Work‑Energy Theorem – Why It Matters
Ever wondered why a cyclist can speed up just by pedalling harder, while a rolling ball slows down on a rough surface? The answer lies in the work‑energy theorem, a handy shortcut that links the forces you apply to the speed you gain.
💡 In Simple Words: The work‑energy theorem says the total work you do on an object equals the change in its kinetic energy – the energy of motion. So if you push harder, you give the object more kinetic energy; if friction steals work away, the object loses kinetic energy.
What the Theorem Actually Says
Work (W) is the product of a force (F) and the distance (d) it moves in the direction of the force: W = F × d × cosθ, where θ is the angle between the force and the displacement.
Kinetic Energy (K) is the energy an object has because it moves: K = ½ m v², with m being mass and v the speed.
The work‑energy theorem ties these together: Net work done on a body = Change in kinetic energy, or W_net = ΔK = K_final – K_initial.
How to Use the Theorem – A Simple Checklist
- Identify the object (the "system") you are interested in.
- List all forces acting on it and decide which do positive work (help it move) and which do negative work (slow it down).
- Calculate the work done by each force using W = F d cosθ.
- Sum the works to get the net work.
- Set net work equal to the change in kinetic energy and solve for the unknown (often speed or distance).
Worked Example 1 – Pulling a Box on a Rough Floor
Imagine a 10 kg box being pulled across a horizontal floor by a constant horizontal force of 50 N. The coefficient of kinetic friction between the box and floor is 0.2. The box starts from rest and moves 5 m. Find its final speed.
Step 1 – Forces: The pulling force (F_pull) = 50 N (positive work). Friction force (F_fric) = μ N, where μ = 0.2 and N = mg = 10 kg × 10 m/s² = 100 N, so F_fric = 0.2 × 100 = 20 N (negative work).
Step 2 – Work done:
- Work by pull: W_pull = 50 N × 5 m × cos0° = 250 J.
- Work by friction: W_fric = 20 N × 5 m × cos180° = -100 J.
Net work: W_net = 250 J – 100 J = 150 J.
Step 3 – Apply the theorem: W_net = ΔK = ½ m v_f² – ½ m v_i². Since the box starts from rest, v_i = 0.
150 J = ½ × 10 kg × v_f² → v_f² = 30 → v_f ≈ 5.48 m/s
So the box finishes the 5 m stretch at about 5.5 m/s.
Worked Example 2 – A Roller Coaster Drop
A coaster car of mass 500 kg descends a friction‑less hill that is 20 m high. It starts from rest at the top. What speed does it have at the bottom?
Because there is no friction, the only work done is by gravity, which can be treated as a change in potential energy (energy stored because of height). The work‑energy theorem still works if we write gravity’s work as the loss of potential energy.
Step 1 – Potential energy loss: ΔU = m g h = 500 kg × 10 m/s² × 20 m = 100 000 J (negative because it goes down).
Step 2 – Net work: W_net = –ΔU = 100 000 J (the minus sign flips because loss of potential becomes positive work).
Step 3 – Apply theorem:
100 000 J = ½ × 500 kg × v_f² → v_f² = 400 → v_f = 20 m/s
The car zooms through the bottom of the hill at 20 m/s.
Quick Comparison – Work, Kinetic Energy, Potential Energy
| Quantity | Formula | What it Describes |
|---|---|---|
| Work (W) | W = F d cosθ | Energy transferred by a force acting through a distance. |
| Kinetic Energy (K) | K = ½ m v² | Energy of motion. |
| Potential Energy (U) | U = m g h (gravitational) or ½ k x² (elastic) | Stored energy due to position or configuration. |
Why the Theorem Saves You Time on Exams
Instead of writing separate equations for each force and then solving a messy system, the work‑energy theorem lets you jump straight from “how much work is done?” to “what’s the speed change?”. That’s why teachers love to ask questions that can be tackled with a single line of algebra after you compute the work.
Key Points to Remember
- Only the component of force along the displacement does work (cosθ factor).
- Positive work adds kinetic energy; negative work removes it.
- If friction or air resistance is present, treat them as forces that do negative work.
- When no non‑conservative forces act, the loss of potential energy equals the gain in kinetic energy.
📝 Likely Exam Questions
- Question: A 2 kg block is pushed across a frictionless table by a 10 N force acting at 30° to the horizontal for 4 m. Find its final speed.
Answer: Work = 10 N × 4 m × cos30° = 34.6 J. ΔK = 34.6 J = ½·2·v² → v ≈ 5.9 m/s. - Question: A 1500 kg car accelerates from rest to 20 m/s on a level road. If the engine supplies a constant force and the road provides a resisting force of 2000 N, what distance does the car travel during this acceleration?
Answer: ΔK = ½·1500·20² = 300 000 J. Net work = F_engine·d – 2000·d = 300 000 J → (F_engine – 2000)·d = 300 000 J. Using F_engine = m·a, a = v²/(2d) → solve to get d ≈ 150 m. - Question: A 0.5 kg ball is thrown vertically upward with speed 15 m/s. Ignoring air resistance, calculate the maximum height reached using the work‑energy theorem.
Answer: At the top, kinetic energy = 0. Net work by gravity = –m g h. So –0.5·10·h = ½·0.5·15² → –5h = 56.25 → h = 11.25 m. - Question: A sled of mass 8 kg slides down a 10 m long incline that makes a 30° angle with the horizontal. The coefficient of kinetic friction is 0.1. Find the sled’s speed at the bottom.
Answer: Work by gravity = m g h = 8·10·(10 sin30°) = 400 J. Friction work = μ m g cos30°·d = 0.1·8·10·cos30°·10 ≈ 69 J (negative). Net work = 331 J = ½·8·v² → v ≈ 9.1 m/s.