Problem 47 - Calculus Single Variable

Question

A force \(F(x)>0\) causes a mass \(m\), initially at rest at \(x=0\), to move along the \(x\) -axis with velocity \(v(x)\). Let \(W(b)\) be the work done in moving the body from \(x=0\) to \(x=b\). Show that \(W(b)=m v(b)^{2} / 2 .\) In other words, the work done is equal to the gain in kinetic energy. Hint: Start from Newton's Law, which involves the derivative of velocity with respect to time. Use the Chain Rule to calculate the derivative of \(v\) with respect to \(x\).

Step-by-Step Solution

Answer

The work done, \( W(b) = m v(b)^2 / 2 \), equals the gain in kinetic energy.

Step 1: Apply Newton's Second Law

Newton's second law states that the force acting on an object is equal to the mass of the object times its acceleration: \[ F(x) = m \frac{dv}{dt} \]where \(dv/dt\) is the derivative of velocity with respect to time, i.e., acceleration.

Step 2: Use the Chain Rule

The Chain Rule allows us to relate the derivative of velocity with respect to time to the derivative of velocity with respect to position. Since the velocity \(v\) depends on position \(x\), we can express:\[ \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v \]where \(dx/dt = v\) is the velocity.

Step 3: Substitute into Newton's Law

Substitute the Chain Rule expression of acceleration back into Newton's Second Law:\[ F(x) = m \cdot \frac{dv}{dx} \cdot v \]

Step 4: Derive the Work-Energy Theorem

The work done, \(W(b)\), moving the mass from \(x = 0\) to \(x = b\) is the integral of force over distance:\[ W(b) = \int_{0}^{b} F(x)\, dx = \int_{0}^{b} m \cdot \frac{dv}{dx} \cdot v \cdot dx \]

Step 5: Change of Variable in Integration

Perform a change of variable for the integral. Since \(dv = \frac{dv}{dx} \cdot dx\), substitute and transform the integration over \(x\) to integration over \(v\):\[ W(b) = m \int_{v(0)}^{v(b)} v \cdot dv \]

Step 6: Evaluate the Integral

Evaluate the integral \( m \int v \, dv \) to find the work:\[ W(b) = m \left[ \frac{v^2}{2} \right]_{v(0)}^{v(b)} = m \left( \frac{v(b)^2}{2} - \frac{v(0)^2}{2} \right) \]

Step 7: Initial Condition and Simplification

The mass is initially at rest, so \(v(0) = 0\). Therefore, simplify the result:\[ W(b) = m \cdot \frac{v(b)^2}{2} \]

Key Concepts

Newton's Second Law

Newton's Second Law forms the foundation of classical mechanics. It defines the relationship between an object's motion and the forces acting upon it. In simple terms, this law states:\[ F = m imes a \]where:

• \(F\) is the force applied to an object,
• \(m\) is the object's mass,
• \(a\) is the object's acceleration.

This equation asserts that force equals mass times acceleration. Accelerating a heavier object requires more force than accelerating a lighter one.
In the context of our problem, we're analyzing how a force \(F(x)\) moves a mass \(m\) along the x-axis. Initially, the object is at rest, so the only force is due to \(F(x)\). Here, Newton's second law is expressed as:\[ F(x) = m \frac{dv}{dt} \] which includes \( \frac{dv}{dt} \) as the acceleration. Understanding this crucial aspect allows us to see how the force influences the change in velocity over time.

Chain Rule in Calculus

The Chain Rule in Calculus is a powerful method for differentiating compositions of functions. When a function depends on another function, the Chain Rule helps to find the derivative of the outer function relative to the inner one. The rule is often expressed as:\[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]In our exercise, the velocity \(v\) depends on position \(x\), and position changes as time \(t\) progresses. Using the Chain Rule simplifies expressions involving these connections.
Let's dissect the solution:

• We have velocity \(v\) as a function of position \(x\).
• To find \(\frac{dv}{dt}\), we use the Chain Rule: \(\frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}\).
• Since \(\frac{dx}{dt} = v\), the Chain Rule gives \( \frac{dv}{dt} = \frac{dv}{dx} \cdot v \).

By substituting back into Newton's law, we convert the relationship to \( F(x) = m \cdot \frac{dv}{dx} \cdot v \). This establishes a bridge between the change in velocity and the force experienced, crucial for further analysis.

Kinetic Energy

Kinetic Energy is the energy associated with an object's motion. Defined classically, kinetic energy \(KE\) for an object with mass \(m\) moving at velocity \(v\) is given by:\[ KE = \frac{1}{2} m v^2 \]This formula illustrates that kinetic energy grows with the square of velocity and is directly proportional to mass. In the context of work and energy, the Work-Energy Theorem connects the work done on an object with its kinetic energy change.
In the problem context:

• Initial condition: The mass starts from rest, i.e., \(v(0) = 0\).
• Work done \(W(b)\) from \(x = 0\) to \(x = b\) equates to the change in kinetic energy.
• Through analysis and integration: \( W(b) = m \int v \, dv \).

Evaluating this integral, \(W(b) = \frac{1}{2} m (v(b)^2 - v(0)^2)\) simplifies to \(W(b) = \frac{1}{2} m v(b)^2\) when \(v(0) = 0\). This expression validates the problem's claim: the work done equals the gain in kinetic energy, aligning with the Work-Energy Theorem principles.