/*! 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 61 A conservative force \(\vec{F}\)... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A conservative force \(\vec{F}\) is in the \(+x\) -direction and has magnitude \(F(x)=\alpha /\left(x+x_{0}\right)^{2},\) where \(\alpha=0.800 \mathrm{~N} \cdot \mathrm{m}^{2}\) and \(x_{0}=0.200 \mathrm{~m}\). (a) What is the potential- energy function \(U(x)\) for this force? Let \(U(x) \rightarrow 0\) as \(x \rightarrow \infty .\) (b) An object with mass \(m=0.500 \mathrm{~kg}\) is released from rest at \(x=0\) and moves in the \(+x\) -direction. If \(\vec{F}\) is the only force acting on the object, what is the object's speed when it reaches \(x=0.400 \mathrm{~m} ?\)

Short Answer

Expert verified
The potential-energy function is \( U(x) = \frac{\alpha}{x + x_0}\) and the object's speed when it reaches \(x=0.400 \)m is calculated by \( v_f = \sqrt{\frac{2K_f}{m}} \)

Step by step solution

01

Derive the potential-energy function

From the concept of the potential energy associated with a conservative force, we know that it is given by \( U(x) = - \int F(x) dx \). So \( U(x) = - \int \frac{\alpha}{ (x+x_{0})^2 } dx = -\alpha \int \frac{1}{ (x+x_{0})^2 } dx\). This integral can be directly computed using the integral similarity \(\int \frac{1}{ x^2 } dx = - \frac{1}{x}\) and obtain \( U(x) = \frac{\alpha}{x + x_0} + C \) where C is the constant of integration.
02

Determine the constant of integration

Given that \( U(x) \rightarrow 0 \) as \( x \rightarrow \infty \), we can determine the constant of integration. Because the only term left does not go to zero, we conclude that the constant must be zero. So the potential-energy function we obtain is: \( U(x) = \frac{\alpha}{x + x_0}\)
03

Compute the initial potential energy

As the object starts from rest at \( x = 0 \), we can compute the initial potential energy, \( U_i = \frac{\alpha}{0 + x_0} = \frac{\alpha}{x_0}\)
04

Compute the final potential energy

When the object is at \( x = 0.400 \)m, we should calculate the final potential energy, \( U_f = \frac{\alpha}{0.400 + x_0}\)
05

Apply conservation of mechanical energy

From the conservation of mechanical energy, we know that the initial kinetic energy plus the initial potential energy equals the final kinetic energy plus the final potential energy. Given that the initial kinetic energy is zero (The object is released from rest), we find the final kinetic energy \( K_f = U_i - U_f \)
06

Calculate the object's speed

We know the final kinetic energy is \( K_f = \frac{1}{2}m v_f^2 \), where \( v_f \) is the object's speed at \( x = 0.400 \)m, and so we can solve for \( v_f = \sqrt{\frac{2K_f}{m}} \)

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.

Potential Energy Function
When discussing forces and motion within physics, one critical concept is the potential energy function, an expression that determines the amount of stored energy in a system. This energy can be thought of as 'potential' because it has the capacity to be converted into other forms of energy, such as kinetic energy, under the right circumstances.

In the context of the provided exercise, we have a conservative force, which means that the work done by the force moving an object from one point to another is independent of the path taken. Because of this, we can define a potential energy function that is solely dependent on the position, not the path. For a force \( \vec{F} \) with a given magnitude in the \(+x\) direction \( F(x)=\frac{\alpha}{(x+x_{0})^{2}} \) where \( \alpha=0.800 \, \text{N} \cdot \text{m}^{2} \) and \( x_{0}=0.200 \, \text{m} \) the potential-energy function is found through the negative integral of the force with respect to distance.

The integration results in the general formula for potential energy function \( U(x) = \frac{\alpha}{x+x_0} \) where \( x \) is the position of the object in the direction of the force. The constant of integration is determined by the boundary condition that when \( x \rightarrow \infty \) the potential energy \( U(x) \rightarrow 0 \) ensuring that at an infinite distance away, the energy stored in the system is zero.
Conservation of Mechanical Energy
The conservation of mechanical energy is a fundamental principle in physics that asserts the total mechanical energy in a closed system (one without external work or non-conservative forces like friction) remains constant. Mechanical energy itself is the sum of potential and kinetic energy in a system. Understanding this principle is key to solving many physics problems, especially those involving motion.

