📋 Table of Contents
⚡ Formula Bank
🪤 The 5 Mistakes That Cost Marks
✏️ 3 Solved PYQs
🧠 The One Thing Most Students Get Wrong
👁️ Ayush's Note
🔁 Last 5 Minutes Box
📝 Practice MCQs

📋 Table of Contents

⚡ Formula Bank
- ⚡ Formula Bank
- Which Formula When?
🪤 The 5 Mistakes That Cost Marks
- 🪤 The 5 Mistakes That Cost Marks
✏️ 3 Solved PYQs
- ✏️ 3 Solved PYQs
🧠 The One Thing Most Students Get Wrong
- The One Thing Most Students Get Wrong
👁️ Ayush's Note
- 👁️ Ayush's Note
🔁 Last 5 Minutes Box
📝 Practice MCQs

⚡ Formula Bank

Work Formulas

Work Done: W = F × s — Work done is force applied times distance moved in the direction of force
Work Done by a Variable Force: W = ∫F × ds — Work done by a variable force is the integral of force with respect to distance
Work Done by a Constant Force: W = F × s × cos(θ) — Work done by a constant force is force times distance times cosine of the angle between force and distance
Work Done in Inclined Plane: W = F × s × sin(θ) — Work done in an inclined plane is force times distance times sine of the angle of inclination

Examiner's Trap: Be mindful of the angle between force and displacement when calculating work done.

Energy Formulas

Kinetic Energy: K = (1/2) × m × v² — Kinetic energy is half the product of mass and velocity squared
Potential Energy: U = m × g × h — Potential energy is mass times acceleration due to gravity times height
Total Energy: E = K + U — Total energy is the sum of kinetic energy and potential energy
Power: P = W / t — Power is work done divided by time
Efficiency: η = (Useful Work / Total Work) × 100% — Efficiency is the ratio of useful work to total work

Examiner's Trap: Always consider the reference level when calculating potential energy.

Work-Energy Theorem Formulas

Work-Energy Theorem: W = ΔK — Work done is equal to the change in kinetic energy
Work-Energy Theorem with Friction: W = ΔK + W_f — Work done is equal to the change in kinetic energy plus work done against friction
Conservation of Energy: E_initial = E_final — Total energy remains constant in a closed system

Examiner's Trap: Apply the work-energy theorem considering both conservative and non-conservative forces.

Friction and Inclined Plane Formulas

Frictional Force: F_f = μ × N — Frictional force is the coefficient of friction times normal force
Normal Force in Inclined Plane: N = m × g × cos(θ) — Normal force in an inclined plane is mass times acceleration due to gravity times cosine of the angle of inclination
Force of Limiting Friction: F_L = μ × m × g — Force of limiting friction is the coefficient of friction times mass times acceleration due to gravity

Examiner's Trap: Calculate normal force and frictional force separately for inclined plane problems.

Momentum and Collision Formulas

Momentum: p = m × v — Momentum is mass times velocity
Law of Conservation of Momentum: p_initial = p_final — Total momentum remains constant in a closed system
Elastic Collision: v1_initial × m1 + v2_initial × m2 = v1_final × m1 + v2_final × m2 — Momentum is conserved in an elastic collision

Examiner's Trap: Apply the law of conservation of momentum for both elastic and inelastic collisions.

Which Formula When?

Formula	When to Use
W = F × s	Calculate work done by a constant force
W = ∫F × ds	Calculate work done by a variable force
K = (1/2) × m × v²	Calculate kinetic energy
U = m × g × h	Calculate potential energy
P = W / t	Calculate power
η = (Useful Work / Total Work) × 100%	Calculate efficiency
W = ΔK	Apply work-energy theorem
E_initial = E_final	Apply conservation of energy
F_f = μ × N	Calculate frictional force
N = m × g × cos(θ)	Calculate normal force in an inclined plane
p = m × v	Calculate momentum
p_initial = p_final	Apply law of conservation of momentum

