Section 1 of 6Kinetic Energy Equation
A moving object can do work as it slows down, and this ability is its kinetic energy.
Definition
- Kinetic energy is the energy a body has because of its motion, measured in joules.
Equation
- Ek = ½mv², where m is mass in kg and v is speed in m s⁻¹.
Square law
- Kinetic energy depends on v², so doubling the speed increases the energy by a factor of four.
Scalar
- Kinetic energy has size but no direction, and it can never be negative.
Link to momentum
- Using p = mv, the energy can also be written as Ek = p²/2m.
| Change in speed | Change in kinetic energy |
|---|---|
| × 2 | × 4 |
| × 3 | × 9 |
| × ½ | × ¼ |
| +10 % | +21 % |
Section 2 of 6Deriving the Equation
The formula follows from the equations of motion and the definition of work.
1
Start with a constant resultant force F accelerating a mass m from rest over a distance s.
2
The work done is W = Fs, and Newton’s second law gives F = ma.
3
For motion from rest, v² = 2as, so s = v²/2a.
4
Substituting, W = ma × v²/2a = ½mv².
5
Because the work done is stored as motion, the kinetic energy gained is ½mv².
Section 3 of 6Work and Kinetic Energy
Work and Kinetic Energy
Work–energy link
- The net work done on an object equals its change in kinetic energy, W = ΔEk.
Braking
- The work done by brakes equals the loss of kinetic energy, so Fd = ½mv² for a car stopping from speed v.
Stopping distance
- For a constant braking force the stopping distance is proportional to v², so doubling speed quadruples the distance.
Energy transfer
- Kinetic energy removed by braking becomes thermal energy in the brake discs and pads.
Section 4 of 6Examples in Context
Vehicle safety
- Crumple zones increase the distance over which kinetic energy is removed, reducing the average force on passengers.
Wind turbines
- Moving air carries kinetic energy, and a turbine converts part of it into electrical energy.
Projectiles
- A thrown ball’s kinetic energy changes with height, but the total mechanical energy is conserved without drag.
Section 5 of 6Worked Example
Question
- A car of mass 1200 kg travels at 25 m s⁻¹.
- (a) Calculate its kinetic energy.
- (b) It brakes to rest in 50 m with a constant force.
- Find the braking force.
- (a) Ek = ½mv² = 0.5 × 1200 × 25² = 3.75 × 10⁵ J.
- (b) Work done by the brakes equals the kinetic energy lost, so F × 50 = 3.75 × 10⁵.
- F = 3.75 × 10⁵ / 50 = 7.5 × 10³ N.
Answer (a) 3.75 × 10⁵ J; (b) 7.5 × 10³ N.
Section 6 of 6Exam Checklist & Tips
3.6.4 – Key Points for the Exam
- Kinetic energy The energy of a body due to its motion, ½mv².
- Work–energy theorem The net work done on a body equals the change in its kinetic energy.
- Scalar A quantity with magnitude only, such as energy and speed.
- Stopping distance The distance travelled while a vehicle is braking to rest.
Common Mistakes
- Forgetting to square the speed and calculating ½mv instead of ½mv².
- Squaring the product mv instead of squaring v alone.
- Using km h⁻¹ without converting to m s⁻¹ first.
- Saying that doubling speed doubles the kinetic energy; it quadruples it.