8.2.1 Gravitational potential energy

The gravitational potential energy store

Definition

Gravitational potential energy

Gravitational potential energy is energy stored by an object because of its position in a gravitational field.

Definition

Gravitational field strength

Gravitational field strength is the force per unit mass acting on an object placed in a gravitational field, measured in newtons per kilogram (N/kg).

Gravitational potential energy is stored whenever an object is raised in a gravitational field, and the store belongs to the object and the Earth together.
The change in this store depends on the object's mass, the gravitational field strength and the change in vertical height.
A heavier object has a larger weight, so more energy has to be transferred to raise it through the same height.
Raising the same object twice as high transfers twice as much energy into the store.
Gravitational field strength at the Earth's surface is taken as $10\ \text{N/kg}$ unless a question gives a different value.
The value of $g$ on the Moon is much smaller, so the same lift through the same height stores far fewer joules there.
Only changes in this store are calculated, because the height counted as zero is chosen rather than fixed by nature.

The change in GPE equation

The change in the store is calculated from $ΔGPE=m×g×Δh\Delta GPE = m \times g \times \Delta h$ .
Here $ΔGPE\Delta GPE$ is the change in gravitational potential energy in joules, $m$ is the mass in kilograms, $g$ is the gravitational field strength in $N/kg\text{N/kg}$ and $Δh\Delta h$ is the change in vertical height in metres.
The symbol $Δ\Delta$ means change in, so $Δh\Delta h$ is the final height minus the starting height.
Rearranging gives $\dfrac{\Delta GPE}{g \times \Delta h}$ for the mass and $Δh=ΔGPEm×g\Delta h = \dfrac{\Delta GPE}{m \times g}$ for the height.
Convert grams into kilograms and centimetres into metres first, so $kg250\ \text{g} = 0.250\ \text{kg}$ and $m40\ \text{cm} = 0.40\ \text{m}$ .
The height $Δh\Delta h$ is measured straight upwards, so an object pushed up a ramp gains the store set by the vertical rise and not by the length of the slope.
The store increases while the object is being raised and decreases while it falls, so a falling object is emptying this store.

Linking the store to work done

The weight of an object is its mass multiplied by the gravitational field strength, measured in newtons, and this is the force a steady lift has to balance.
The work done in that lift is $\times d = m \times g \times \Delta h$ , which is the same expression as the change in the store.
Lifting a mass therefore transfers the work done straight into the gravitational potential energy store, which is why the two equations agree.
Work done against friction on a ramp is extra to $\times g \times \Delta h$ , so a ramp needs more energy in total than a straight lift to the same height.

Example

Lifting a box onto a shelf

A shelf stacker raises a box of mass $kg25\ \text{kg}$ through a vertical height of $m3.0\ \text{m}$ , where $10\ \text{N/kg}$ .
The equation is $ΔGPE=m×g×Δh\Delta GPE = m \times g \times \Delta h$ .
Substituting gives $ΔGPE=25×10×3.0\Delta GPE = 25 \times 10 \times 3.0$ .
The change in the store is $J\Delta GPE = 750\ \text{J}$ , and the store increases because the box has been raised.

Example

Pushing a wheelbarrow up a ramp

A wheelbarrow of mass $kg40\ \text{kg}$ is pushed $m5.0\ \text{m}$ up a ramp that rises $m1.2\ \text{m}$ vertically, with $10\ \text{N/kg}$ .
The vertical rise is the value needed, so $m\Delta h = 1.2\ \text{m}$ and the $m5.0\ \text{m}$ measured along the slope is not used.
Substituting gives $ΔGPE=40×10×1.2\Delta GPE = 40 \times 10 \times 1.2$ .
The gravitational potential energy store increases by $J\Delta GPE = 480\ \text{J}$ .

Example

Finding a height from the energy stored

A crane transfers $J36\,000\ \text{J}$ into the gravitational store of a $kg300\ \text{kg}$ girder, with $10\ \text{N/kg}$ .
Rearranging gives $Δh=ΔGPEm×g\Delta h = \dfrac{\Delta GPE}{m \times g}$ .
Substituting gives $000300×10\Delta h = \dfrac{36\,000}{300 \times 10}$ .
The girder has been raised through $m\Delta h = 12\ \text{m}$ .

Exam technique

Getting the GPE values right

Write $ΔGPE=m×g×Δh\Delta GPE = m \times g \times \Delta h$ first, then list the three values you are about to substitute.
Take the value of $g$ from the question, and use $N/kg10\ \text{N/kg}$ only when no value is given.
Read the diagram carefully for the vertical height, because ramp lengths and slope distances are printed there to catch you out.
State whether the store increases or decreases when a description is wanted as well as a number.
Answer in joules, and convert to kilojoules only when the question asks for them.

