Work-Energy Theorem
The work-energy theorem relating net work to change in kinetic energy.
Includes derivation, formulas, and worked examples.
The Theorem
The net work done on an object equals the change in its kinetic energy. If positive work is done, the object speeds up. If negative work is done, the object slows down.
Related Formulas
Kinetic Energy: KE = ½mv²
Power: P = W / t = F · v
Variables
| Symbol | Meaning | Unit |
|---|---|---|
| W | Work done | Joules (J) |
| F | Force applied | Newtons (N) |
| d | Distance (displacement) | Meters (m) |
| θ | Angle between force and displacement | Degrees or radians |
| KE | Kinetic energy | Joules (J) |
| m | Mass | Kilograms (kg) |
| v | Velocity | m/s |
| P | Power | Watts (W) |
| t | Time | Seconds (s) |
Example 1 — Finding Final Speed
A 2 kg block starts at rest. A net force does 100 J of work on it. What is the final speed?
W_net = ½mv₂² − ½mv₁²
100 = ½(2)(v₂²) − 0 (starts at rest, v₁ = 0)
100 = v₂²
v₂ = 10 m/s
Example 2 — Work Done by Friction
A 1,500 kg car traveling at 20 m/s brakes to a stop. How much work did friction do?
W_net = ½mv₂² − ½mv₁²
W = ½(1500)(0²) − ½(1500)(20²)
W = 0 − 300,000
W = −300,000 J (negative = energy removed by friction)
Example 3 — Calculating Work from Force and Distance
A 50 N force pushes a box 4 m along the floor at a 30° angle. How much work is done?
W = F · d · cos(θ)
W = 50 × 4 × cos(30°)
W = 200 × 0.866
W = 173.2 J
When to Use It
- Finding the speed of an object after a force acts over a distance
- Calculating the work done to stop a moving object
- Problems where you know force and distance but not time or acceleration
- Energy conservation problems involving kinetic energy changes