Lesson 5: Work, Energy & Conservation

Weeks 5–6 — The power of energy methods

🎯 Learning Objectives

Define work as force times displacement
Distinguish kinetic energy, potential energy, and mechanical energy
Apply conservation of energy to solve problems

🎯 Intuition First: Energy as Currency

Think of energy like money. You can exchange it between different forms — but the total amount stays the same (if no "tax" like friction).

Kinetic energy = energy of motion (like cash in hand)
Potential energy = stored energy (like money in a bank)
Conservation of energy = total wealth never changes (in a closed system)

💡 Example:

A roller coaster at the top of a hill has maximum potential energy (PE) and zero kinetic energy (KE). As it drops, PE converts to KE. At the bottom, it has maximum KE and minimum PE. The total (PE + KE) stays constant if we ignore friction!

📋 Work

Work: W = F · d · cos(θ)

Units: Joules (J) = N·m

Where θ is the angle between the force and the displacement.

📝 Key Points:

If force is parallel to motion (θ = 0°): W = Fd (maximum positive work)
If force is perpendicular to motion (θ = 90°): W = 0 (no work! Carrying something horizontally does no work against gravity)
If force opposes motion (θ = 180°): W = −Fd (negative work — energy removed)

📋 Kinetic and Potential Energy

Kinetic Energy: KE = ½mv²

Gravitational Potential Energy: PE = mgh

Elastic (Spring) Potential Energy: PEₛ = ½kx²

📋 The Work-Energy Theorem

W_net = ΔKE = KE_f − KE_i

The net work done on an object equals its change in kinetic energy.

📋 Conservation of Mechanical Energy

When only conservative forces (gravity, spring) do work:

KE_i + PE_i = KE_f + PE_f

or: ½mv₁² + mgh₁ = ½mv₂² + mgh₂

📝 Worked Example: Roller Coaster

A 500-kg roller coaster starts from rest at the top of a 40-m hill. What is its speed at the bottom? (Ignore friction.)

Step 1 — Energy at the top:

KE₁ = 0 (starts from rest)

PE₁ = mgh = 500 × 9.8 × 40 = 196,000 J

Total E = 196,000 J

Step 2 — Energy at the bottom (h = 0):

PE₂ = 0, KE₂ = Total E = 196,000 J

Step 3 — Solve for v:

½(500)v² = 196,000

250v² = 196,000

v² = 784

v = 28 m/s (about 101 km/h!)

✅ Speed at bottom = 28 m/s

📝 Worked Example: Spring Launch

A spring (k = 200 N/m) is compressed 0.15 m. A 0.5-kg ball is placed against it. When released, what speed does the ball have?

Step 1 — Spring PE at start:

PEₛ = ½kx² = ½(200)(0.15)² = 100 × 0.0225 = 2.25 J

Step 2 — Convert to KE:

½mv² = 2.25 J

½(0.5)v² = 2.25

0.25v² = 2.25

v² = 9

v = 3 m/s

✅ Launch speed = 3 m/s

✅ Check Your Understanding

Q1: Is work done when you hold a heavy weight still above your head?

Click to reveal answer

No. Work = Fd·cos(θ). The weight doesn't move (d = 0), so W = 0. Even though you're tired, physics says no work is done on the weight.

Q2 (mini-problem): A car (1500 kg) is traveling at 20 m/s. The driver slams on the brakes. If the braking force is 6000 N, how far does the car skid?

Click to reveal answer

Initial KE = ½(1500)(20²) = 300,000 J

Work by friction = Fd (negative) = ΔKE

−6000 × d = 0 − 300,000

d = 300,000/6000 = 50 m