Common Mistake

Do not put a weight in newtons into the $m$ position, because that position needs a mass in kilograms.
Do not use the distance measured along a slope as $Δh\Delta h$ , because only the vertical rise changes this store.
Do not call $g$ gravity, because the quantity is gravitational field strength measured in $N/kg\text{N/kg}$ .
Do not leave a mass in grams, because the equation needs kilograms.

Self review

State what gravitational potential energy is and which two objects share the store.
Write the equation for the change in gravitational potential energy.
Give the unit of each quantity in that equation.
Explain why the vertical rise is used for a load pushed up a ramp.
Calculate the change in the store when a $kg2.0\ \text{kg}$ book is raised $m1.5\ \text{m}$ with $10\ \text{N/kg}$ .

8.2.2 Kinetic energy

The kinetic energy store

Definition

Kinetic energy

Kinetic energy is energy stored by an object because it is moving.

Kinetic energy is stored by an object because it is moving, so every moving object holds some.
The size of the store depends on the object's mass and on its speed.
Speed matters more than mass, because the store depends on the square of the speed.
A stationary object has a speed of $m/s0\ \text{m/s}$ , so its kinetic energy store holds $J0\ \text{J}$ .
The store fills as an object speeds up and empties as it slows down.
Kinetic energy is measured in joules, $J\text{J}$ , like every other store.

The kinetic energy equation

The store is calculated from $\dfrac{1}{2} \times m \times v^{2}$ .
Here $K E$ is the kinetic energy in joules, $m$ is the mass in kilograms and $v$ is the speed in metres per second, $m/s\text{m/s}$ .
Square the speed before multiplying, so a speed of $m/s6.0\ \text{m/s}$ gives $m2/s2v^{2} = 36\ \text{m}^{2}/\text{s}^{2}$ .
Writing the working as $\times m \times v^{2}$ makes the factor of one half harder to forget.
Rearranging for speed gives $\sqrt{\dfrac{2 \times KE}{m}}$ , and rearranging for mass gives $\dfrac{2 \times KE}{v^{2}}$ .
Doubling the mass at the same speed doubles the store, because the mass is not squared.
Doubling the speed at the same mass makes the store four times larger, since $2^{2} = 4$ .
Tripling the speed makes the store nine times larger, since $3^{2} = 9$ .
Convert grams into kilograms, and convert any speed given in $km/h\text{km/h}$ into $m/s\text{m/s}$ , before substituting.

Kinetic energy in falling and braking

A falling object empties its gravitational potential energy store and fills its kinetic energy store, so $12×m×v2=m×g×Δh\dfrac{1}{2} \times m \times v^{2} = m \times g \times \Delta h$ when resistive forces are small.
Cancelling the mass gives $\sqrt{2 \times g \times \Delta h}$ , which shows that the landing speed does not depend on how heavy the object is.
A vehicle's kinetic energy store has to be emptied before it stops, and the work done by the braking force equals that store, so $\times d = \dfrac{1}{2} \times m \times v^{2}$ .
Because the speed is squared, doubling a car's speed makes its kinetic energy store four times larger and so needs about four times the distance to stop with the same braking force.
Those joules are transferred into the thermal energy stores of the brakes, tyres and road, which is why brake discs become hot.

Example

Kinetic energy of a cyclist

A cyclist and bicycle have a combined mass of $kg80\ \text{kg}$ and travel at $m/s6.0\ \text{m/s}$ .
The equation is $\dfrac{1}{2} \times m \times v^{2}$ .
Squaring the speed gives $v^{2} = 6.0^{2} = 36$ .
Substituting gives $\times 80 \times 36$ .
The kinetic energy store holds $1440\ \text{J}$ .

Example

A mass given in grams

A cricket ball of mass $g160\ \text{g}$ is thrown at $m/s18\ \text{m/s}$ .
Converting the mass gives $kg160\ \text{g} = 0.160\ \text{kg}$ .
Squaring the speed gives $v^{2} = 18^{2} = 324$ .
Substituting gives $\times 0.160 \times 324$ .
The kinetic energy store holds $25.9\ \text{J}$ to three significant figures.

Example

Speed of a falling stone

A stone is dropped from a bridge $m20\ \text{m}$ above a river, with $10\ \text{N/kg}$ and air resistance small enough to ignore.
The gravitational store emptied equals the kinetic store filled, so $12×m×v2=m×g×Δh\dfrac{1}{2} \times m \times v^{2} = m \times g \times \Delta h$ .
Cancelling the mass and rearranging gives $\sqrt{2 \times g \times \Delta h}$ .
Substituting gives $\sqrt{2 \times 10 \times 20} = \sqrt{400}$ .
The stone reaches the water at $20\ \text{m/s}$ .

