Kinetic and potential energies

What you'll learn

How to calculate kinetic energy using $Ek=12mv2E_k = \frac{1}{2}mv^2$ .
How to calculate gravitational potential energy in a uniform gravitational field using $E_p = mgh$ .
How both equations can be derived from work done.
How gravitational potential energy and kinetic energy exchange during motion.

Energy and work: the starting point

Before kinetic and potential energy, remember the key idea from work, energy and power: energy is transferred when work is done.

Definition

Work done

Work done is the energy transferred when a force moves an object through a displacement. For a constant force acting in the direction of motion, $W = F x$ , where $W$ is work done in joules, $F$ is force in newtons, and $x$ is displacement in metres.

One joule is one newton metre: 1 J = 1 N m. Energy is also measured in joules, because work done is a transfer of energy.

Kinetic energy

Definition

Kinetic energy

Kinetic energy, $E_k$ , is the energy an object has because it is moving.

For an object of mass $m$ moving at speed $v$ :

Ek=12mv2E_k = \frac{1}{2}mv^2

where:

$E_k$ is kinetic energy in joules, J
$m$ is mass in kilograms, kg
$v$ is speed in metres per second, m s $^{-1}$

The square on speed is very important. Doubling the speed makes the kinetic energy four times larger.

Key Idea

Speed matters a lot

Kinetic energy is proportional to $v^2$ , not $v$ . A small increase in speed can produce a much larger increase in kinetic energy.

Example

Calculating kinetic energy

A 0.060 kg tennis ball moves at 25 m s $^{-1}$ . Calculate its kinetic energy.

Choose the kinetic energy equation because the object has a known mass and speed:
$Ek=12mv2E_k = \frac{1}{2}mv^2$
Substitute the values, keeping the units with the quantities:
$s−1)2E_k = \frac{1}{2} \times 0.060\,\text{kg} \times \left(25\,\text{m s}^{-1}\right)^2$
Square the speed first, then multiply:
$JE_k = 0.030 \times 625 = 18.75\,\text{J}$
Quote a sensible final answer:
$JE_k \approx 19\,\text{J}$

Common Mistake

Forgetting to square the speed

Do not calculate $12mv\frac{1}{2}mv$ . The equation is $12mv2\frac{1}{2}mv^2$ , so the speed must be squared before multiplying by the mass.

Deriving kinetic energy from first principles

OCR expects you to be able to recall $Ek=12mv2E_k = \frac{1}{2}mv^2$ and derive it from first principles.

“First principles” here means starting from basic equations you already know, rather than simply quoting the result.

We start with:

work done by a force: $W = F x$
Newton’s second law: $F = ma$
constant-acceleration motion: $v^2 = u^2 + 2ax$

For an object starting from rest, $u = 0$ , so:

v^2 = 2ax

Rearrange for displacement:

\frac{v^2}{2a}

Now substitute $F = ma$ and $\frac{v^2}{2a}$ into $W = F x$ :

W=FxW=ma×v22aW=12mv2\begin{aligned} W &= Fx \\ W &= ma \times \frac{v^2}{2a} \\ W &= \frac{1}{2}mv^2 \end{aligned}

That work done becomes the object’s kinetic energy, so:

Ek=12mv2E_k = \frac{1}{2}mv^2

Tip

What the derivation is really saying

A resultant force does work on an object while accelerating it. The energy transferred by that work appears as kinetic energy.

Gravitational potential energy

Definition

Gravitational potential energy

Gravitational potential energy, $E_p$ , is the energy an object has because of its position in a gravitational field.

In a uniform gravitational field, the gravitational field strength $g$ is constant. Near the Earth’s surface, we usually take $9.81\,\text{N kg}^{-1}$ , which is equivalent to $s−29.81\,\text{m s}^{-2}$ .

For an object of mass $m$ raised through a height $h$ :

E_p = mgh

where:

$E_p$ is gravitational potential energy in joules, J
$m$ is mass in kilograms, kg
$g$ is gravitational field strength in newtons per kilogram, N kg $^{-1}$
$h$ is height in metres, m

Strictly, this equation gives the change in gravitational potential energy relative to a chosen reference level.