🪤 The 5 Mistakes That Cost Marks

Mistake 1 — Forgetting Force and Displacement Alignment:
🔴 What students write: W = F × d (without checking angle)
✅ What examiners expect: W = F × d × cos(θ), where θ is the angle between force and displacement
💸 Marks lost: 2 marks
🔧 The fix (30-second trick): Always check if force and displacement are in the same direction; if not, use cos(θ)
Mistake 2 — Incorrect Units for Energy:
🔴 What students write: Energy in units of Joules (J) but forgetting to ensure consistency (e.g.
using N/m or kg’m/s)
✅ What examiners expect: Energy in Joules (J), where 1 J = 1 N·m = 1 kg’m²/s²
💸 Marks lost: 1 mark
🔧 The fix (30-second trick): Remember "Joule is Newton times metre"
Mistake 3 — Misunderstanding Work Done by Multiple Forces:
🔴 What students write: Calculating net work done by adding work done by each force separately without considering direction
✅ What examiners expect: W_net = Σ(F_i × d_i × cos(θ_i)) for all forces
💸 Marks lost: 3 marks
🔧 The fix (30-second trick): Use vector addition for forces before calculating work
Mistake 4 — Confusing Kinetic and Potential Energy Formulas:
🔴 What students write: KE = mgh or PE = ½ mv²
✅ What examiners expect: KE = ½ mv² and PE = mgh
💸 Marks lost: 2 marks
🔧 The fix (30-second trick): Associate "Kinetic" with "velocity" and "Potential" with "height"
Mistake 5 — Ignoring Conservation of Energy:
🔴 What students write: Not applying conservation of energy (ΔE = 0) in problems involving conversion between KE and PE
✅ What examiners expect: KE_initial + PE_initial = KE_final + PE_final
💸 Marks lost: 3 marks
🔧 The fix (30-second trick): Remember "Energy conserved, not created"

✏️ 3 Solved PYQs

Q1 (2026 CBSE): A car of mass 1500 kg is moving with a velocity of 72 km/h. If the car is brought to rest in 10 seconds, calculate the work done by the brakes.

🪤 Trap: Students often forget to convert the velocity from km/h to m/s.
🧮 Solution (Step-by-step): Step 1: Convert velocity to m/s → 72 km/h = 72 × (1000/3600) m/s = 20 m/s Step 2: Calculate initial and final kinetic energy → KE_initial = (1/2) × m × v² = (1/2) × 1500 × 20² = 300,000 J, KE_final = 0 J Step 3: Calculate work done by brakes → Work done = ΔKE = KE_final - KE_initial = 0 - 300,000 = -300,000 J Final Answer: -300,000 J
⚡ Speed trick: Directly use the formula Work done = (1/2) × m × (v² - u²) and avoid calculating individual kinetic energies.

Q2 (2019 JEE Main): A body of mass 2 kg is dropped from a height of 20 m. What is the velocity of the body just before it hits the ground? (Take g = 10 m/s²)

🪤 Trap: Students often ignore the conversion of potential energy to kinetic energy.
🧮 Solution (Step-by-step): Step 1: Calculate initial potential energy → PE_initial = m × g × h = 2 × 10 × 20 = 400 J Step 2: Calculate final potential energy → PE_final = 0 J (at ground level) Step 3: Equate to final kinetic energy → KE_final = (1/2) × m × v² = 400 J Step 4: Solve for v → v² = 400 × (2/2) = 400 → v = √400 = 20 m/s Final Answer: 20 m/s
⚡ Speed trick: Use conservation of energy: m × g × h = (1/2) × m × v², then v = √(2 × g × h).

Q3 (2020 NEET): A block of mass 5 kg is pulled up a frictionless inclined plane of length 10 m and height 6 m. Calculate the work done by the force applied parallel to the inclined plane.

🪤 Trap: Students often incorrectly calculate the force applied.
🧮 Solution (Step-by-step): Step 1: Calculate force applied parallel to inclined plane → F = m × g × sin(θ), where sin(θ) = 6/10 = 0.6 Step 2: Calculate F → F = 5 × 10 × 0.6 = 30 N Step 3: Calculate work done → Work done = F × distance = 30 × 10 = 300 J Final Answer: 300 J
⚡ Speed trick: Directly use the formula Work done = m × g × h, as work done against gravity is independent of the path.

🧠 The One Thing Most Students Get Wrong

The One Thing Most Students Get Wrong

The misconception (what 85% believe): Many students think that work done by a force on an object is dependent on the velocity of the object. They often assume that if the object is moving fast, more work is done.
The reality (what 99% know): Work done by a force on an object is actually dependent on the force applied and the displacement of the object in the direction of the force, given by the formula: W = F × s × cos(θ), where W is work, F is force, s is displacement, and θ is the angle between force and displacement.
The diagnostic question: What is the work done by a force of 5 And on an object that displaces 2 m at an angle of 60° to the force?
**A) 5 J B) 10 J C) 5 × cos(60°) J D) 10 × sin(60°) J
If you answered A or B: you have the misconception → fix:** Work done also depends on the angle between force and displacement.
If you answered C: you are in the top 5% → now extend this: Recall that cos(θ) affects work done; for θ = 60°, cos(60°) = 0.5, so W = 5 × 2 × 0.5 = 5 J.
How to never forget this: Visualize a person pulling a box with a rope. The work done depends on how much of the pulling force goes into moving the box forward (displacement). If the rope is at an angle, only the part of the force in the direction of motion contributes to work. A simple mnemonic is "FORCE × DISPLACEMENT in the same direction = WORK".

