The Hook: Why This is a Game-Changer
Situation: A car brakes and stops. You need to find braking distance.
The Old Way (Newton’s Laws):
- Draw FBD
- Find acceleration: $a = \frac{F}{m}$
- Use kinematics: $v² = u² + 2as$
- Three steps minimum!
The Work-Energy Way:
$$\text{Work by friction} = \text{Change in KE}$$ $$-\mu mg \cdot s = 0 - \frac{1}{2}mv²$$ONE equation. Done.
This is why Work-Energy theorem is called the “shortcut formula” of mechanics!
The Core Concept
The Theorem (Remember This!)
$$\boxed{W_{net} = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2}$$In simple terms: “Do work ON an object → it speeds up. Work done AGAINST motion → it slows down.”
What It Really Means
| Net Work | What Happens | Example |
|---|---|---|
| Positive | Speed ↑ (KE increases) | Pushing a cart |
| Negative | Speed ↓ (KE decreases) | Braking a car |
| Zero | Speed unchanged | Circular motion at constant speed |
When Dom accelerates his car:
- Engine does positive work → car speeds up
- When he brakes, friction does negative work → car slows down
The total work equals the change in kinetic energy. Always!
Interactive: Work-Energy in Action
See how applied force and friction affect kinetic energy:
The Power Formula
Memory Trick: “WE” = Work = Energy change
$$W_{ALL \, FORCES} = KE_{final} - KE_{initial}$$Expand it:
$$W_1 + W_2 + W_3 + ... = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$$W_{net}$ = Work by ALL forces added together
- Gravity
- Friction
- Applied force
- Normal force (usually 0, perpendicular)
- Spring force
- Tension
Add up work done by EVERY force!
When to Use (vs Newton’s Laws)
What is the question asking for?
| If asking about… | Use |
|---|---|
| Final/initial speed | Work-Energy |
| Distance to stop | Work-Energy |
| Acceleration | Newton’s Laws |
| Time taken | Newton’s Laws |
| Force at a moment | Newton’s Laws |
Rule: Speed/Distance → Work-Energy | Time/Acceleration → Newton
The Master Problem-Solving Template
Step 1: Define initial and final states
- “Starting from rest” → $v_i = 0$
- “Comes to rest” → $v_f = 0$
Step 2: List ALL forces and their work
| Force | Work Formula | Sign |
|---|---|---|
| Gravity | $mgh$ | + (if moving down) |
| Friction | $-\mu N \cdot s$ | Always − |
| Applied force | $Fs\cos\theta$ | Depends on direction |
| Normal | $0$ | (perpendicular to motion) |
| Spring | $\frac{1}{2}kx_i^2 - \frac{1}{2}kx_f^2$ | Depends |
Step 3: Add all work = ΔKE
$$W_1 + W_2 + W_3 = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$Step 4: Solve!
Pattern Recognition: JEE Favorite Types
Pattern 1: Block Stops Due to Friction
Instant Solution:
$$-\mu mg \cdot s = 0 - \frac{1}{2}mv^2$$ $$\mu = \frac{v^2}{2gs}$$Memory: “$\mu$ = v-square by 2gs”
Pattern 2: Bullet Penetrates Block
Instant Solution:
$$-F \cdot d = 0 - \frac{1}{2}mv^2$$ $$F = \frac{mv^2}{2d}$$Pattern 3: Spring Launches Block
Instant Solution:
$$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$$ $$v = x\sqrt{\frac{k}{m}}$$Memory: “v = x root (k/m)”
Pattern 4: Block Slides Down Rough Incline
Instant Solution: Length of incline = $\frac{h}{\sin\theta}$
$$mgh - \mu mg\cos\theta \cdot \frac{h}{\sin\theta} = \frac{1}{2}mv^2$$ $$v = \sqrt{2gh(1 - \mu\cot\theta)}$$Common Mistakes to Avoid
Usually: Normal is perpendicular → Work = 0
Exception: On a MOVING incline, normal force CAN do work!
Rule: Work = 0 only when force ⟂ displacement
For connected bodies:
- Tension does +ve work on one body
- Tension does −ve work on the other
- Net work by tension = 0 (for inextensible string)
But individually, tension DOES work on each body!
Spring work = $\frac{1}{2}kx_i^2 - \frac{1}{2}kx_f^2$
NOT $\frac{1}{2}kx^2$ always!
If spring goes from compressed ($x_i$) to natural length ($x_f = 0$): Work = $\frac{1}{2}kx_i^2 - 0 = \frac{1}{2}kx_i^2$ (positive, gives energy)
Practice Problems (Difficulty-wise)
Level 1: Direct Application
A 2 kg block moving at 10 m/s stops after 25 m on rough surface. Find μ. (g = 10 m/s²)
Solution:
$$-\mu(2)(10)(25) = 0 - \frac{1}{2}(2)(10)^2$$ $$-500\mu = -100$$ $$\mu = \boxed{0.2}$$Level 2: Two Forces
A 5 kg block slides from rest down a rough 30° incline of height 3 m. If μ = 0.2, find speed at bottom.
Solution: Work by gravity = $mgh = 5 \times 10 \times 3 = 150$ J
Incline length = $\frac{h}{\sin 30°} = 6$ m
Work by friction = $-\mu mg\cos30° \times 6 = -0.2 \times 50 \times 0.866 \times 6 = -52$ J
Total work = $150 - 52 = 98$ J
$$98 = \frac{1}{2}(5)v^2$$ $$v = \sqrt{39.2} = \boxed{6.26 \text{ m/s}}$$Level 3: Variable Force (Integration)
Force $F = (4 - 2x)$ N acts on 1 kg body at rest at origin. Find speed at x = 3 m.
Solution:
$$W = \int_0^3 (4-2x)dx = [4x - x^2]_0^3 = 12 - 9 = 3 \text{ J}$$ $$3 = \frac{1}{2}(1)v^2$$ $$v = \sqrt{6} = \boxed{2.45 \text{ m/s}}$$Quick Revision Box
| Situation | Formula |
|---|---|
| Block stops on friction | $\mu = \frac{v^2}{2gs}$ |
| Spring launch | $v = x\sqrt{\frac{k}{m}}$ |
| Fall from height h | $v = \sqrt{2gh}$ |
| Rough incline | $v = \sqrt{2gh(1-\mu\cot\theta)}$ |
| Bullet penetration | $F = \frac{mv^2}{2d}$ |
Teacher’s Summary
Work-Energy = Best Friend for Speed Problems
- No need for acceleration, time, or FBD complexity
NET Work = Sum of Work by ALL Forces
- Don’t forget any force!
Sign Matters!
- Force in direction of motion → +ve work
- Force against direction of motion → −ve work
“When you need to find speed or distance, think Work-Energy first!”
Related Topics
Within Work, Energy & Power
- Conservation of Energy — When only conservative forces act
- Potential Energy — Stored energy in position/configuration
- Collisions — Applying WE theorem to impacts
Connected Chapters
- Kinematic Equations — Alternative approach (when you need time)
- Newton’s Laws — Force-based approach to same problems
- Rotational Motion — WE theorem for rotating bodies
- Integration — For variable force problems
Cross-Subject Connection
- Chemical Thermodynamics — Energy conservation in chemical reactions