Problem 25 - Thomas Calculus

Question

Kinetic energy If a variable force of magnitude \(F(x)\) moves a body of mass \(m\) along the \(x\) -axis from \(x_{1}\) to \(x_{2}\) , the body's velocity \(v\) can be written as \(d x / d t\) (where \(t\) represents time). Use Newton's second law of motion \(F=m(d v / d t)\) and the Chain Rule $$ \frac{d v}{d t}=\frac{d v}{d x} \frac{d x}{d t}=v \frac{d v}{d x} $$ to show that the net work done by the force in moving the body from \(x_{1}\) to \(x_{2}\) is $$ W=\int_{x_{1}}^{x_{2}} F(x) d x=\frac{1}{2} m v_{2}^{2}-\frac{1}{2} m v_{1}^{2} $$ where \(v_{1}\) and \(v_{2}\) are the body's velocities at \(x_{1}\) and \(x_{2}\) . In physics, the expression \((1 / 2) m v^{2}\) is called the kinetic energy of a body of mass \(m\) moving with velocity \(v .\) Therefore, the work done by the force equals the change in the body's kinetic energy, and we can find the work by calculating this change.

Step-by-Step Solution

Answer

The net work done is equal to the change in the body's kinetic energy.

Step 1: Interpret Newton's Second Law with Chain Rule

Begin by applying Newton's second law, which is given by \( F = m \frac{dv}{dt} \). Use the chain rule to substitute \( \frac{dv}{dt} = v \frac{dv}{dx} \) into this equation to yield:\[F = m \left(v \frac{dv}{dx}\right)\] Rearrange to get:\[F dx = m v dv\] This equation expresses the differential form of work done over a small displacement \( dx \) as \( F dx \), and relates it to the change in velocity \( v dv \) multiplied by mass \( m \).

Step 2: Integrate Both Sides to Find Work Done

To find the total work done \( W \) by the force as the object moves from \( x_1 \) to \( x_2 \), integrate both sides over the corresponding intervals:\[\int_{x_1}^{x_2} F(x) dx = \int_{v_1}^{v_2} m v dv\] This integral on the right side simplifies to:\[\int_{v_1}^{v_2} m v \, dv = m \left[ \frac{v^2}{2} \right]_{v_1}^{v_2} = \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2\] Thus, we have shown that the work done \( W \) is \[W = \int_{x_1}^{x_2} F(x) \, dx = \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2\]

Step 3: Interpret Result as Change in Kinetic Energy

Recognize that the right side of the equation represents a change in kinetic energy of the body. Thus, the result illustrates the work-energy principle, which states that the work done by the force on the body is equal to the change in its kinetic energy:\[\Delta KE = \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2\] The left side, \( \int_{x_1}^{x_2} F(x) dx \), represents the net work done as the force moves the body from \( x_1 \) to \( x_2 \).

Key Concepts

Kinetic Energy

Kinetic energy is a concept used to describe the energy a body possesses because of its motion. It is given by the formula \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. This formula tells us how the speed and mass of an object determine its energy while moving.
When a force acts on a body, its velocity changes, which means its kinetic energy also changes. The difference between the kinetic energy at two different points of motion gives us a way to quantify the work done by the force. This change in kinetic energy due to work is the foundation of the work-energy principle, which relates the force applied to the displacement of a body and its corresponding change in motion.- **Important Points:**

• Kinetic energy deals with motion-related energy.
• It depends on both the mass and the velocity of the object.
• The change in kinetic energy equates to the work done by the force.

Newton's Second Law

Newton's Second Law serves as the bridge between force, mass, and acceleration. It is commonly expressed as \( F = m a \), where \( F \) is the force applied, \( m \) is the mass, and \( a \) is the acceleration produced. However, acceleration can also be transformed into changes in velocity over time, \( a = \frac{dv}{dt} \), leading us to the equation \( F = m \frac{dv}{dt} \). This version shows us how a force affects the velocity of an object over time.
In the context of the exercise, Newton's Second Law supports the connection between force and the motion of the object. By utilizing this relation, it establishes the foundation for calculating work done and understanding kinetic energy changes. It's crucial for decoding situations where force causes a body's movement.-**Key Ideas:**

• Force equals mass times acceleration.
• The law connects force with motion changes.
• It is central to explaining motion dynamics due to force.

Variable Force

When dealing with forces that vary along a path or over time, we call these variable forces. Unlike constant forces, whose magnitude and direction remain steady, a variable force changes depending on the position or other factors, like time.
In the exercise, the force \( F(x) \) is not the same over the range from \( x_1 \) to \( x_2 \). This requires us to calculate the work done by integrating the force over the path it acts upon. The integral \( \int_{x_1}^{x_2} F(x) dx \) captures the contribution of the varying force over the distance, thereby giving the total work done on the object.-**Main Takeaways:**

• Variable forces change with position or time.
• Integrating helps find the work done by these forces over a path.
• Allows for calculating work where force isn't constant.

Chain Rule

The chain rule is a vital tool for differential calculus, enabling us to differentiate composite functions. It helps when a variable depends on another variable that is itself a function of yet another variable.
In this task, the chain rule allows the transition from time-based derivatives to space-based ones. Specifically, it provides the link needed to replace \( \frac{dv}{dt} \) with \( v \frac{dv}{dx} \). Here, the velocity \( v = \frac{dx}{dt} \) acts as a bridge, making it feasible to transform the derivative of velocity with respect to time into a function of space. This transformation is pivotal in facilitating the integration process necessary for computing work via kinetic energy difference.
-**Core Insights:**

• The chain rule helps handle derivatives of composite functions.
• Useful for swapping derivatives between different variable contexts.
• Essential for bridging different types of rates of change.