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LearnYPH11WPH11 · Unit 1: Mechanics and MaterialsMechanicsKinetic and potential energy
Mechanics · Subtopic 6 of 8

Kinetic and Potential Energy

Kinetic energy Ek=12mv2E_k=\tfrac12mv^2 comes from the work done to accelerate a mass from rest. The change in gravitational potential energy near the Earth is ΔEp=mgΔh\Delta E_p=mg\Delta h, from the work done lifting it. Energy problems become bookkeeping: initial energy == final energy ++ energy dissipated against friction.

Cambridge 9702 Paper 1Cambridge 9702 Paper 2Edexcel YPH11 WPH11AQA 7408 Paper 1≈ 35 minRevision 5 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Derive Ek=12mv2E_k=\tfrac12mv^2 from the equations of motion.
  • Derive ΔEp=mgΔh\Delta E_p=mg\Delta h from W=FsW=Fs.
  • Use Ek=p22mE_k=\dfrac{p^2}{2m}.
  • Solve energy problems including work done against resistive forces.
Pause and think

Two balls, one of mass 11 kg and one of 1010 kg, are dropped from the same height. Which hits the ground faster? Which has more kinetic energy?

Section 2 of 14

Core idea

Start hereWhere did the energy go?

A roller coaster starts 30 m up and reaches the bottom slower than it "should". The missing energy hasn't vanished; it's become heat and sound through friction. Energy conservation always balances once every store is counted.

The big idea

Ek=12mv2E_k=\frac12mv^2; ΔEp=mgΔh\Delta E_p=mg\Delta h (near Earth). Energy is conserved: loss in one store = gain in others (including work against friction).

1. Deriving the formulas

Kinetic and potential energy
  1. W=Fs=masW=Fs=mas
    Work to accelerate from rest.
  2. v2=2as⇒as=12v2v^2=2as\Rightarrow as=\tfrac12v^2
    suvat with u=0u=0.
  3. Ek=12mv2E_k=\tfrac12mv^2
    Substitute.
  4. ΔEp=mgΔh\Delta E_p=mg\Delta h
    Work against weight mgmg over height Δh\Delta h (uniform field).
A pendulum: energy swaps between kinetic and potential
−1−0.50.510.20.40.60.811.20position across the swingenergy (J)potentialkinetictotal
At the ends: all potential, momentarily at rest. In the middle: all kinetic, moving fastest. The total stays the same.
Worked exampleA pendulum

Released 0.20 m above its lowest point.

Goal
Speed at the bottom.
Strategy
PE → KE.
  1. mg(0.20)=12mv2mg(0.20)=\tfrac12mv^2
    No losses; mm cancels.
  2. v=2(9.81)(0.20)=1.98 m s−1v=\sqrt{2(9.81)(0.20)}=1.98\text{ m s}^{-1}
    Solve.
Answer1.981.98 m s−1^{-1}.
✓ CheckIndependent of mass and string length ✓.
Worked exampleRoller coaster with friction

500 kg from rest at 30 m reaches the bottom at 22 m s−1^{-1}.

Goal
Work against friction.
Strategy
PE lost − KE gained.
  1. 500(9.81)(30)=147150 J500(9.81)(30)=147150\text{ J}
    PE lost.
  2. 12(500)(222)=121000 J\tfrac12(500)(22^2)=121000\text{ J}
    KE gained.
  3. 26150 J26150\text{ J}
    Difference.
AnswerAbout 2626 kJ.
✓ CheckWithout friction: v=2(9.81)(30)=24.3>22v=\sqrt{2(9.81)(30)}=24.3>22 ✓.
Common misconceptionmghmgh far from Earth mgΔhmg\Delta h assumes a uniform field: fine near the surface, not for orbits.

Summary

  • EkE_k, ΔEp\Delta E_p derived from work.
  • Conservation plus work against friction.
Key words
Conservation of energy
Total energy constant.
HistoryÉmilie du Châtelet's v². In the 17401740s, Émilie du Châtelet argued (using experiments dropping balls into clay) that the "living force" of a moving body depends on v2v^2, not vv: twice the speed made a dent four times deeper. She also translated Newton's Principia into French.
Fun factA meteor's energy. Because kinetic energy depends on v2v^2, small fast objects carry huge energy. A 11 kg meteorite at 2020 km s⁻¹ has 2×1082\times10^8 J of kinetic energy, about as much as 5050 kg of TNT.
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