Energy Conservation & Applications Work, Energy & Power

LAW OF CONSERVATION OF ENERGY

"Energy cannot be created or destroyed — it can only be converted from one form to another."

KE₁ + PE₁ = KE₂ + PE₂ (no friction)

With friction: KE₁ + PE₁ − W_friction = KE₂ + PE₂

Spring-Mass Systems

Spring problems are among the top 3 most-tested topics in this chapter for JEE Main and Advanced. The key is identifying who does work on whom.

In spring problems, ALWAYS apply energy conservation between clearly defined initial and final states. Never try to work with spring forces directly unless the problem demands acceleration.

Case 1: Block Attached to Spring on Frictionless Surface

Block of mass m attached to spring (k), compressed by x₀ and released from rest. Find maximum speed.

1

Initial state: Block at rest, spring compressed by x₀

KE₁ = 0, PE₁ = ½kx₀²
2

Maximum speed: When spring returns to natural length (PE = 0)

KE₂ = ½mv²_max, PE₂ = 0
3

Energy conservation: KE₁ + PE₁ = KE₂ + PE₂

0 + ½kx₀² = ½mv²_max + 0

v_max = x₀√(k/m)

v_max = x₀√(k/m) is the standard result. Know it. Also: v_max occurs at equilibrium position (natural length of spring).

Case 2: Block on Rough Surface + Spring

Block (mass m, μ) slides distance d on rough surface and compresses spring by x. Find v when spring compressed by x/2.

Include friction loss in energy equation. Energy lost to friction = μmg × total distance traveled. This is where students go wrong — they forget friction acts on the FULL path, not just part of it.

Energy Equation

½mv² = KE_initial − ½kx² − μmg(d + x)

Spring Combinations (JEE Advanced)

🔴 Two Springs, Same Block ▾

Block between two springs (k₁ and k₂), displaced by x. Both springs share deformation.

Effective spring constant (parallel)

k_eff = k₁ + k₂

PE = ½(k₁ + k₂)x²

In series: same force, extensions add. In parallel: same extension, forces add. This determines PE = ½k_eff x².

🔴 Spring Cut in Half ▾

If a spring of spring constant k is cut into n equal parts, each part has spring constant nk.

k_part = n × k_original

JEE: "A spring of k = 100 N/m is cut into 4 equal parts. A block is attached to 2 of these parts in parallel. Find new k." Answer: Each part k = 400 N/m. Two in parallel: k_eff = 800 N/m.

🔴 Spring-Mass Maximum Compression ▾

Block hits stationary spring with velocity v. Find maximum compression.

At max compression: Block velocity = 0 (momentarily at rest)

½mv² = ½kx²_max → x_max = v√(m/k)

🎯 JEE Advanced Spring Problem Type

Two blocks m₁ and m₂ connected by spring. m₁ moving with velocity v collides with m₂ (at rest). Find maximum compression of spring.

At maximum compression: Both blocks move with SAME velocity (common velocity). Use momentum conservation first: v_common = m₁v/(m₁+m₂). Then energy: ½m₁v² − ½(m₁+m₂)v_common² = ½kx²_max.

Energy Conservation on Inclined Planes

For inclined plane problems, ALWAYS resolve height, not distance. h = L sinα where L = length along incline, α = inclination angle. This is where 40% of students lose marks.

Case: Block sliding down rough incline

Block slides from top to bottom of rough incline

½mv² = mgh − μmg cosα × L

= mgL sinα − μmgL cosα = mgL(sinα − μcosα)

Common error: Using μmg as friction force on incline. Correct friction force = μN = μmg cosα (not μmg). The normal force reduces on an incline.

Critical Angle and Stopping Conditions

When does block slide at all?

Block slides only if: mg sinα > μmg cosα → tan α > μ → α > arctan(μ)

If tan α = μ: Block is in equilibrium (just about to slide). If tan α < μ: Block does not slide. This is tested in NEET multiple times.

Sliding Up Then Back Down

A block is given initial velocity v₀ up a rough incline. Find velocity when it returns to start.

1

Going up: Both gravity and friction oppose motion. Max height h reached:

½mv₀² = mgh + μmg cosα × h/sinα
2

Coming down: Gravity helps, friction opposes. Final speed:

mgh − μmg cosα × h/sinα = ½mv²
3

Result: v < v₀. Energy lost = 2×μmg cosα × L

On a round trip: Energy lost = 2 × W_friction (friction acts both ways, always opposing motion).

Vertical Circular Motion

A classic application of energy conservation. The condition for completing a vertical loop is one of the most frequently tested concepts in this chapter.

Minimum Speed at TOP of Loop

v_top = √(gR)

At top: centripetal force = mg (minimum condition). Normal force = 0.

