Statement and Proof of the Work-Energy Theorem

In the study of classical mechanics, the Work-Energy Theorem serves as a fundamental principle that bridges the gap betw

In the study of classical mechanics, the Work-Energy Theorem serves as a fundamental principle that bridges the gap between force, displacement, and the changing state of motion of an object. While Newton's Laws of Motion provide a vector-based approach to understanding dynamics, the Work-Energy Theorem offers a scalar perspective that often simplifies complex problems. By relating the cumulative effect of forces (work) to the change in an object's motion (kinetic energy), this theorem provides a powerful tool for analyzing physical systems and lays the essential groundwork for the Law of Conservation of Energy.

Fundamental Concepts

To understand the theorem, we must first define its two primary constituents: Kinetic Energy and Work.

  • Kinetic Energy ($E_k$): This is the energy possessed by an object due to its motion. For a particle with a mass $m$ moving at a velocity $v$, the kinetic energy is defined by the expression:
    $$E_k = \frac{1}{2}mv^2$$
    It is a scalar quantity, meaning it has magnitude but no direction, and it is always non-negative.

  • Work ($W$): Work represents the transfer of energy that occurs when a force is applied to an object over a certain displacement.

    • For a constant force $\mathbf{F}$ acting over a displacement $\Delta \mathbf{r}$, work is calculated as the dot product: $W = \mathbf{F} \cdot \Delta \mathbf{r}$.
    • For a variable force, work is defined as the line integral of the force along the path of motion: $W = \int \mathbf{F} \cdot d\mathbf{s}$.

Statement of the Theorem

The Work-Energy Theorem states that the net work done by all external forces acting on a particle is equal to the change in its kinetic energy.

Mathematically, this relationship is expressed as:
$$W_{\text{net}} = \Delta E_k = E_{k, \text{final}} - E_{k, \text{initial}}$$

Expanding this using the definition of kinetic energy, we get:
$$W_{\text{net}} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$$

Where:

  • $W_{\text{net}}$ is the total work done by the sum of all forces acting on the object.
  • $v_1$ is the initial velocity.
  • $v_2$ is the final velocity.

Mathematical Derivation

To provide a rigorous proof of this theorem, we begin with Newton's Second Law of Motion, which relates the net force acting on a mass to its acceleration.

  1. Starting Point: According to Newton's Second Law, the net force $\mathbf{F}{\text{net}}$ acting on a mass $m$ is:
    $$\mathbf{F}    {\text{net}} = m\mathbf{a} = m\frac{d\mathbf{v}}{dt}$$

  2. Infinitesimal Work: Consider an infinitesimal displacement $d\mathbf{s}$ occurring in a time interval $dt$. The infinitesimal work $dW$ done by the net force during this interval is:
    $$dW = \mathbf{F}_{\text{net}} \cdot d\mathbf{s}$$
    Since $d\mathbf{s} = \mathbf{v} dt$, we can substitute this into the equation:
    $$dW = \left( m \frac{d\mathbf{v}}{dt} \right) \cdot (\mathbf{v} dt) = m \mathbf{v} \cdot d\mathbf{v}$$

  3. Integration: To find the total work done as the object moves from an initial state (velocity $v_1$) to a final state (velocity $v_2$), we integrate the expression along the path:
    $$W_{\text{net}} = \int_{1}^{2} dW = \int_{v_1}^{v_2} m \mathbf{v} \cdot d\mathbf{v}$$

  4. Final Result: In the context of the magnitude of velocity (speed), the dot product $\mathbf{v} \cdot d\mathbf{v}$ simplifies to $v , dv$. Thus:
    $$W_{\text{net}} = m \int_{v_1}^{v_2} v , dv = m \left[ \frac{1}{2}v^2 \right]{v_1}^{v_2}$$
    $$W    {\text{net}} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2 = \Delta E_k$$

This derivation confirms that the theorem holds true not only for linear motion but also for curvilinear motion, as the vector notation accounts for changes in direction.

Practical Application: An Illustrative Example

One of the greatest advantages of the Work-Energy Theorem is its ability to bypass the need to calculate instantaneous acceleration or time, which is often required when using Newton's Second Law.

Problem Scenario:
A block with a mass of $m = 2,\text{kg}$ is placed on a rough horizontal surface. A constant horizontal pulling force of $F = 10,\text{N}$ is applied to the block. The block starts from rest and moves a distance of $s = 5,\text{m}$. The kinetic friction force acting against the motion is $f = 2,\text{N}$. What is the velocity of the block at the end of the $5,\text{m}$ displacement?

Solution Steps:

  1. Identify the work done by each force:

    • Work done by the pulling force ($W_F$): Since the force and displacement are in the same direction:
      $$W_F = F \cdot s = 10,\text{N} \times 5,\text{m} = 50,\text{J}$$
    • Work done by friction ($W_f$): Since friction acts in the opposite direction of displacement:
      $$W_f = -f \cdot s = -2,\text{N} \times 5,\text{m} = -10,\text{J}$$
    • Work done by gravity and normal force: Because these forces are perpendicular to the displacement, they do zero work ($W_g = 0, W_N = 0$).

  2. Calculate the net work ($W_{\text{net}}$):
    $$W_{\text{net}} = W_F + W_f = 50,\text{J} - 10,\text{J} = 40,\text{J}$$

  3. Apply the Work-Energy Theorem:
    Since the block starts from rest, the initial kinetic energy $E_{k1} = 0$.
    $$W_{\text{net}} = \frac{1}{2}mv_2^2 - 0$$
    $$40 = \frac{1}{2}(2,\text{kg}) \cdot v_2^2$$
    $$40 = v_2^2$$
    $$v_2 = \sqrt{40} \approx 6.32,\text{m/s}$$

Conclusion:
The block's final velocity is approximately $6.32,\text{m/s}$. By using the Work-Energy Theorem, we directly linked the forces and displacement to the final speed, avoiding the intermediate step of calculating the net acceleration.