Lesson 5 – Work, Energy and Power Advanced / IIT-JEE Numerical Problems
These problems reflect IIT-JEE patterns where using the energy method is faster than Newton’s laws. Each problem highlights the correct method choice.
Problem 1: Variable Force (IIT Favorite)
Q1. A force F = 3x² N acts on a particle along x-direction. Find the work done in moving the particle from x = 0 to x = 2 m.
Method: Energy / integration
Work done:
W = ∫ F dx = ∫ 3x² dx
W = 3 ∫ x² dx = 3 [x³ / 3]₀² = 2³ = 8 J
Answer: Work done = 8 J
IIT Tip: Variable force → NEVER use F = ma directly.
Problem 2: Energy Conservation with Height
Q2. A body of mass 2 kg is dropped from a height of 5 m. Find its speed just before hitting the ground. (Take g = 10 m/s²)
Method: Conservation of energy
Potential energy = mgh = 2 × 10 × 5 = 100 J
Kinetic energy = ½ mv²
½ × 2 × v² = 100 → v² = 100 → v = 10 m/s
Answer: Speed = 10 m/s
Shortcut: Time & acceleration avoided completely.
Problem 3: Friction + Energy Loss
Q3. A block of mass 1 kg slides on a rough surface with initial speed 6 m/s. If friction force is 3 N, find the distance travelled before it stops.
Method: Work–energy theorem
Initial KE = ½ mv² = 18 J
Work done by friction = −Fs = −3s
Using work–energy theorem:
−3s = −18 → s = 6 m
Answer: Distance travelled = 6 m
IIT Trap: Friction breaks energy conservation.
Problem 4: Power at an Instant
Q4. A body moves with velocity 5 m/s under a force of 4 N. Find the instantaneous power.
Formula: P = F · v
P = 4 × 5 = 20 W
Answer: Power = 20 W
Key Idea: Power depends on velocity, not displacement.
Problem 5: Constant Velocity Trap
Q5. A body moves with constant velocity on a horizontal surface. What is the net power acting on it?
At constant velocity → Net force = 0
Hence, Power = F · v = 0
Answer: Net power = 0
IIT Trap: Individual forces may do work, but net power is zero.
Problem 6: Energy Method vs Newton’s Law
Q6. A particle moves under a force such that its kinetic energy changes from 9 J to 25 J. Find the work done.
Work done = Change in kinetic energy = 25 − 9 = 16 J
Answer: Work done = 16 J
Toppers’ Trick: KE change → no need of mass, velocity, or force.
Advanced Problem-Solving Strategy
- Check if energy method applies first
- Avoid time whenever possible
- Use integration for variable forces
- Account for energy losses carefully
“Choose the right method — half the problem is solved.”
Work, Energy & Power – Complete Physics Library
This is the MASTER LIBRARY PAGE for the complete chapter Work, Energy and Power, prepared for Intermediate, IIT-JEE (Main & Advanced), NEET and competitive exams.
All concepts are explained from basic to IIT level, including theory, derivations, numerical problems, objective questions, previous year questions, tough IIT problems, tricks and cautions.
📚 Complete Lesson Index (Part 1 – Part 39)
- Part 1 – Introduction to Work
- Part 2 – Types of Work
- Part 3 – Variable Force & Graphs
- Part 4 – Kinetic Energy
- Part 5 – Work–Energy Theorem
- Part 6 – Potential Energy
- Part 7 – Conservative Forces
- Part 8 – Mechanical Energy
- Part 9 – Power
- Part 10 – Power Applications
- Part 11
- Part 12
- Part 13
- Part 14
- Part 15
- Part 16
- Part 17
- Part 18
- Part 19
- Part 20
- Part 21
- Part 22
- Part 23
- Part 24
- Part 25
- Part 26
- Part 27
- Part 28
- Part 29
- Part 30
- Part 31
- Part 32
- Part 33
- Part 34
- Part 35
- Part 36
- Part 37
- Part 38
- Part 39 – Final IIT Tough Problems & Solutions
Prepared by: Shaktimatha Learning 🌱
Strong Concepts • Smart Practice • Exam Success
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