In our exercise, the conservation of mechanical energy is applied to determine the speed of an object released from a certain point under the influence of a conservative force. Since there are no non-conservative forces at play, mechanical energy will be conserved. Initially, the object has potential energy \( U_i \) and no kinetic energy since it’s at rest. As the object moves to some final position \( x_f = 0.400 \, \text{m} \) it gains kinetic energy while its potential energy decreases. The total mechanical energy (sum of potential and kinetic) at the initial and final positions remains constant. Through this, we can calculate the final kinetic energy \( K_f \) as the difference between initial and final potential energies, thus allowing us to solve for the object's speed.
Kinetic Energy Calculation
Kinetic energy is the energy of motion. It can be calculated if the mass \( m \) and the velocity \( v \) of an object are known using the formula \( K = \frac{1}{2}mv^2 \). Whenever an object moves with a certain velocity, it has kinetic energy proportional to its mass and the square of its velocity.

In the case of our exercise, the object is set into motion by a conservative force, and as it moves, its kinetic energy increases. By conserving mechanical energy, we deduced the kinetic energy at a particular position \( x \) to be \( K_f = U_i - U_f \), with the initial potential energy \( U_i \) and the final potential energy \( U_f \). The final step to find the object's speed is rearranging the kinetic energy formula to solve for velocity, \( v_f = \sqrt{\frac{2K_f}{m}} \). This equation provides the means to calculate the speed of the object at position \( x = 0.400 \, \text{m} \) by taking the square root of twice the final kinetic energy divided by its mass.

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 \(2.00 \mathrm{~kg}\) block is pushed against a spring with negligible mass and force constant \(k=400 \mathrm{~N} / \mathrm{m}\), compressing it \(0.220 \mathrm{~m}\). When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope \(37.0^{\circ}\) (Fig. \(\mathbf{P 7 . 4 0}\) ). (a) What is the speed of the block as it slides along the horizontal surface after having left the spring? (b) How far does the block travel up the incline before starting to slide back down?

BIO Human Energy vs. Insect Energy. For its size, the common flea is one of the most accomplished jumpers in the animal world. A 2.0 -mm-long, \(0.50 \mathrm{mg}\) flea can reach a height of \(20 \mathrm{~cm}\) in a single leap. (a) Ignoring air drag, what is the takeoff speed of such a flea? (b) Calculate the kinetic energy of this flea at takeoff and its kinetic energy per kilogram of mass. (c) If a \(65 \mathrm{~kg}, 2.0-\mathrm{m}\) -tall human could jump to the same height compared with his length as the flea jumps compared with its length, how high could the human jump, and what takeoff speed would the man need? (d) Most humans can jump no more than \(60 \mathrm{~cm}\) from a crouched start. What is the kinetic energy per kilogram of mass at takeoff for such a \(65 \mathrm{~kg}\) person? (e) Where does the flea store the energy that allows it to make sudden leaps?

While a roofer is working on a roof that slants at \(36^{\circ}\) above the horizontal, he accidentally nudges his \(85.0 \mathrm{~N}\) toolbox, causing it to start sliding downward from rest. If it starts \(4.25 \mathrm{~m}\) from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is \(22.0 \mathrm{~N}\) ?

\(\mathrm{A} 2.50 \mathrm{~kg}\) 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 wooden block with mass \(1.50 \mathrm{~kg}\) is placed against a compressed spring at the bottom of an incline of slope \(30.0^{\circ}\) (point \(A\) ). When the spring is released, it projects the block up the incline. At point \(B,\) a distance of \(6.00 \mathrm{~m}\) up the incline from \(A\), the block is moving up the incline at \(7.00 \mathrm{~m} / \mathrm{s}\) and is no longer in contact with the spring. The coefficient of kinetic friction between the block and the incline is \(\mu_{k}=0.50\) The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring.

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.