to the final position:\[ W = \text{Work done} = \text{int}_{x_1 = 1.00 \, \text{m}}^{x_2 = 5.00 \, \text{m}} F_x dx = \text{int}_{1}^{5} (2x + 4) dx \]
02

Evaluate the Work Integral

Evaluate the definite integral to find the work:\[ W = \text{int}_{1}^{5} (2x + 4) dx = [x^2 + 4x]_{1}^{5} = (5^2 + 4\cdot5) - (1^2 + 4\cdot1) \]
03

Calculate the Change in Potential Energy

Since the force is conservative, the work done by the force is equal to the negative change in potential energy (conservation of energy principle).\[ \Delta U = -W \]
04

Determine the Kinetic Energy at x = 5.00 m

Using the work-energy theorem, the work done is equal to the change in kinetic energy.\[ \Delta KE = KE_{x=5} - KE_{x=1} \]Find the initial kinetic energy at x = 1.00 m: \[ KE_{x=1} = \frac{1}{2} m v^2 = \frac{1}{2} \cdot 5 \text{kg} \cdot (3 \text{m/s})^2 \]Then, use the work done to find the final kinetic energy at x = 5.00 m: \[ KE_{x=5} = KE_{x=1} + W \]

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 91影视!

Key Concepts

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

Conservative Force
The concept of a conservative force is pivotal in understanding various energy transformations within a physical system. A conservative force is a special type of force that has the property of conserving mechanical energy. This means that the work done by a conservative force on an object moving from one point to another is independent of the path taken; it only depends on the initial and final positions of the object.

For example, gravitational and electrostatic forces are conservative. In the context of the textbook exercise, the force given by the equation \( F_{x}=(2x+4) \, \mathrm{N} \) is also conservative. This is proven by the fact that the work done by this force can be calculated using a simple integral which only requires the knowledge of the starting and ending points.

A key result of the presence of a conservative force is that the total mechanical energy (sum of potential and kinetic energy) of a system remains constant if only conservative forces are doing work. When a conservative force does work, it results in a transfer of energy between potential and kinetic forms without any loss to the system.
Potential Energy
Potential energy refers to the stored energy an object has by virtue of its position in a force field, such as the gravitational field of the Earth, or an elastic force within a spring. This type of energy is a scalar quantity and is denoted typically by \( U \) or \( PE \).

In the given exercise, when the conservative force acts on the particle as it moves from \( x=1.00 \, \mathrm{m} \) to \( x=5.00 \, \mathrm{m} \), the potential energy of the system changes. According to the work-energy principle, the work done by this force on the particle is equal to the negative change in potential energy, expressed as \( \Delta U = -W \).

This equation emphasizes that if the work done is positive, the potential energy decreases, as the particle would be moving in the direction of the force. Conversely, if the work done is negative, it indicates that the potential energy has increased because the particle moved against the force.
Kinetic Energy
Kinetic energy is the energy associated with the motion of an object. Mathematically, it is defined for a particle of mass \( m \) and velocity \( v \) as \( KE = \frac{1}{2} m v^2 \). Kinetic energy is a form of mechanical energy and is also a scalar quantity.

The textbook exercise involves calculating the kinetic energy of a particle at two different positions along the x-axis. Understanding that kinetic energy represents the capability of a moving object to do work due to its velocity is crucial here. As part of the problem solution, we calculate the initial kinetic energy of the particle using its initial speed. Then, by applying the work-energy theorem which states that the work done by all forces acting on an object is equal to the change in kinetic energy, \( \Delta KE = KE_{x=5} - KE_{x=1} \), we can find the kinetic energy at the final position.

It is important to note that the work-energy theorem is a restatement of the law of conservation of energy, and in this scenario points towards the intimate relationship between the work done by the conservative force and the change in kinetic energy of the particle.

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

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 1050000 children of average mass 36.0 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 over-seas or on the seismograph of the Wolverton Seismic Vault, Hampshire.)

A block of mass 0.500 \(\mathrm{kg}\) is pushed against a horizontal spring of negligible mass until the spring is compressed a distance \(x\) (Fig. P 8.61\()\) . The force constant of the spring is 450 \(\mathrm{N} / \mathrm{m}\) . When it is released, the block travels along a frictionless, horizontal surface to point \(B,\) the bottom of a vertical circular track of radius \(R=1.00 \mathrm{m},\) and continues to move up the track. The speed of the block at the bottom of the track is \(v_{B}=12.0 \mathrm{m} / \mathrm{s},\) and the block experiences an average friction force of 7.00 \(\mathrm{N}\) while sliding up the track. (a) What is \(x ?\) (b) What speed do you predict for the block at the top of the track? (c) Does the block actually reach the top of the track, or does it fall off before reaching the top?

A ball whirls around in a vertical circle at the end of a string. If the total energy of the ball-Earth system remains constant, show that the tension in the string at the bottom is greater than the tension at the top by six times the weight of the ball.

A potential-energy function for a two-dimensional force is of the form \(U=3 x^{3} y-7 x\) . Find the force that acts at the point \((x, y) .\)

At time \(t_{i}\) , the kinetic energy of a particle is 30.0 \(\mathrm{J}\) and the potential energy of the system to which it belongs is 10.0 \(\mathrm{J}\) . At some later time \(t\) , the kinetic energy of the particle is 18.0 \(\mathrm{J}\) . (a) If only conservative forces act on the particle, what are the potential energy and the total energy at time \(t_{f} ?\) (b) If the potential energy of the system at time \(t_{f}\) is 5.00 \(\mathrm{J}\) , are there any nonconservative forces acting on the particle? Explain.
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Fields in Physics

Read Explanation

Physics of Motion

Read Explanation

Rotational Dynamics

Read Explanation

Translational Dynamics

Read Explanation

Classical Mechanics

Read Explanation
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.