IIT PHASE–2 (Advanced) – PART II Advanced Solved Problems

This section contains advanced IIT-level problems where energy methods alone are insufficient. Students must intelligently combine energy, force constraints, and logic.


Problem 1: Block + Rough Surface + Energy Loss

Question:
A block of mass m slides down a rough incline of angle θ with coefficient of friction μ. Find its speed after moving a distance s.

Step 1: Identify forces

  • Gravity → conservative
  • Friction → non-conservative

Step 2: Energy balance

Initial KE + PE − Work by friction = Final KE

0 + mg(s sinθ) − (μmg cosθ)s = ½ mv²

Step 3: Simplify

v² = 2gs (sinθ − μcosθ)

Answer:
v = √[2gs (sinθ − μcosθ)]

IIT Insight: Acceleration never needed.


Problem 2: Pulley System (Constraint + Energy)

Question:
Two blocks of masses m and 2m are connected by a light string over a smooth pulley. Find the speed after the system moves distance h.

Step 1: Constraint

  • Both blocks have same speed magnitude

Step 2: Energy change

  • 2m moves down → loses PE = 2mgh
  • m moves up → gains PE = mgh

Net energy gain:

ΔE = mgh

Step 3: KE of system

½ mv² + ½ (2m)v² = 3/2 mv²

Final equation:

mgh = 3/2 mv²

Answer:
v = √(2gh/3)

Topper move: Avoid tension & acceleration completely.


Problem 3: Variable Force + Work-Energy

Question:
A particle moves under a force F = kx. Find its speed when it moves from x = 0 to x = a. Mass of particle is m.

Step 1: Work done

W = ∫₀ᵃ kx dx = ½ka²

Step 2: Work–energy theorem

½ka² = ½ mv²

Answer:
v = √(ka²/m)

Examiner trick: Acceleration is not constant → equations of motion fail.


Problem 4: Energy + Circular Motion

Question:
A particle slides down a smooth track and enters a vertical loop of radius R. Find the minimum height from which it must start to complete the loop.

Step 1: Condition at top

  • Minimum speed → N = 0
  • v² = gR

Step 2: Energy balance

mgh = mg(2R) + ½m(gR)

Solve:

h = 5R/2

Answer:
Minimum height = 2.5R

Classic IIT favourite problem.


Problem 5: Energy as Inequality

Question:
A particle moves in a potential field U(x). Under what condition is motion possible?

Key condition:

Total Energy ≥ Potential Energy

Meaning:

  • If E < U → motion impossible
  • Turning points occur at E = U

Advanced idea: Motion limits are decided by energy, not forces.


Problem 6: Energy + Sudden Change

Question:
A moving particle suddenly experiences a loss of energy ΔE. How does speed change?

Method:

  • KE₁ = ½mv₁²
  • KE₂ = KE₁ − ΔE

Answer:

v₂ = √(v₁² − 2ΔE/m)

IIT Application: Used in explosion & impact models.


Phase–2 Problem-Solving Discipline

  • Never rush to equations
  • Draw rough energy flow diagram
  • Identify constraints early
  • Break motion into stages

“Phase–2 problems reward patience and structure.”

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.

Prepared by: Shaktimatha Learning 🌱

Strong Concepts • Smart Practice • Exam Success

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