Ever wondered why a roller‑coaster speeds up after the first drop? The secret lies in the work‑energy theorem.
💡 In Simple Words: When you do work on an object, you change its kinetic energy. The amount of work you put in equals the increase (or decrease) in the object's motion energy.
What is the Work‑Energy Theorem?
The work‑energy theorem says: Net work done on a body = change in its kinetic energy. “Net work” means the total work from all forces acting together. “Kinetic energy” (KE) is the energy an object has because it’s moving, given by the formula KE = ½ mv² (m = mass, v = speed).
Deriving the Formula
Start with the definition of work: work = force × displacement × cosθ, where θ is the angle between the force direction and the displacement. If the force is constant and acts along the line of motion, θ = 0° and cosθ = 1, so work = force × displacement.
Newton’s second law tells us force = mass × acceleration (F = ma). Replace force in the work expression and use the kinematic relation a = dv/dt. After a few algebra steps you end up with work = ½ m(v² – u²), which is exactly the change in kinetic energy (final KE minus initial KE).
How to Apply the Theorem – Step by Step
Whenever a question asks for speed, distance, or force, think “work‑energy”. Follow these four moves:
- Identify the system (the object whose speed you want).
- List every force that does work on the system.
- Calculate the work each force does (use W = F d cosθ).
- Set the total work equal to the change in kinetic energy and solve.
Worked Example 1: Pulling a Sled
A child pulls a 30 kg sled across a frictionless snow surface with a constant horizontal force of 60 N. The sled starts from rest and moves 5 m. Find its speed after the pull.
Step 1: System = sled.
Step 2: Only the pulling force does work (no friction).
Step 3: Work = F d = 60 N × 5 m = 300 J.
Step 4: Initial KE = 0 (starts from rest). Set 300 J = ½ m v² → 300 = ½ × 30 × v² → 300 = 15 v² → v² = 20 → v ≈ 4.47 m/s.
So the sled is moving about 4.5 m/s after the pull.
Worked Example 2: Braking a Car
A 1500 kg car is traveling at 20 m/s. The driver applies brakes that produce a constant retarding force of 4000 N until the car stops. How far does the car travel while stopping?
Step 1: System = car.
Step 2: The braking force does negative work (it removes energy).
Step 3: Work = F d cosθ. Here θ = 180° because force opposes motion, so cosθ = –1. Thus, work = –4000 N × d.
Step 4: Change in KE = final KE – initial KE = 0 – ½ m v² = –½ × 1500 × 20² = –300 000 J.
Set –4000 d = –300 000 → d = 75 m.
The car slides 75 m before it halts.
Quick Comparison Table
| Situation | Force Direction | Work Sign | Effect on KE |
|---|---|---|---|
| Pulling a sled forward | Same as motion | Positive | KE increases |
| Braking a car | Opposite to motion | Negative | KE decreases |
| Car moving on level road with friction | Opposite to motion | Negative | KE stays constant if engine supplies equal positive work |
📝 Likely Exam Questions
- State the work‑energy theorem and write its mathematical expression.
Answer: Net work done on a body equals the change in its kinetic energy; W_net = ΔKE = ½ m(v² – u²). - A 2 kg block is pushed with a constant force of 10 N over 3 m. Find its final speed if it started from rest.
Answer: Work = 10×3 = 30 J; 30 = ½×2×v² → v = √(30) ≈ 5.48 m/s. - Explain why a falling object speeds up even though gravity does positive work while air resistance does negative work.
Answer: Gravity adds kinetic energy (positive work). Air resistance removes some of that energy (negative work). The net work is still positive, so KE increases. - Calculate the stopping distance of a 1000 kg train moving at 15 m/s if the brakes apply a constant 5000 N force.
Answer: Initial KE = ½×1000×15² = 112 500 J. Work by brakes = –5000 d. Set –5000 d = –112 500 → d = 22.5 m. - List the four steps to solve a work‑energy problem.
Answer: Identify the system, list all forces, compute each force’s work, set total work equal to change in kinetic energy and solve.