Problem 23 - Thomas Calculus

Question

Kinetic energy 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 $$\frac{d v}{d t}=\frac{d v d x}{d x d t}=v \frac{d v}{d x}$$ 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

Work done by the force equals the change in object's kinetic energy.

Step 1: Understand and Define the Problem

We are asked to prove that the work done by a variable force on an object is equivalent to the change in its kinetic energy. This relationship can be stated as \(W = \frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2}\).

Step 2: Use Newton's Second Law

Newton's second law states that \(F = m \frac{dv}{dt}\). We can equate this to the force function, \(F(x)\). Therefore, \(F(x) = m \frac{dv}{dt}\).

Step 3: Apply the Chain Rule

Using the chain rule, we have \(\frac{dv}{dt} = v \frac{dv}{dx}\). Substitute this into Newton's second law to get \(F(x) = m \left( v \frac{dv}{dx} \right)\).

Step 4: Relate Force and Work

The work done by the force when the object moves from \(x_1\) to \(x_2\) is given by the integral \(W = \int_{x_{1}}^{x_{2}} F(x) \, dx\). Substitute \(F(x)\) from the previous step: \(W = \int_{x_{1}}^{x_{2}} m \left( v \frac{dv}{dx} \right) \, dx \).

Step 5: Integrate to Find Work

Recognize that \( \frac{dv}{dx} \, dx = dv\). The expression inside the integral can be rewritten as \(W = m \int_{v_1}^{v_2} v \, dv\). Integrate this to get \(W = m \left[ \frac{v^2}{2} \right]_{v_1}^{v_2}\).

Step 6: Evaluate the Integral

Evaluating the integral, we get \(W = \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2\). This shows that the work done by the force is equal to the change in the object's kinetic energy.

Key Concepts

Newton's Second Law

Newton's Second Law is a fundamental principle in physics that links force, mass, and acceleration. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration, formulated as \( F = m \frac{dv}{dt} \). This equation implies that any change in an object's velocity over time will result in a change in the force acting on it.

For the problem at hand, we start by expressing force as a function of position \( F(x) \) and relate it to the object's mass \( m \) and its acceleration \( \frac{dv}{dt} \). Here, Newton's Second Law allows us to express force in terms of velocity change as \( F(x) = m \frac{dv}{dt} \).

This expression is crucial for linking force with kinetic energy as we further explore the relationships among force, velocity, and displacement.

Chain Rule

The Chain Rule is a calculus concept used to differentiate composite functions. In this context, it relates the rate of change of velocity with respect to time and position. The Chain Rule states that \( \frac{dv}{dt} = v \frac{dv}{dx} \).

This expression enables us to connect the object's velocity to its spatial displacement. By substituting \( \frac{dv}{dt} = v \frac{dv}{dx} \) into the force equation \( F(x) = m \frac{dv}{dt} \), we reformulate it as \( F(x) = m \left( v \frac{dv}{dx} \right) \).

This step effectively bridges the gap between force and velocity, setting the stage to calculate work done as an integral of force over a path. Thus, the Chain Rule assists in transforming the analysis of dynamic systems into a solvable integral.

Work-Energy Principle

The Work-Energy Principle is a powerful concept connecting mechanical work and kinetic energy. It states that the work done by a force on an object equals the change in its kinetic energy.

In the given exercise, the work done by a variable force moving an object from position \( x_1 \) to \( x_2 \) is calculated by the integral \( W = \int_{x_1}^{x_2} F(x) \, dx \). By substituting our expression of force \( F(x) = m \left( v \frac{dv}{dx} \right) \), the integral becomes \( W = m \int_{x_1}^{x_2} v \frac{dv}{dx} \, dx \).

Recognizing that \( \frac{dv}{dx} \, dx = dv \), we transform this to \( W = m \int_{v_1}^{v_2} v \, dv \). Integrating this expression results in \( W = m \left[ \frac{v^2}{2} \right]_{v_1}^{v_2} \), which simplifies to \( W = \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2 \).

This final equation encapsulates the Work-Energy Principle by showing that the net work done by the force is equal to the change in kinetic energy, emphasizing how movement and force interact and transform within a physical system.