Minimum Speed at BOTTOM to complete loop

v_bottom = √(5gR)

Derivation: Use energy conservation from bottom to top. Height gained = 2R. v²_bottom = v²_top + 4gR. At minimum: v_top = √(gR). So v²_bottom = gR + 4gR = 5gR → v_bottom = √(5gR). Memorize the result, but understand the derivation.

Speed at Any Point in Loop

At angle θ from bottom (height h = R - R cosθ)

v² = v²_bottom − 2g·R(1 − cosθ)

📊 Vertical Loop — Key Points

Position	v_min	N (Normal)
Bottom	√(5gR)	6mg
Middle (side)	√(3gR)	3mg
Top	√(gR)	0 (minimum)

This is where most students go wrong: At the TOP, "N + mg = mv²/R" (both N and mg point towards center). At the BOTTOM, "N − mg = mv²/R". Wrong sign → wrong answer.

NEET pattern: "Minimum speed at bottom for complete revolution" = √(5gR). Appears almost every 2 years. At JEE, this gets extended to: bob on string vs rod vs track — all have different conditions.

Collisions and Energy

Collisions combine momentum conservation (always) with energy conservation (only elastic). This makes collision problems a mandatory JEE topic.

Elastic Collision

• Momentum conserved ✅
• KE conserved ✅
• e = 1 (coefficient of restitution)

Inelastic Collision

• Momentum conserved ✅
• KE NOT conserved ❌
• e < 1

Elastic Collision Formulas

Velocities After Elastic Collision (1D)

v₁ = (m₁−m₂)u₁/(m₁+m₂) + 2m₂u₂/(m₁+m₂)

v₂ = 2m₁u₁/(m₁+m₂) + (m₂−m₁)u₂/(m₁+m₂)

Special Cases — Memorize These

Equal masses (m₁ = m₂), u₂ = 0 ▾

v₁ = 0, v₂ = u₁. Bodies exchange velocities. Newton's cradle!

Used in Newton's cradle, billiards, nuclear moderation.

m₁ >> m₂ (heavy hits light), u₂ = 0 ▾

v₁ ≈ u₁ (heavy barely changes), v₂ ≈ 2u₁ (light bounces with double speed)

m₂ >> m₁ (light hits heavy wall), u₂ = 0 ▾

v₁ ≈ −u₁ (bounces back with same speed), v₂ ≈ 0 (wall stays)

Ball bouncing off a wall. The wall doesn't move. Ball reverses direction.

Perfectly Inelastic Collision

Common velocity after perfectly inelastic collision

v_common = (m₁u₁ + m₂u₂)/(m₁ + m₂)

KE Lost

ΔKE = ½ × (m₁m₂/(m₁+m₂)) × (u₁−u₂)²

This is always positive → always energy lost in inelastic collision.

The quantity m₁m₂/(m₁+m₂) is the reduced mass (μ). It appears in many advanced contexts. Remember it.

Coefficient of Restitution (e)

Definition

e = (v₂ − v₁)/(u₁ − u₂) = relative speed after / relative speed before

e = 1 → Elastic collision
e = 0 → Perfectly inelastic (stick together)
0 < e < 1 → Partially inelastic

Ball dropped from height H

Ball dropped from H. After bouncing with e, height reached:

h₁ = e²H

After n bounces: hₙ = e²ⁿH

Potential Energy Curves (JEE Advanced)

Reading PE vs position graphs is a pure JEE Advanced skill. Master this and you can solve problems that NEET students cannot even attempt.

From any U(x) curve, you can read off: (1) Force at any point (F = −slope), (2) Equilibrium positions (where slope = 0), (3) Type of equilibrium (stable/unstable/neutral), (4) Turning points (where E = U), (5) Forbidden regions (where E < U).

Reading a PE Curve — Step by Step

1

Draw horizontal line at total energy E. Where it intersects U(x): KE = 0 (turning points). Below the line: KE > 0 (motion allowed).
2

Find equilibrium: Where dU/dx = 0 (slope = 0). These are points where F = 0.
3

Determine stability: Minimum of U → stable (d²U/dx² > 0). Maximum of U → unstable.
4

Find force: F = −dU/dx = −(slope of U-x graph). Steep slope → large force.

📊 Annotated PE Curve

🟢 Stable Eq. 🔴 Unstable Eq. 🟣 Turning Points 🔵 Total Energy E

The region where U(x) > E is the classically forbidden region — the particle cannot enter it (insufficient energy). This concept is fundamental to quantum mechanics and appears in JEE Advanced theoretical questions.

JEE Advanced MCQ pattern: "Which of the following is NOT correct about the given PE curve?" Always check all four options systematically using the 4 reading steps above.