/*! 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

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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 \( v_f = \sqrt{\frac{2K_f}{m}} \)

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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.

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