Problem 23 - Thomas Calculus in SI Units

Question

If a variable force of magnitude \(F(x)\) moves an object of mass \(m\) along the \(x\) -axis from \(x_{1}\) to \(x_{2},\) the object'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 object 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 object's velocities at \(x_{1}\) and \(x_{2} .\) In physics, the expression \((1 / 2) m v^{2}\) is called the kinetic energy of an object of mass \(m\) moving with velocity \(v\). Therefore, the work done by the force equals the change in the object's kinetic energy, and we can find the work by calculating this change.

Step-by-Step Solution

Answer

Net work done is the change in kinetic energy: \(\frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2\).

Step 1: Understanding Newton's Second Law and Chain Rule

Start by recalling Newton's second law: \(F = m \frac{dv}{dt}\). Apply the chain rule to express \(\frac{dv}{dt}\) in terms of velocity \(v\) and position \(x\) as \(\frac{dv}{dt} = v \frac{dv}{dx}\). This allows reformulating \(F\) as \(m v \frac{dv}{dx}\).

Step 2: Express the Force in Terms of Kinematic Variables

Since \(F(x) = m\frac{dv}{dt}\) and using the chain rule we derived \(F(x) = m v \frac{dv}{dx}\), rewrite the force in terms of velocity and position: \(F(x)\,dx = m v \frac{dv}{dx} \times dx \). Simplifying, this leads to: \(F(x)\,dx = m v\, dv\).

Step 3: Integrate Both Sides to Find Work Done

To find the net work done, integrate both sides from \(x_1\) to \(x_2\): \[ W = \int_{x_1}^{x_2} F(x) \, dx = \int_{v_1}^{v_2} m v \, dv. \]This evaluates the work as the change in the integrated velocity term.

Step 4: Evaluate the Integral

Calculate the right-hand side integral:\[ \int_{v_1}^{v_2} m v \, dv = \left[ \frac{1}{2} m v^2 \right]_{v_1}^{v_2} = \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2.\]Therefore, the net work done is expressed as the change in kinetic energy.

Key Concepts

Newton's Second Law of Motion

Newton's Second Law of Motion is fundamental in understanding how forces affect the motion of objects. The law states that the force applied to an object, denoted as \( F \), is equal to the mass of the object \( m \) multiplied by the acceleration of the object \( a \). Mathematically, it is expressed as: \[ F = ma \]Acceleration is the rate of change of velocity with respect to time. In our exercise, the law is used in the form \( F = m \frac{dv}{dt} \), where \( \frac{dv}{dt} \) represents the derivative of velocity with respect to time — an expression for acceleration.
The connection between force and motion becomes clearer when considering this relationship. As a force acts on an object, the object's velocity changes, and this change in velocity is directly related to the work done on the object, which we explore further using the Work-Energy Theorem.
In simpler terms, Newton's Second Law helps us quantify how much an object's velocity changes when an external force is applied, and is crucial for solving problems involving the Work-Energy Theorem.

Kinetic Energy

Kinetic energy is a measure of the energy an object possesses due to its motion. For an object with mass \( m \) moving at a velocity \( v \), the kinetic energy \( KE \) is given by the formula: \[ KE = \frac{1}{2} mv^2 \]Kinetic energy is always positive, reflecting the fact that energy must be non-negative.
Understanding kinetic energy is crucial when dealing with problems involving motion and work. In our exercise, the net work done on the object by a force results in a change in its kinetic energy.
The idea behind this is simple: when work is done on an object, energy is transferred to it, allowing it to move faster and thus increasing its kinetic energy.

• The initial kinetic energy is \( \frac{1}{2} mv_1^2 \)
• The final kinetic energy is \( \frac{1}{2} mv_2^2 \)
• Work done equals the change in kinetic energy: \( \Delta KE = \frac{1}{2} mv_2^2 - \frac{1}{2} mv_1^2 \)

This relationship, which equates work done and change in kinetic energy, is known as the Work-Energy Theorem.

Chain Rule

The chain rule is an essential concept from calculus used to differentiate compositions of functions. In physics, particularly when analyzing motion, the chain rule helps us express rates of change in different terms.
In our given exercise, the chain rule is applied to relate the change in velocity over time \( \frac{dv}{dt} \) to changes in velocity with respect to position \( \frac{dv}{dx} \). Mathematically:\[ \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \cdot \frac{dv}{dx} \]Here, \( \frac{dx}{dt} = v \) gives us the velocity, which simplifies applying Newton’s second law in terms of kinetic variables.
The chain rule enables us to rewrite the expression for force, transforming it into an integration problem that directly connects force and changes in velocity — crucial for deriving the expression for work.

• Converts the differential equation from force to a more integrable form.
• Allows linking of position, velocity, and time seamlessly.

The elegance of the chain rule simplifies complex problems, such as proving that the work done by a force results in a change in kinetic energy.