Why the Work‑Energy Theorem matters
Ever wondered why a roller‑coaster speeds up after a steep drop? The answer lies in the work‑energy theorem – a neat shortcut that lets you trade force calculations for energy changes.
💡 In Simple Words: The work‑energy theorem says the total work a net force does on an object equals the change in that object’s kinetic energy. In other words, if you push something and it speeds up, the push’s work turned into motion energy.
What is "work" in physics?
Work is the effort you do when a force moves something. Imagine pushing a shopping cart: the force you apply times how far the cart rolls in the direction of your push gives the work done. Mathematically, work = force × displacement × cosθ, where θ is the angle between the force and the direction of motion.
What is "kinetic energy"?
Kinetic energy (KE) is the energy an object has because it’s moving. The faster it goes, the more KE it carries. The formula is KE = ½ mv², where m is mass and v is speed.
The Work‑Energy Theorem
The theorem ties the two ideas together:
Net work done on an object = change in its kinetic energy. In symbols, W_net = ΔKE = KE_final – KE_initial.
Why “net” work? Because several forces can act at once – gravity, friction, a push – and we add up the work each does. If the total is positive, the object speeds up; if it’s negative, it slows down.
When does the theorem shine?
- When forces are messy but you know the start‑ and end‑speeds.
- When you need a quick check on energy conservation.
- In exam problems that ask for final speed after a push or drop.
Worked Example 1: Pushing a crate
A student pushes a 15 kg crate across a friction‑less floor with a constant horizontal force of 60 N for 4 m. What is the crate’s speed after the push if it started from rest?
Step 1 – Find the work done. Since the force is parallel to the motion, θ = 0°, cosθ = 1.
Work = F × d × cosθ = 60 N × 4 m × 1 = 240 J (joules).
Step 2 – Apply the work‑energy theorem. Initial KE = 0 (starts from rest). So ΔKE = 240 J = KE_final.
Step 3 – Solve for final speed. KE_final = ½ m v² → 240 = ½ × 15 × v² → v² = 240 × 2 / 15 = 32 → v ≈ 5.66 m/s.
The crate ends up moving at about 5.7 m/s.
Worked Example 2: A falling ball
A 0.2 kg ball is dropped from a height of 5 m. Ignoring air resistance, find its speed just before it hits the ground using the work‑energy theorem.
Step 1 – Identify the work. The only force doing work is gravity. The work done by gravity equals weight (mg) times the vertical drop (h): W = mg h.
Plug in numbers: m = 0.2 kg, g ≈ 9.8 m/s², h = 5 m → W = 0.2 × 9.8 × 5 = 9.8 J.
Step 2 – Relate to kinetic energy. The ball starts from rest, so initial KE = 0. Therefore, ΔKE = 9.8 J = KE_final.
Step 3 – Find the speed. 9.8 = ½ × 0.2 × v² → v² = 9.8 × 2 / 0.2 = 98 → v ≈ 9.9 m/s.
So the ball hits the ground at roughly 10 m/s.
Work‑Energy Theorem vs. Conservation of Mechanical Energy
| Aspect | Work‑Energy Theorem | Conservation of Mechanical Energy |
|---|---|---|
| What it relates | Net work ↔ change in kinetic energy | Sum of kinetic + potential energy stays constant (if no non‑conservative forces) |
| When to use | When external forces do work (e.g., push, friction) | When only conservative forces (gravity, spring) act |
| Handles friction | Yes – friction does negative work | No – friction violates conservation unless accounted as energy loss |
| Typical formula | W_net = ΔKE | KE_i + PE_i = KE_f + PE_f |
Key Points to Remember
- Work is a scalar (just a number), not a vector.
- Positive work increases kinetic energy; negative work decreases it.
- If the net work is zero, the object’s speed doesn’t change (Newton’s first law).
- Always keep track of the direction of forces when calculating work (use cosθ).
📝 Likely Exam Questions
- Question: A 10 kg crate is pulled 3 m up a rough incline by a 50 N force parallel to the slope. The coefficient of kinetic friction is 0.2. Find the crate’s speed after the pull if it started from rest.
Answer: Work by pull = 50 × 3 = 150 J. Work by friction = –μ mg cosθ × d (θ≈30°) ≈ –0.2 × 10 × 9.8 × cos30° × 3 ≈ –50 J. Net work ≈ 100 J = ΔKE → v = √(2 W/m) ≈ √(20) ≈ 4.5 m/s. - Question: A 2 kg block slides down a frictionless 4‑m hill from rest. Using the work‑energy theorem, calculate its speed at the bottom.
Answer: Work by gravity = mg h = 2 × 9.8 × 4 = 78.4 J = ΔKE. v = √(2 W/m) = √(78.4) ≈ 8.86 m/s. - Question: Explain why a skydiver reaches a constant terminal speed despite gravity continuously doing work.
Answer: Gravity does positive work, but air resistance does equal negative work. Net work becomes zero, so kinetic energy (and speed) stops increasing – that’s terminal velocity.