Problem 61 - University Physics with Modern Physics

Question

A conservative force \(\overrightarrow{F}\) is in the \(+x\)-direction and has magnitude \(F(x) = a/(x + x_0)^2\), where \(\alpha = 0.800\) N \(\cdot\) m\(^2\) and \(x_0 = 0.200\) 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 \(m = 0.500\) kg is released from rest at \(x = 0\) and moves in the \(+x\)-direction. If \(\overrightarrow{F}\) is the only force acting on the object, what is the object's speed when it reaches \(x = 0.400\) m?

Step-by-Step Solution

Answer

(a) \( U(x) = \frac{0.800}{x + 0.200} \). (b) Speed at \( x = 0.400 \) m is \( 3.27 \) m/s.

Step 1: Understanding the Force

The force given is \( F(x) = \frac{a}{(x + x_0)^2} \), with \(a = 0.800\) N·m² and \(x_0 = 0.200\) m. This is a conservative force, meaning there exists a potential energy function \( U(x) \) associated with it.

Step 2: Determine the Potential Energy Function

The potential energy function \( U(x) \) for a conservative force can be determined from the relation \( F(x) = -\frac{dU}{dx} \). To find \( U(x) \), integrate \( F(x) \) w.r.t. \( x \) with the condition \( U(x) \rightarrow 0 \) as \( x \rightarrow \infty \).

Step 3: Integrate to Find U(x)

Integrate \( F(x) = \frac{a}{(x + x_0)^2} \) to find \( U(x) \).\[ U(x) = -\int F(x) \, dx = -\int \frac{a}{(x + x_0)^2} \, dx \]The integral is:\[ U(x) = \frac{a}{x+x_0} + C \]Given \( U(x) \rightarrow 0 \) as \( x \rightarrow \infty \), \( C = 0 \) because as \( x \rightarrow \infty \), \( \frac{a}{x + x_0} \rightarrow 0 \). Thus, \( U(x) = \frac{a}{x + x_0} \).

Step 4: Calculate U(x) at x = 0 and x = 0.400

We need the potential energy values at the initial and final positions. Calculate:At \( x = 0 \):\[ U(0) = \frac{0.800}{0 + 0.200} = 4.0 \, \text{J} \]At \( x = 0.400 \):\[ U(0.400) = \frac{0.800}{0.400 + 0.200} = 1.333 \, \text{J} \]

Step 5: Apply Conservation of Energy

Since the force is conservative and the object is released from rest, use the conservation of energy principle:\[ \frac{1}{2}mv^2 + U(x) = U(x=0) \]Substitute known values to find \( v \):\[ 0 + 4.0 = \frac{1}{2}(0.500)v^2 + 1.333 \]\[ 4.0 = 0.25v^2 + 1.333 \]\[ 2.667 = 0.25v^2 \]\[ v^2 = \frac{2.667}{0.25} = 10.668 \]\[ v = \sqrt{10.668} \approx 3.27 \, \text{m/s} \]

Step 6: Conclusion: Speed at x = 0.400

The object's speed when it reaches \( x = 0.400 \) m is approximately 3.27 m/s.

Key Concepts

Potential Energy

Potential energy, often denoted as \( U(x) \), is a form of energy that is stored in an object due to its position within a force field, like a gravitational or electromagnetic field. In the realm of physics, particularly when discussing conservative forces, potential energy is crucial.

Conservative forces, such as gravity or the electrostatic force, have potential energy associated with them. The neat part about these forces is that the total mechanical energy (kinetic plus potential) of a closed system remains constant. This means you can transform potential to kinetic energy and vice versa, but the overall 'energy count' stays the same.

For this exercise, the potential energy function \( U(x) \) is derived from the conservative force \( F(x) = -\frac{dU}{dx} \). This relationship tells us that the force is the negative gradient of the potential energy, basically showing how energy changes spatially.

The solution finds \( U(x) = \frac{a}{x+x_0} \), ensuring that \( U(x) \rightarrow 0 \) as \( x \rightarrow \infty \). This represents a standard approach for calculating potential energy in systems driven by conservative forces.

Integration

Integration is a fundamental concept in calculus used to find accumulations and the 'against the grain' measure, such as areas under curves or finding potential energy from force functions.

In the context of this problem, integrating the force function \( F(x) = \frac{a}{(x + x_0)^2} \) was used to derive the potential energy function. The integration is an antidote to differentiation. If differentiation tells you the rate of change of a function, integration essentially compiles these little changes to describe the whole picture.

The integral \( -\int \frac{a}{(x + x_0)^2} \, dx \) was solved to find \( U(x) = \frac{a}{x + x_0} \), leaving us with the potential energy function. Understanding this process requires seeing how integration can "reverse" the calculus act, moving backward from a force function to a potential energy graph.

Conservation of Energy

The principle of the conservation of energy asserts that in an isolated system, the total energy remains constant. Energy cannot be created or destroyed, only transformed from one form to another.

In this exercise, the idea of conservation of energy plays a key role. Initially, all energy is stored as potential energy because the object is at rest. When the object moves due to a conservative force, this stored (potential) energy converts into kinetic energy, which is the energy of motion.

Using the conservation of energy principle, \( \frac{1}{2}mv^2 + U(x) = U(x=0) \), the object's speed was found when it reaches \( x = 0.400 \) m. Here, the change in potential energy was transformed into kinetic energy, allowing us to calculate the velocity and showing energy transformation in action.

Kinematics

Kinematics is a branch of physics that describes the motion of objects without considering the forces that cause these motions. It's all about tracking position, velocity, and acceleration.

In this problem, kinematics isn't explored in depth, but it underlines the context where you need to find the velocity of the object.

The object is initially released from rest, so its initial velocity is zero. When it is allowed to move under the conservative force, its velocity changes. By using kinematic principles like the equations of motion, coupled with energy considerations, we find out how fast it goes. The equation \( v = \sqrt{10.668} \approx 3.27 \, \text{m/s} \) gives us this speed, showing the vital link between forces, energy, and motion.

Kinematics provides the vocabulary for these calculations, even when the primary focus remains on energy transformations.