1. The Physics Definition of Work
In everyday language, “work” implies physical or mental effort. In physics, work has a highly specific, mathematically rigorous definition: Work is done when a force causes a displacement, and there is a component of the force in the direction of the displacement.
If a constant force $F$ acts on an object while it undergoes a straight-line displacement $s$, the work $W$ done by the force is:
$$W = Fs \cos \theta$$
Where:
- $F$ is the magnitude of the constant force (in Newtons, N).
- $s$ is the magnitude of the displacement (in meters, m).
- $\theta$ is the angle between the force vector and the displacement vector.
- The SI unit of work is the Joule (J), where $1 \text{ J} = 1 \text{ N} \cdot \text{m}$.
2. Work as a Vector Dot Product
For JEE-Mains, students must be extremely comfortable moving away from the scalar formula and into vector algebra. The formula $W = Fs \cos \theta$ is the geometric definition of the dot product (or scalar product) of two vectors.
Therefore, work is the dot product of the force vector $\vec{F}$ and the displacement vector $\vec{s}$:
$$W = \vec{F} \cdot \vec{s}$$
The Geometric Interpretation (Projection):
The dot product can be grouped in two conceptual ways, both mathematically identical but physically distinct:
- $W = (F \cos \theta)s$: The product of the displacement and the component of the force parallel to the displacement.
- $W = F(s \cos \theta)$: The product of the force and the component of the displacement parallel to the force.
3. The Three Regimes of Work
The angle $\theta$ entirely dictates the sign of the work done, representing whether energy is being added to or removed from the system.
| Angle (θ) | Value of cosθ | Work Sign | Physical Meaning | Example |
| $0^\circ \le \theta < 90^\circ$ | Positive | Positive ($W > 0$) | Force adds kinetic energy to the system. | Pushing a block forward. |
| $\theta = 90^\circ$ | Zero | Zero ($W = 0$) | Force does not affect the speed of the object. | Tension in a pendulum string; Centripetal force. |
| $90^\circ < \theta \le 180^\circ$ | Negative | Negative ($W < 0$) | Force removes kinetic energy (slows the object down). | Kinetic friction; Air resistance. |
4. Cartesian (3D) Coordinate Form
In JEE problems, forces and displacements are frequently given in unit vector notation.
If $\vec{F} = F_x \hat{i} + F_y \hat{j} + F_z \hat{k}$ and
the displacement is
$\vec{s} = \Delta x \hat{i} + \Delta y \hat{j} + \Delta z \hat{k}$,
the dot product expands algebraically. Because the dot product of perpendicular unit vectors is zero ($\hat{i} \cdot \hat{j} = 0$) and parallel unit vectors is one ($\hat{i} \cdot \hat{i} = 1$), the calculation simplifies to:
$$W = F_x \Delta x + F_y \Delta y + F_z \Delta z$$
Finding Displacement from Coordinates:
If a particle moves from an initial point $A(x_1, y_1, z_1)$ to a final point $B(x_2, y_2, z_2)$, the displacement vector is the difference between their position vectors:
$$\vec{s} = \vec{r}_B – \vec{r}_A = (x_2 – x_1)\hat{i} + (y_2 – y_1)\hat{j} + (z_2 – z_1)\hat{k}$$
5. Superposition (Multiple Forces)
Work is a scalar quantity. If multiple forces act on a body simultaneously, the net work done can be calculated in two ways:
- Scalar Addition: Calculate the work done by each force individually, then sum them up:
$W_{net} = W_1 + W_2 + \dots$ - Vector Addition: Find the net force vector first, then take the dot product with the displacement:
$W_{net} = \vec{F}_{net} \cdot \vec{s}$
JEE-Mains Practice Worksheet
Question 1: Vector Displacement
A constant force $\vec{F} = (3\hat{i} + 4\hat{j} – 5\hat{k}) \text{ N}$ acts on a particle. The particle is displaced from an initial coordinate $A(1, 2, 3) \text{ m}$ to a final coordinate $B(4, 5, -2) \text{ m}$. Calculate the work done by the force.
Solution:
First, find the displacement vector $\vec{s}$:
$\vec{s} = \vec{r}_B – \vec{r}_A$
$\vec{s} = (4 – 1)\hat{i} + (5 – 2)\hat{j} + (-2 – 3)\hat{k}$
$\vec{s} = (3\hat{i} + 3\hat{j} – 5\hat{k}) \text{ m}$
Next, apply the dot product $W = \vec{F} \cdot \vec{s}$:
$W = (3)(3) + (4)(3) + (-5)(-5)$
$W = 9 + 12 + 25$
$W = 46 \text{ J}$
Question 2: Angle and Component Formulation
A block of mass $5 \text{ kg}$ is pulled across a rough horizontal surface by a constant applied force of $50 \text{ N}$ directed at an angle of $37^\circ$ above the horizontal. The block is displaced by $10 \text{ m}$. What is the work done by the applied force? (Take $\sin 37^\circ = \frac{3}{5}, \cos 37^\circ = \frac{4}{5}$)
Solution:
Since we only need the work done by the applied force, we ignore friction and normal force for this specific calculation.
$W = Fs \cos \theta$
$W = 50 \times 10 \times \cos(37^\circ)$
$W = 500 \times \left(\frac{4}{5}\right)$
$W = 400 \text{ J}$
Question 3: Superposition of Forces
Two constant forces $\vec{F}_1 = (2\hat{i} – \hat{j} + 3\hat{k}) \text{ N}$ and $\vec{F}_2 = (-\hat{i} + 3\hat{j} – \hat{k}) \text{ N}$ act simultaneously on a body. Under the influence of these forces, the body is displaced from the origin to the position $\vec{r} = (4\hat{i} + \hat{j} + 2\hat{k}) \text{ m}$. Calculate the net work done on the body.
Solution:
Method: Find the net force vector first.
$\vec{F}_{net} = \vec{F}_1 + \vec{F}_2$
$\vec{F}_{net} = (2 – 1)\hat{i} + (-1 + 3)\hat{j} + (3 – 1)\hat{k}$
$\vec{F}_{net} = (\hat{i} + 2\hat{j} + 2\hat{k}) \text{ N}$
The displacement from the origin is just the final position vector:
$\vec{s} = (4\hat{i} + \hat{j} + 2\hat{k}) \text{ m}$
Calculate net work via dot product:
$W_{net} = \vec{F}_{net} \cdot \vec{s}$
$W_{net} = (1)(4) + (2)(1) + (2)(2)$
$W_{net} = 4 + 2 + 4$
$W_{net} = 10 \text{ J}$