The big idea: Kick a football and it races off — while it's moving it carries kinetic energy, the energy of motion. Stop it and that energy is gone.
The faster and heavier it is, the more it has: Ek = ½mv² (in joules).
The speed is squared, so a small speed-up is a big jump in energy.
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Spot it — speed is SQUARED: Because of the v², speed matters a lot: double the speed ⇒ four times the kinetic energy.
That's why a car at 60 km/h has 4× the energy (and a far longer stopping distance) than at 30 km/h.
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The kinetic-energy formula is given in the data booklet. The right-hand part, p²/2m, is just the same energy written using the momentum p = mv — handy when you're given the momentum instead of the speed.
- kinetic energy — the energy of motion (J, joules)
- mass of the object (kg)
- speed of the object (m s⁻¹)
- momentum, p = mv (kg m s⁻¹)
Don't forget to square the speed: The most common slip is multiplying instead of squaring. Square v first, then halve and multiply by the mass.
Keep the order: v² → × m → × ½.
A 1500 kg car travels at 20 m s⁻¹. Find its kinetic energy.
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How this is tested — kinetic energy turns up wherever motion and energy meet:
Paper 1A
- A quick calculation of Ek.
- Or the ×4 'double the speed' idea.
Paper 2
- The work-energy principle: a force acts over a distance, then is removed.
- Find how much further the object slides before stopping.
The classic trap: The object stops because friction does negative work that removes all the kinetic energy. Set friction force × stopping distance = Ek, then solve for the distance.
The work-energy principle: The net work done on an object equals its change in kinetic energy: Wnet = ΔEk.
A push speeds it up (positive work, gains Ek). Friction slows it down (negative work, loses Ek). When all the kinetic energy is used up, it stops.
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Sliding to rest against friction: While an object slides to rest, friction takes away all its kinetic energy:
friction force × distance slid = Ek at the start of the slide.
Rearrange for the distance: s = Ek ÷ friction force.
A 4.0 kg box starts from rest on a rough floor. A constant horizontal force of 18 N pushes it through 5.0 m; friction is a steady 6.0 N. (a) Find the box's speed at the end of the push. (b) The push is then removed, with only friction still acting. Find how much further the box slides before it stops.
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