Question

Problem 2.7 is about a class of one-dimensional problems that can always be reduced to doing an integral. Here is another. Show that if the net force on a one-dimensional particle depends only on position, \(F=F(x),\) then Newton's second law can be solved to find \(v\) as a function of \(x\) given by $$v^{2}=v_{\mathrm{o}}^{2}+\frac{2}{m} \int_{x_{0}}^{x} F\left(x^{\prime}\right) d x^{\prime}$$ [Hint: Use the chain rule to prove the following handy relation, which we could call the " \(v\) dv/d \(x\) rule": If you regard \(v\) as a function of \(x,\) then $$\dot{v}=v \frac{d v}{d x}=\frac{1}{2} \frac{d v^{2}}{d x}$$ Use this to rewrite Newton's second law in the separated form \(m d\left(v^{2}\right)=2 F(x) d x\) and then integrate from \(x_{\mathrm{o}}\) to \(x .\) ] Comment on your result for the case that \(F(x)\) is actually a constant. (You may recognise your solution as a statement about kinetic energy and work, both of which we shall be discussing in Chapter 4.)

Short answer

Velocity squared relates to work done by force: \( v^2 = v_0^2 + \frac{2}{m} \int_{x_0}^{x} F(x') dx' \). For constant \(F\), work-energy theorem applies.

Step-by-step solution

Understand the Problem Statement

We need to find how velocity \(v\) as a function of position \(x\) is derived from Newton's second law when force \(F\) depends only on position. We are given a form of \(v^2\) involving an integral that relates force to velocity for one-dimensional motion.

Use the Chain Rule for Velocity

To start, we will use the hint provided, which involves the chain rule. For a function \(v(x)\): \( \dot{v} = v \frac{dv}{dx} = \frac{1}{2} \frac{d(v^2)}{dx} \). This transformation allows us to relate acceleration and velocity squared.

Rewrite Newton's Second Law

Newton's second law in one dimension is \( F = m \dot{v} \). Substituting from Step 2, we get \( F = m \cdot v \frac{dv}{dx} = m \cdot \frac{1}{2} \frac{d(v^2)}{dx} \). Thus, \( m \cdot \frac{1}{2} \frac{d(v^2)}{dx} = F(x) \).

Separate Variables

Separate the variables to prepare for integration: \(m \cdot d(v^2) = 2 F(x) dx\). Now, both sides can be integrated with respect to their respective variables.

Integrate Both Sides

Integrate from initial position \(x_0\) to \(x\), and from initial velocity squared \(v_0^2\): \( \int_{v_0^2}^{v^2} m \cdot d(v^2) = \int_{x_0}^{x} 2 F(x') dx' \). Evaluating the left side gives \(m(v^2 - v_0^2)\).

Solve for v^2(x)

After integration, the result is: \[ m(v^2 - v_0^2) = 2 \int_{x_0}^{x} F(x') dx' \]. Simplifying this, we solve for \(v^2\): \[ v^2 = v_0^2 + \frac{2}{m} \int_{x_0}^{x} F(x') dx' \]. This is the required expression.

Consider Constant Force

If \(F(x)\) is constant \(F_0\), the integral simplifies to \(F_0(x - x_0)\). Thus, \( v^2 = v_0^2 + \frac{2 F_0}{m}(x - x_0) \), reflecting the principle of work-energy where kinetic energy change equals the work done by a constant force.

Key concepts

Understanding the Chain Rule

The chain rule is a fundamental concept in calculus that allows us to differentiate a function that is affected indirectly by another variable. In this context, when we examine velocity as a function of position, it's crucial to see how position influences velocity and subsequently acceleration. The chain rule helps bridge these relationships through differentiation.
For instance, if we perceive velocity, \(v\), as a function of position, \(x\), we can express the acceleration, \(\dot{v}\), as \(v \frac{dv}{dx}\). This expression derives from the chain rule by setting up the derivative \(\frac{d}{dt}\), which reads as the change in velocity over time. However, since we are considering velocity as dependent on position immediately, we convert via the chain rule:

• Differentiate velocity with respect to position: \(\frac{dv}{dx}\).
• Multiply by velocity: \(v \cdot \frac{dv}{dx}\).

This transformation is particularly useful as it helps us convert Newton's second law, which initially deals with forces and accelerations, into a form that aligns with velocity and position, enabling a straightforward integration solution.

Velocity as a Function of Position

In one-dimensional motion, velocity can be described as a function of position rather than time. This becomes important when force depends only on position, \(F = F(x)\). Newton's second law can be combined with the chain rule to pivot focus from time dependence to position dependence.
To achieve this, we rewrite Newton's second law, \(F = m \dot{v}\), into a format that incorporates \(\frac{dv}{dx}\), yielding \(m v \frac{dv}{dx} = F(x)\). We then transform this into the functional form \(m \frac{1}{2} \frac{d(v^2)}{dx} = F(x)\), effectively reducing the complexity by one step because it relates directly to \(v^2\).

By manipulating the equation to separate variables, we arrange it in an integrable form:

• Multiply and rearrange: \(m \cdot d(v^2) = 2 F(x) dx\).
• Integrate both sides to find the relationship between velocity squared, position, and force.

This gives us \(v^2 = v_0^2 + \frac{2}{m} \int_{x_0}^{x} F(x') dx'\), a powerful expression that presents velocity directly as a function of position.

Kinetic Energy and Work

Kinetic energy and work are closely linked in physics, often related through Newton's laws in equations involving motion. When considering a particle in one-dimensional motion with force \(F(x)\), the derivation of velocity as a function of position inherently touches upon the work-energy principle.
In the equation \(v^2 = v_0^2 + \frac{2}{m} \int_{x_0}^{x} F(x') dx'\), we encounter concepts where:

• Kinetic energy change is represented by the term \(v^2 - v_0^2\).
• The work done by the force is represented by the integral \(\frac{2}{m} \int_{x_0}^{x} F(x') dx'\).

This framework illustrates that kinetic energy change corresponds to the work done by a variable or constant force. Specifically, when \(F(x)\) is constant, the integral simplifies and we recognize kinetic energy gain equaling work done:
\[ v^2 = v_0^2 + \frac{2 F_0}{m}(x - x_0) \].

This direct relationship serves as an example of the work-energy theorem, stating that work done on an object results in an equivalent change in its kinetic energy.