The big idea: Kinetic energy is the energy an object has because it is moving. The faster it goes, the more it has.
It depends on the mass and the speed squared: E_k = ½mv².
Unit: the joule (J), like all energy.
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.
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.
Worked example — kinetic energy of a moving object
A 1500 kg car travels at 20 m s⁻¹. Find its kinetic energy.
Solution
- Start with the given formula:
- Put in the numbers (m = 1500, v = 20) — square the speed first:
- Work it out (20² = 400) — keep the unit:
Final answer
Ek = 300 000 J = 3.0 × 10⁵ J (300 kJ).
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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 classic work-energy principle — a force acts over a distance, then is removed, and you find how much further the object slides before stopping.
Classic trap: the object stops because friction does negative work that removes all the kinetic energy. Set friction force × stopping distance = E_k, then solve for the distance.
The work-energy principle: The net work done on an object equals its change in kinetic energy: W_{net} = ΔE_k.
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.
Sliding to rest against friction: While an object slides to rest, friction takes away all its kinetic energy:
friction force × distance slid = E_k at the start of the slide.
Rearrange for the distance: s = E_k ÷ friction force.
IB-style question — (a) speed after a push
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 on the box is a steady 6.0 N. Find the box's speed at the end of the push.
Solution
- Net force = push − friction (friction opposes the motion):
- Net work done = net force × distance (this is the kinetic energy gained):
- It started from rest, so this 60 J is all of its kinetic energy. Use the given formula:
- Put in the numbers (Ek = 60, m = 4.0) and solve for v:
- Take the square root — keep the unit:
Final answer
speed at the end of the push = 5.5 m s⁻¹ (to 2 s.f.).
IB-style question — (b) how much further it slides
The 18 N push is now removed. With only the 6.0 N friction still acting, find how much further the box slides before it stops.
Solution
- It has 60 J of kinetic energy. Friction must do enough negative work to remove all of it, so set friction × distance = Ek:
- Put in the numbers (Ff = 6.0, Ek = 60) and solve for s:
- Divide — keep the unit:
Final answer
the box slides 10 m further before stopping — friction (6.0 N) does −60 J of work, exactly cancelling its 60 J of kinetic energy.