Category: Work, Energy and Power

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:

1. $W = (F \cos \theta)s$: The product of the displacement and the component of the force parallel to the displacement.
2. $W = F(s \cos \theta)$: 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:

1. Scalar Addition: Calculate the work done by each force individually, then sum them up:
   $W_{net} = W_1 + W_2 + \dots$
2. 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:

$W_{net} = \vec{F}_{net} \cdot \vec{s}$

$W_{net} = (1)(4) + (2)(1) + (2)(2)$

$W_{net} = 4 + 2 + 4$

$W_{net} = 10 \text{ J}$