Work Done by Constant Force and Variable Force

In classical mechanics, the concept of work serves as a vital bridge between mechanical forces and energy transformation

In classical mechanics, the concept of work serves as a vital bridge between mechanical forces and energy transformation, quantifying the cumulative effect of a force acting over a distance. Depending on whether the applied force remains steady or changes dynamically, the computation of work branches into two primary categories: work done by a constant force and work done by a variable force. Mastering both approaches forms the cornerstone of solving mechanics problems and sets the stage for understanding the work-energy theorem.
When the magnitude and direction of a force acting on an object remain unchanged throughout its motion, the force is classified as a constant force.

The fundamental algebraic formula for calculating the work done by a constant force is expressed as:
$$W = F s \cos\theta$$

Where:

  • $W$ represents the work done, measured in joules (J).
  • $F$ denotes the magnitude of the constant force, measured in newtons (N).
  • $s$ stands for the displacement magnitude of the object, measured in meters (m).
  • $\theta$ is the angle between the force vector $\vec{F}$ and the displacement vector $\vec{s}$.

The sign of the work carries distinct physical significance depending on the angle $\theta$:

  • When $0 \le \theta < 90^\circ$, $\cos\theta > 0$, resulting in $W > 0$. This indicates that the force does positive work, thereby increasing the kinetic energy of the object.
  • When $\theta = 90^\circ$, $\cos\theta = 0$, resulting in $W = 0$. The force does no work on the object.
  • When $90^\circ < \theta \le 180^\circ$, $\cos\theta < 0$, resulting in $W < 0$. This signifies negative work (often described as the object overcoming the force), which decreases the kinetic energy.

In a Cartesian coordinate system, work can also be elegantly represented via the dot product of vectors:
$$W = \vec{F} \cdot \vec{s} = F_x x + F_y y + F_z z$$

The Calculus Approach: Work Done by a Variable Force

In real-world physical scenarios, forces frequently fluctuate in magnitude or direction as an object moves through space. Because such forces are non-constant, simple multiplication falls short. Instead, we must employ integral calculus by breaking down the complex trajectory into an infinite number of infinitesimal segments.

Imagine an object traveling along a path from point $A$ to point $B$. We can partition this pathway into numerous infinitesimal displacement vectors $d\vec{r}$. Within any single microscopic displacement $d\vec{r}$, the varying force $\vec{F}$ can be locally approximated as a constant force.

  • The elementary work $dW$ done over a single infinitesimal step is given by: $dW = \vec{F} \cdot d\vec{r}$
  • The total work over the entire path is subsequently obtained through a line integral:

$$W = \int_{A}^{B} \vec{F} \cdot d\vec{r} = \int_{A}^{B} (F_x dx + F_y dy + F_z dz)$$

For an object restricted to one-dimensional motion (such as along the $x$-axis), this simplifies to:
$$W = \int_{x_1}^{x_2} F(x) dx$$
In graphical terms, this means the work done by a variable-magnitude force equals the net area under the force-displacement curve $F(x)$.

Classic Case Studies

To solidify the distinction between these two calculation methods, let us examine two quintessential physics examples.

Case 1: Gravity (Constant Force)

Consider an object of mass $m$ descending from an initial height $h_1$ to a final height $h_2$. Gravity acts as a uniform force both in magnitude and direction, expressed as $\vec{F} = -mg\hat{j}$, while the displacement vector is $\Delta y = (h_2 - h_1)\hat{j}$.

Substituting these into the constant force formula yields:
$$W_g = \vec{F} \cdot \Delta y = (-mg) \cdot (h_2 - h_1) = mg(h_1 - h_2)$$

This demonstrates that gravitational work depends exclusively on the vertical elevation change between the initial and final positions, remaining entirely independent of the specific trajectory taken.

Case 2: Spring Force (Variable Force)

According to Hooke's law, the restorative force exerted by an ideal spring is proportional to its displacement $x$ from equilibrium and acts in the opposite direction, formulated as $F(x) = -kx$, where $k$ represents the spring constant. This is a textbook example of a variable force.

To calculate the work done by the spring force as it is stretched from an initial deformation $x_1$ to $x_2$:
$$W_s = \int_{x_1}^{x_2} (-kx) dx = \left[ -\frac{1}{2} kx^2 \right]_{x_1}^{x_2} = \frac{1}{2} k x_1^2 - \frac{1}{2} k x_2^2$$

If the spring is stretched or compressed directly from its unstretched length ($x_1 = 0$) to a final displacement $x$:
$$W_s = -\frac{1}{2} kx^2$$
This confirms that the work done by a spring force is path-independent and directly tied to changes in elastic potential energy.

Problem-Solving Strategies

When tackling mechanics problems, correctly identifying the nature of the applied force is the critical first step:

  1. Evaluate Force Characteristics: Carefully analyze whether the magnitude and direction of the force remain constant or vary with position and time.
  2. Select the Appropriate Mathematical Tool:
    • For a constant force, directly apply the algebraic formula $W = Fs\cos\theta$ or vector dot products.
    • For a variable force, derive the functional relationship $F(x)$ and set up an integration model, or alternatively, invoke the work-energy theorem ($W_{\text{net}} = \Delta E_k$) to calculate work indirectly through kinetic energy changes.
  3. Keep Track of Signs: Pay close attention to the directional orientation between the force and displacement vectors to ensure your calculated work accurately reflects physical reality.