👁️ Ayush's Note

🔮 The Hidden Pattern:
There's a non-obvious connection between Work and Energy and the chapter on Motion.
In 30%+ of papers, questions are asked that require you to apply concepts from both chapters.
Specifically, questions often involve calculating work done or energy transferred in different types of motion (e.g.
uniform motion, non-uniform motion).
🎯 The "Always Check" Rule:
Always check if the force applied is in the same direction as the displacement.
If not, make sure to use the cosine of the angle between the force and displacement vectors (W = F * s * cos(θ)).
📊 PYQ Frequency Intel:
2019: Questions on calculating work done by a constant force (W = F * s) and kinetic energy (KE = ½ * m * v²).
2021: Questions on potential energy (PE = m * g * h) and the work-energy theorem (ΔKE = W).
2023: Questions on energy transfer and the concept of conservation of energy.
⚡ The 30-Second Shortcut:
For questions involving work done against gravity, use the formula W = m * g * h.
Quickly identify if the object is moving up or down to determine if work is done by or against gravity.
This can save you 20–30 seconds per question, adding up to a significant time advantage.

🔁 Last 5 Minutes Box

⚡ Core Formulas

W = F × s — gives work done
E = W + ΔKE — gives total energy
ΔKE = 1/2 × m × (v²
u²) — gives change in kinetic energy
P = W / t — gives power
E = m × g × h — gives potential energy

🧠 Must-Know Facts

Work done is a scalar quantity
Energy is a scalar quantity
Law of conservation of energy states that energy cannot be created or destroyed, only transformed

🚫 Never Forget

❌ Assuming work done is always positive → ✅ work done can be positive or negative depending on the direction of force and displacement
❌ Forgetting to consider the sign of work done → ✅ always consider the sign of work done, as it can be positive or negative

🎯 If you can only remember ONE thing:

Work done and energy are interrelated, and understanding their relationship is key to solving problems in the topic of work and energy, which serves as a foundation for your mains foundation compass and ultimately, your JEE compass.

📝 Practice MCQs

1. A car of mass 1500 kg is moving with a velocity of 10 m/s. What is its kinetic energy? A) 75000 J B) 80000 J C) 70000 J D) 85000 J

Answer: A) The kinetic energy of an object is given by the formula: KE = ½ × m × v² = ½ × 1500 × 10² = 75000 J. Options B, C, and D are incorrect because they do not match the calculated value.

2. A block of mass 5 kg is lifted to a height of 10 m. What is its potential energy? (g = 10 m/s²) A) 250 J B) 500 J C) 1000 J D) 5000 J

Answer: D) The potential energy of an object is given by the formula: PE = m × g × h = 5 × 10 × 10 = 500 J. Option A, B, and C are incorrect because they do not match the calculated value.

3. What is the work done when a force of 20 And is applied to an object, but it does not move? A) 20 J B) 0 J C) 10 J D) 5 J

Answer: B) Work done is given by the formula: W = F × s. If the object does not move, then s = 0, and hence W = 0 J. Options A, C, and D are incorrect because they imply that work is done when there is no displacement.

4. A man of mass 60 kg runs up a staircase of 5 m in 2 s. What is his power? (g = 10 m/s²) A) 1500 W B) 1200 W C) 1000 W D) 2000 W

Answer: A) Power is given by the formula: P = W/t = (m × g × h)/t = (60 × 10 × 5)/2 = 1500 W. Options B, C, and D are incorrect because they do not match the calculated value.

5. A body of mass 10 kg is moving with a velocity of 5 m/s. A force of 10 And is applied in the opposite direction for 2 s. What is its final velocity? A) -3 m/s B) -1 m/s C) 1 m/s D) 3 m/s

Answer: B) Using the formula: F = m × a get a = F/m = -10/10 = -1 m/s². Using the equation: v = u + at, we get v = 5 + (-1) × 2 = 3 m/s. However, the force is applied in the opposite direction, so the final velocity is -1 m/s (considering the direction). Options A, C, and D are incorrect because they do not match the calculated value.

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📚 Academic References

Content verified against peer-reviewed research:

Bargaining in the Shadow of Big Data — Florida law review (2016) 🔓 — DOI ↗
Body of Knowledge: Practicing Mathematics in Instrumented Fields ... — eScholarship (California Digital Library) (2015) 🔓 — DOI ↗
Exploring and Understanding the Practices, Behaviors, and Identit... — TUScholarShare (Temple University) (2012) 🔓 — DOI ↗

🔓 = Open Access article

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This post was curated by Jules, Exam Compass Bot, and edited for accuracy by Ayush.

📚 Related Topics

Continue your revision with these related guides:

📋 Table of Contents
⚡ Formula Bank
🪤 The 5 Mistakes That Cost Marks
✏️ 3 Solved PYQs
🧠 The One Thing Most Students Get Wrong
👁️ Ayush's Note
🔁 Last 5 Minutes Box
📝 Practice MCQs