Definition

Reference level

A reference level is the height where you choose $E_p = 0$ . Heights above this level have positive gravitational potential energy relative to it.

Example

Calculating gravitational potential energy

A 72 kg climber gains 450 m in height. Calculate the increase in gravitational potential energy. Use $9.81\,\text{N kg}^{-1}$ .

Choose $E_p = mgh$ because the question gives mass, gravitational field strength and vertical height gained.
Substitute the values:
$mE_p = 72\,\text{kg} \times 9.81\,\text{N kg}^{-1} \times 450\,\text{m}$
Calculate the energy increase:
$JE_p = 317844\,\text{J}$
Write the answer in standard form to a sensible number of significant figures:
$JE_p \approx 3.2 \times 10^5\,\text{J}$

Common Mistake

Using distance instead of vertical height

In $E_p = mgh$ , $h$ is the vertical change in height, not the distance travelled along a slope or path.

Deriving gravitational potential energy from first principles

To lift an object at constant speed in a uniform gravitational field, the upward force you apply must equal the object’s weight.

Weight is given by:

W = m g

Here $W$ means weight, measured in newtons. Be careful: the same letter $W$ is also used for work done in some contexts.

The work done lifting the object through a vertical height $h$ is:

force\text{work done} = \text{force} \times \text{distance moved in direction of force}

So:

done=mgh\begin{aligned} \text{work done} &= mgh \end{aligned}

This work done is stored as gravitational potential energy, so:

E_p = mgh

Tip

Unit check for gravitational potential energy

$m g$ is a force in newtons. Multiplying by height in metres gives newton metres, which are joules.

Energy exchange between $E_p$ and $E_k$

When an object moves up or down in a gravitational field, energy can be transferred between gravitational potential energy and kinetic energy.

Schematic showing gravitational potential energy converting into kinetic energy as a ball falls in a uniform gravitational field

If air resistance is negligible, the object’s total mechanical energy remains constant:

Ek+Ep=constantE_k + E_p = \text{constant}

So when an object falls:

$E_p$ decreases because height decreases
$E_k$ increases because speed increases
the loss of $E_p$ equals the gain in $E_k$

For an object released from rest from height $h$ , with negligible air resistance:

\frac{1}{2}mv^2

The mass cancels:

\frac{1}{2}v^2

So:

\sqrt{2gh}

This means that, neglecting air resistance, objects dropped from the same height reach the same speed, regardless of mass.

Common Mistake

Mechanical energy is not always constant

If air resistance or friction is significant, some mechanical energy is transferred to internal energy of the surroundings and the object. Total energy is still conserved, but $E_k + E_p$ decreases.

Example

Finding speed from a fall height

A stone is dropped from rest from a height of 12 m. Calculate its speed just before reaching the ground. Ignore air resistance and use $9.81\,\text{m s}^{-2}$ .

Identify the energy transfer: the stone loses gravitational potential energy and gains kinetic energy.
$\frac{1}{2}mv^2$
Cancel the mass, because it appears on both sides:
$\frac{1}{2}v^2$
Rearrange for $v$ :
$\sqrt{2gh}$
Substitute the values:
$\sqrt{2 \times 9.81\,\text{m s}^{-2} \times 12\,\text{m}}$
Calculate the speed:
$\sqrt{235.44\,\text{m}^2\text{s}^{-2}} = 15.3\,\text{m s}^{-1}$

Example

Finding maximum height from launch speed

A ball is thrown vertically upwards at 18 m s $^{-1}$ . Ignore air resistance. Calculate the maximum height reached above the launch point.

At launch, the ball has kinetic energy. At maximum height, its speed is zero, so its kinetic energy is zero.
Set initial kinetic energy equal to the gain in gravitational potential energy:
$12mv2=mgh\frac{1}{2}mv^2 = mgh$
Cancel the mass and rearrange for height:
$\frac{v^2}{2g}$
Substitute the values:
$\frac{\left(18\,\text{m s}^{-1}\right)^2}{2 \times 9.81\,\text{m s}^{-2}}$
Calculate:
$\frac{324}{19.62}\,\text{m} = 16.5\,\text{m}$

Energy methods versus SUVAT

Many vertical motion problems can be solved using either energy equations or constant-acceleration equations. Energy is often quicker when the question asks about speeds and height changes but does not ask about time.

