Gravitational potential energy and kinetic energy

8 minAS Level4.5.2

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can define and calculate gravitational potential energy ( $\text{GPE} = mgh$ ) and kinetic energy ( $\text{KE} = \tfrac12mv^2$ ), and understand energy as the capacity to do work.

Key to the symbols

Symbol	Meaning	Unit
$m$	mass	kg
$g$	$10\ \text{m s}^{-2}$
$h$	height above a chosen zero level	m
$v$	speed	$\text{m s}^{-1}$
$W$	work done	J
$\Delta$	"the change in", so $\Delta h$ means the change in height
GPE, KE	gravitational potential energy, kinetic energy	J

Energy and work

Energy is the capacity to do work. When work is done on an object, its energy increases. When an object does work, its energy decreases. Energy and work share the same unit: the joule (J).

Two fundamental forms of mechanical energy in this topic:

Gravitational potential energy (GPE) — energy due to position in a gravitational field
Kinetic energy (KE) — energy due to motion

Definition

Gravitational potential energy (GPE)

For a mass $m$ raised a vertical height $h$ above a reference level: $\text{GPE} = mgh$ where $g = 10\text{ m s}^{-2}$ (as specified in the 9709 syllabus).

GPE is relative to the chosen reference level (zero point).
Only changes in GPE are physically meaningful.
Gaining height → GPE increases. Losing height → GPE decreases.

Definition

Kinetic energy (KE)

For a particle of mass $m$ moving with speed $v$ : $\text{KE} = \frac{1}{2}mv^2$

KE is always $\ge 0$ (speed squared).
KE is a scalar — it depends on speed, not velocity direction.
If speed doubles, KE quadruples.

Where do these formulas come from?

GPE = mgh: Lifting a mass $m$ through height $h$ at constant velocity requires an upward force $F = mg$ . Work done by the lifting force: $W = Fh = mgh$ . This work is stored as GPE.

KE = ½mv²: Accelerate a mass $m$ from rest to speed $v$ using a constant force $F$ . Starting from rest ( $u = 0$ ), $v^2 = u^2 + 2as$ gives $v^2 = 2as$ , so the distance moved is $s = \dfrac{v^2}{2a}$ . With $F = ma$ : $W = Fs = ma \cdot \frac{v^2}{2a} = \frac{1}{2}mv^2$ The work done equals the kinetic energy gained. This is the work–energy principle in action.

Worked example: Calculating GPE and KE

A ball of mass $0.2\text{ kg}$ is thrown vertically upwards from ground level with an initial speed of $15\text{ m s}^{-1}$ . Taking $g = 10\text{ m s}^{-2}$ , find: (a) its initial kinetic energy (b) its maximum height (c) its GPE at maximum height

Have a go first?

Worked example: Speed from height loss

A skier of mass $70\text{ kg}$ starts from rest at the top of a slope $50\text{ m}$ high. Ignoring friction and air resistance, find the speed at the bottom.

Have a go first?

Check yourself

CalculateQuestion 1 · Recall

Answer the 3 questions above to continue.