Work — Revision Notes

⚡ 30-Second Revision

Definition: — $W = \vec{F} \cdot \vec{d} = Fd \cos\theta$
Units: — Joule (J), $1\text{ J} = 1\text{ N} \cdot \text{m}$
Scalar Quantity: — Work has magnitude only, no direction.
Positive Work: — $0^\circ \le \theta < 90^\circ$ (force aids motion)
Negative Work: — $90^\circ < \theta \le 180^\circ$ (force opposes motion)
Zero Work: — $\theta = 90^\circ$ (force perpendicular to displacement) or $d=0$.
Variable Force: — $W = \int_{x_i}^{x_f} F(x) \, dx$ (Area under F-x graph)
Work-Energy Theorem: — $W_{net} = \Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$
Work by Gravity: — $W_g = \pm mgh$ (positive if falling, negative if rising)
Work by Spring: — $W_s = -\frac{1}{2}kx^2$ (by spring from equilibrium $x=0$ to $x$)
Work by Friction: — Always negative, $W_f = -f_k d$

2-Minute Revision

Work is the transfer of energy due to a force causing displacement. It's a scalar quantity, measured in Joules. The fundamental formula for constant force is $W = Fd \cos\theta$, where $\theta$ is the angle between force and displacement.

Work can be positive (force helps motion), negative (force opposes motion, like friction), or zero (force perpendicular to motion or no displacement). For forces that vary with position, work is calculated by integrating the force over displacement, or by finding the area under the Force-displacement graph.

The Work-Energy Theorem is crucial: the net work done on an object equals its change in kinetic energy ($W_{net} = \Delta K$). Remember specific cases: work by gravity is $mgh$ (with appropriate sign), and work by a spring (from equilibrium) is $-\frac{1}{2}kx^2$.

Always convert units to SI (e.g., cm to m) and pay attention to signs in calculations. This topic is frequently tested in NEET, often in conjunction with energy conservation and Newton's laws.

5-Minute Revision

Work is a fundamental concept in physics, defining how energy is transferred to or from an object. It's a scalar quantity, meaning it only has magnitude, and its SI unit is the Joule (J). The most basic definition for a constant force $\vec{F}$ causing a displacement $\vec{d}$ is $W = \vec{F} \cdot \vec{d} = Fd \cos\theta$.

Here, $F$ and $d$ are magnitudes, and $\theta$ is the angle between the force and displacement vectors. This angle is critical: if $\theta = 0^\circ$, work is maximum positive ($Fd$); if $\theta = 90^\circ$, work is zero; if $\theta = 180^\circ$, work is maximum negative ($-Fd$).

When the force is not constant but varies with position, the work done is found by integration: $W = \int_{x_i}^{x_f} F(x) \, dx$ for one-dimensional motion. Graphically, this corresponds to the area under the Force-displacement ($F-x$) curve. Remember that areas below the x-axis represent negative work.

The Work-Energy Theorem is a cornerstone: $W_{net} = \Delta K = K_f - K_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$. This theorem states that the net work done by all forces on an object equals the change in its kinetic energy. It's a powerful tool for solving problems without direct use of Newton's laws and kinematics.

Specific forces and their work:

Gravity: — $W_g = -mgh$ when an object is lifted (displacement up, force down) and $W_g = +mgh$ when it falls (displacement down, force down). Work done by gravity is path-independent.
Spring Force: — For an ideal spring with constant $k$, stretched or compressed by $x$ from equilibrium, the work done *by* the spring is $W_s = -\frac{1}{2}kx^2$. The work done *by an external agent* to stretch/compress it is $W_{ext} = +\frac{1}{2}kx^2$.
Friction: — Kinetic friction always opposes motion, so the work done by friction is always negative: $W_f = -f_k d$.

Example: A $2\text{ kg}$ block is pulled $3\text{ m}$ by a $15\text{ N}$ force at $60^\circ$ above horizontal. Friction force is $5\text{ N}$.

1
Work by applied force: $W_F = (15\text{ N})(3\text{ m})\cos(60^\circ) = 45 \times 0.5 = 22.5\text{ J}$.
2
Work by friction: $W_f = -(5\text{ N})(3\text{ m}) = -15\text{ J}$.
3
Work by gravity/normal force: $0\text{ J}$ (perpendicular to displacement).
4
Net work: $W_{net} = 22.5 - 15 = 7.5\text{ J}$.

This net work would equal the change in the block's kinetic energy. Always be mindful of signs, units, and applying the correct formula for constant vs. variable forces.

Prelims Revision Notes

Work (W) is a scalar quantity, representing energy transfer. Its SI unit is Joule (J), where $1\text{ J} = 1\text{ N} \cdot \text{m}$.

1. Work by Constant Force:

$W = \vec{F} \cdot \vec{d} = Fd \cos\theta$, where $\theta$ is the angle between $\vec{F}$ and $\vec{d}$.
Positive Work: — $0^\circ \le \theta < 90^\circ$. Force component is in direction of motion. Example: Pushing a box forward.
Negative Work: — $90^\circ < \theta \le 180^\circ$. Force component opposes motion. Example: Friction, gravity on a rising object.
Zero Work: — $\theta = 90^\circ$ or $d=0$. Force is perpendicular to displacement (e.g., normal force, centripetal force) or no displacement occurs (e.g., pushing a wall).

2. Work by Variable Force:

$W = \int_{x_i}^{x_f} F(x) \, dx$ for 1D motion. This is the area under the Force-displacement (F-x) graph.
For a spring, $F_s = -kx$. Work done *by* the spring from equilibrium ($x=0$) to $x$ is $W_s = -\frac{1}{2}kx^2$. Work done *by external agent* is $W_{ext} = +\frac{1}{2}kx^2$.

3. Work-Energy Theorem:

$W_{net} = \Delta K = K_f - K_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$.
The net work done by all forces on an object equals the change in its kinetic energy.

4. Work Done by Specific Forces:

Gravity: — $W_g = mgh$ (if displacement is downwards) or $W_g = -mgh$ (if displacement is upwards). It's a conservative force, so work is path-independent.
Friction: — Always does negative work, $W_f = -f_k d$, as it always opposes relative motion.

5. Key Points for NEET:

Always draw a free-body diagram to identify all forces.
Carefully determine the angle $\theta$ for each force.
Convert all units to SI before calculation.
Be proficient in calculating areas of basic shapes (triangles, rectangles, trapezoids) for F-x graphs.
The Work-Energy Theorem is often a shortcut for problems involving changes in speed.
Distinguish between work done by individual forces and net work done.
Conceptual questions often test understanding of positive, negative, and zero work scenarios.

Vyyuha Quick Recall

Work: For Displacement, Consider Output Sign.

Force
Displacement
Consider Output Sign (for $W = Fd \cos\theta$, where $\cos\theta$ determines the sign of work).