For example, the energy result:

\sqrt{2gh}

matches the SUVAT equation $v^2 = u^2 + 2as$ for a drop from rest, where $u = 0$ , $a = g$ , and $s = h$ .

Key Idea

Energy focuses on start and finish

Energy methods compare states: initial height and speed against final height and speed. They often avoid needing the time taken.

Choosing the right equation

Use $Ek=12mv2E_k = \frac{1}{2}mv^2$ when motion and speed are central.

Use $E_p = mgh$ when height in a uniform gravitational field is central.

Use conservation of mechanical energy when energy is being exchanged and resistive forces are negligible:

Ek\text{loss in } E_p = \text{gain in } E_k

Ek,initial+Ep,initial=Ek,final+Ep,finalE_{k,\text{initial}} + E_{p,\text{initial}} = E_{k,\text{final}} + E_{p,\text{final}}

Exam technique

In the exam

Check whether the question is about speed, height, or an energy transfer, then choose $Ek=12mv2E_k = \frac{1}{2}mv^2$ , $E_p = mgh$ , or conservation of mechanical energy.
Keep units in SI: mass in kilograms, speed in metres per second, height in metres, and energy in joules.
If using energy conservation, state the assumption: air resistance and friction are negligible, so mechanical energy is conserved.

Self review

Check yourself

Why does doubling an object’s speed make its kinetic energy four times larger?
In $E_p = mgh$ , why must $h$ be a vertical height change rather than distance along a slope?
A ball falls from rest with negligible air resistance. What happens to $E_p$ , $E_k$ , and total mechanical energy as it falls?

What you'll learn

How to calculate kinetic energy using $Ek=12mv2E_k = \frac{1}{2}mv^2$ .
How to calculate gravitational potential energy in a uniform gravitational field using $E_p = mgh$ .
How both equations can be derived from work done.
How gravitational potential energy and kinetic energy exchange during motion.

Energy and work: the starting point

Before kinetic and potential energy, remember the key idea from work, energy and power: energy is transferred when work is done.

Definition

Work done

One joule is one newton metre: 1 J = 1 N m. Energy is also measured in joules, because work done is a transfer of energy.

Kinetic energy

Definition

Kinetic energy

Kinetic energy, $E_k$ , is the energy an object has because it is moving.

For an object of mass $m$ moving at speed $v$ :

Ek=12mv2E_k = \frac{1}{2}mv^2

where:

$E_k$ is kinetic energy in joules, J
$m$ is mass in kilograms, kg
$v$ is speed in metres per second, m s $^{-1}$

The square on speed is very important. Doubling the speed makes the kinetic energy four times larger.

Key Idea

Speed matters a lot

Kinetic energy is proportional to $v^2$ , not $v$ . A small increase in speed can produce a much larger increase in kinetic energy.

Example

Calculating kinetic energy

A 0.060 kg tennis ball moves at 25 m s $^{-1}$ . Calculate its kinetic energy.

Choose the kinetic energy equation because the object has a known mass and speed:
$Ek=12mv2E_k = \frac{1}{2}mv^2$
Substitute the values, keeping the units with the quantities:
$s−1)2E_k = \frac{1}{2} \times 0.060\,\text{kg} \times \left(25\,\text{m s}^{-1}\right)^2$
Square the speed first, then multiply:
$JE_k = 0.030 \times 625 = 18.75\,\text{J}$
Quote a sensible final answer:
$JE_k \approx 19\,\text{J}$

Common Mistake

Forgetting to square the speed

Do not calculate $12mv\frac{1}{2}mv$ . The equation is $12mv2\frac{1}{2}mv^2$ , so the speed must be squared before multiplying by the mass.

Deriving kinetic energy from first principles

OCR expects you to be able to recall $Ek=12mv2E_k = \frac{1}{2}mv^2$ and derive it from first principles.

“First principles” here means starting from basic equations you already know, rather than simply quoting the result.

We start with:

work done by a force: $W = F x$
Newton’s second law: $F = ma$
constant-acceleration motion: $v^2 = u^2 + 2ax$

For an object starting from rest, $u = 0$ , so:

v^2 = 2ax

Rearrange for displacement:

\frac{v^2}{2a}

Now substitute $F = ma$ and $\frac{v^2}{2a}$ into $W = F x$ :

W=FxW=ma×v22aW=12mv2\begin{aligned} W &= Fx \\ W &= ma \times \frac{v^2}{2a} \\ W &= \frac{1}{2}mv^2 \end{aligned}

That work done becomes the object’s kinetic energy, so:

Ek=12mv2E_k = \frac{1}{2}mv^2

Tip

What the derivation is really saying

A resultant force does work on an object while accelerating it. The energy transferred by that work appears as kinetic energy.

Gravitational potential energy

Definition

Gravitational potential energy

Gravitational potential energy, $E_p$ , is the energy an object has because of its position in a gravitational field.

For an object of mass $m$ raised through a height $h$ :

E_p = mgh

where:

$E_p$ is gravitational potential energy in joules, J
$m$ is mass in kilograms, kg
$g$ is gravitational field strength in newtons per kilogram, N kg $^{-1}$
$h$ is height in metres, m

Strictly, this equation gives the change in gravitational potential energy relative to a chosen reference level.

Definition

Reference level

A reference level is the height where you choose $E_p = 0$ . Heights above this level have positive gravitational potential energy relative to it.

Example

Calculating gravitational potential energy

A 72 kg climber gains 450 m in height. Calculate the increase in gravitational potential energy. Use $9.81\,\text{N kg}^{-1}$ .

Choose $E_p = mgh$ because the question gives mass, gravitational field strength and vertical height gained.
Substitute the values:
$mE_p = 72\,\text{kg} \times 9.81\,\text{N kg}^{-1} \times 450\,\text{m}$
Calculate the energy increase:
$JE_p = 317844\,\text{J}$
Write the answer in standard form to a sensible number of significant figures:
$JE_p \approx 3.2 \times 10^5\,\text{J}$

Common Mistake

Using distance instead of vertical height

In $E_p = mgh$ , $h$ is the vertical change in height, not the distance travelled along a slope or path.

Deriving gravitational potential energy from first principles

To lift an object at constant speed in a uniform gravitational field, the upward force you apply must equal the object’s weight.

Weight is given by:

W = m g

Here $W$ means weight, measured in newtons. Be careful: the same letter $W$ is also used for work done in some contexts.

The work done lifting the object through a vertical height $h$ is:

force\text{work done} = \text{force} \times \text{distance moved in direction of force}

So:

done=mgh\begin{aligned} \text{work done} &= mgh \end{aligned}

This work done is stored as gravitational potential energy, so:

E_p = mgh

Tip

Unit check for gravitational potential energy

$m g$ is a force in newtons. Multiplying by height in metres gives newton metres, which are joules.

Energy exchange between $E_p$ and $E_k$

When an object moves up or down in a gravitational field, energy can be transferred between gravitational potential energy and kinetic energy.

Schematic showing gravitational potential energy converting into kinetic energy as a ball falls in a uniform gravitational field

If air resistance is negligible, the object’s total mechanical energy remains constant:

Ek+Ep=constantE_k + E_p = \text{constant}

So when an object falls:

$E_p$ decreases because height decreases
$E_k$ increases because speed increases
the loss of $E_p$ equals the gain in $E_k$

For an object released from rest from height $h$ , with negligible air resistance:

\frac{1}{2}mv^2

The mass cancels:

\frac{1}{2}v^2

So:

\sqrt{2gh}

This means that, neglecting air resistance, objects dropped from the same height reach the same speed, regardless of mass.

Common Mistake

Mechanical energy is not always constant

If air resistance or friction is significant, some mechanical energy is transferred to internal energy of the surroundings and the object. Total energy is still conserved, but $E_k + E_p$ decreases.

Example

Finding speed from a fall height

A stone is dropped from rest from a height of 12 m. Calculate its speed just before reaching the ground. Ignore air resistance and use $9.81\,\text{m s}^{-2}$ .

Identify the energy transfer: the stone loses gravitational potential energy and gains kinetic energy.
$\frac{1}{2}mv^2$
Cancel the mass, because it appears on both sides:
$\frac{1}{2}v^2$
Rearrange for $v$ :
$\sqrt{2gh}$
Substitute the values:
$\sqrt{2 \times 9.81\,\text{m s}^{-2} \times 12\,\text{m}}$
Calculate the speed:
$\sqrt{235.44\,\text{m}^2\text{s}^{-2}} = 15.3\,\text{m s}^{-1}$

Example

Finding maximum height from launch speed

A ball is thrown vertically upwards at 18 m s $^{-1}$ . Ignore air resistance. Calculate the maximum height reached above the launch point.

At launch, the ball has kinetic energy. At maximum height, its speed is zero, so its kinetic energy is zero.
Set initial kinetic energy equal to the gain in gravitational potential energy:
$12mv2=mgh\frac{1}{2}mv^2 = mgh$
Cancel the mass and rearrange for height:
$\frac{v^2}{2g}$
Substitute the values:
$\frac{\left(18\,\text{m s}^{-1}\right)^2}{2 \times 9.81\,\text{m s}^{-2}}$
Calculate:
$\frac{324}{19.62}\,\text{m} = 16.5\,\text{m}$

Energy methods versus SUVAT

For example, the energy result:

\sqrt{2gh}

matches the SUVAT equation $v^2 = u^2 + 2as$ for a drop from rest, where $u = 0$ , $a = g$ , and $s = h$ .

Key Idea

Energy focuses on start and finish

Energy methods compare states: initial height and speed against final height and speed. They often avoid needing the time taken.

Choosing the right equation

Use $Ek=12mv2E_k = \frac{1}{2}mv^2$ when motion and speed are central.

Use $E_p = mgh$ when height in a uniform gravitational field is central.

Use conservation of mechanical energy when energy is being exchanged and resistive forces are negligible:

Ek\text{loss in } E_p = \text{gain in } E_k

Ek,initial+Ep,initial=Ek,final+Ep,finalE_{k,\text{initial}} + E_{p,\text{initial}} = E_{k,\text{final}} + E_{p,\text{final}}

Exam technique

In the exam

Check whether the question is about speed, height, or an energy transfer, then choose $Ek=12mv2E_k = \frac{1}{2}mv^2$ , $E_p = mgh$ , or conservation of mechanical energy.
Keep units in SI: mass in kilograms, speed in metres per second, height in metres, and energy in joules.
If using energy conservation, state the assumption: air resistance and friction are negligible, so mechanical energy is conserved.

Self review

Check yourself

Why does doubling an object’s speed make its kinetic energy four times larger?
In $E_p = mgh$ , why must $h$ be a vertical height change rather than distance along a slope?
A ball falls from rest with negligible air resistance. What happens to $E_p$ , $E_k$ , and total mechanical energy as it falls?

What you'll learn

Energy and work: the starting point

Kinetic energy

Deriving kinetic energy from first principles

Gravitational potential energy

Deriving gravitational potential energy from first principles

Energy exchange between $E_p$ and $E_k$

Energy methods versus SUVAT

Choosing the right equation

Review Kinetic and potential energies by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Kinetic and potential energies

What you'll learn

Energy and work: the starting point

Kinetic energy

Deriving kinetic energy from first principles

Gravitational potential energy

Deriving gravitational potential energy from first principles

Energy exchange between $E_p$ and $E_k$

Energy methods versus SUVAT

Choosing the right equation

Review Kinetic and potential energies by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Kinetic and potential energies

What you'll learn

Energy and work: the starting point

Kinetic energy

Deriving kinetic energy from first principles

Gravitational potential energy

Deriving gravitational potential energy from first principles

Energy exchange between EpE_pEp​ and EkE_kEk​

Energy methods versus SUVAT

Choosing the right equation

Review Kinetic and potential energies by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Kinetic and potential energies Revision Guide

Kinetic and potential energies

What you'll learn

Energy and work: the starting point

Kinetic energy

Deriving kinetic energy from first principles

Gravitational potential energy

Deriving gravitational potential energy from first principles

Energy exchange between EpE_pEp​ and EkE_kEk​

Energy methods versus SUVAT

Choosing the right equation

Review Kinetic and potential energies by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Kinetic and potential energies Revision Guide

Energy exchange between $E_p$ and $E_k$

Energy exchange between $E_p$ and $E_k$