Example

Finding a speed from the store

A trolley of mass $kg2.5\ \text{kg}$ holds $J45\ \text{J}$ in its kinetic energy store.
Rearranging gives $\sqrt{\dfrac{2 \times KE}{m}}$ .
Substituting gives $\sqrt{\dfrac{2 \times 45}{2.5}} = \sqrt{36}$ .
The trolley is moving at $6.0\ \text{m/s}$ .

Exam technique

Working with the squared speed

Write $\dfrac{1}{2} \times m \times v^{2}$ and square the speed on its own line so the working can be followed.
Square only the speed, never the mass and never the one half.
Keep the full value in the calculator when a rearranged answer is wanted, then take the square root last.
Compare two kinetic energy stores by comparing the masses and the squares of the speeds rather than the speeds themselves.
Round to the same number of significant figures as the data in the question, which is usually two or three.

Common Mistake

Do not multiply by the speed once instead of squaring it, because that badly underestimates the store.
Do not leave the mass in grams, because using grams in place of kilograms makes the answer a thousand times too large.
Do not forget the factor of $12\dfrac{1}{2}$ , which is the most commonly dropped mark in this calculation.
Do not say that a heavier object always holds more kinetic energy, because a light object moving quickly can hold more.

Self review

State what kinetic energy is and give its unit.
Write the equation for kinetic energy and name each quantity in it.
Explain what happens to the kinetic energy store when the speed triples.
Calculate the kinetic energy of a $kg1200\ \text{kg}$ car travelling at $m/s15\ \text{m/s}$ .
Explain why the landing speed of a dropped object does not depend on its mass.

8.2.3 Dissipation of energy

Dissipation of energy

Definition

Dissipation

Dissipation is the spreading of transferred energy into the thermal energy stores of the surroundings, which makes the energy less useful for further transfers.

Every real change in a system transfers some energy into stores that were not wanted, and that energy is said to be dissipated.
Dissipated energy almost always ends up in the thermal energy stores of the object and its surroundings.
Conservation of energy still holds, so all of the joules are still there and none have been destroyed.
Dissipated energy is described as less useful because it has been shared out between huge numbers of particles in the surroundings.
Spread that thinly, the energy raises the temperature of the surroundings by only a tiny amount and cannot be gathered back to drive the change again.
Whether a transfer counts as useful depends on the job the device is meant to do, so heating is the useful transfer in a kettle and an unwanted one in a drill.
Sound waves also carry energy away from a system, and those joules finish in the thermal energy store of the surroundings as well.

Why mechanical processes waste energy

A mechanical process is one in which forces do work, such as a motor turning a shaft or a wheel rolling along a road.
Whenever two surfaces slide or rub across each other, friction acts between them and work is done against it.
That work transfers energy into the thermal energy stores of both surfaces, so their temperature rises.
A mechanical process becomes wasteful when it causes this rise in temperature, because the energy is then dissipated into the surroundings by heating.
Moving through air or water means doing work against air resistance or drag, which warms both the fluid and the moving object.
The rise in temperature is the signature of the waste, so a machine that runs hot is dissipating a large share of the energy supplied to it.
The unwanted transfer also depends on how long the process runs, because a longer run dissipates more joules.

Tracking dissipation in real systems

A bouncing ball rebounds to a lower height each time, because energy is dissipated at every bounce.
During a bounce the ball squashes and then recovers, and the work done inside the rubber warms both the ball and the ground.
The bounce also makes a sound, so a further share of the energy leaves the ball as sound waves.
Less energy is therefore returned to the gravitational potential energy store each time, so each bounce is lower than the one before.
A car left in neutral rolls to a stop because friction in the bearings and air resistance move its kinetic energy store into thermal energy stores.
A filament lamp transfers most of the energy supplied to it by heating the surroundings rather than by light, so most of that transfer is unwanted.
Brakes are the case that proves the rule, because they are designed to raise a thermal energy store, yet once those joules are spread through the brakes and air they can no longer move the car.

Example

Bounce heights of a dropped ball

A tennis ball is dropped from $m1.0\ \text{m}$ onto a hard playground surface and rebounds to about $m0.55\ \text{m}$ .
Falling empties the ball's gravitational potential energy store and fills its kinetic energy store.
At the moment of impact the ball squashes, so work is done inside the rubber and against the ground.
That work transfers energy into the thermal energy stores of the ball, the ground and the air, and some energy leaves as sound waves.
Only the energy still in the ball's kinetic store after the bounce can refill its gravitational store, so the ball rises to a lower height.
Every later bounce dissipates more energy, so the heights keep falling until the ball stays on the ground.

Example

Dissipation in an electric drill

A drill is used to make a hole in a wall, and its casing feels warm afterwards.
Energy is transferred electrically from the supply and increases the kinetic energy store of the rotating bit, which is the useful transfer.
Friction acts in the bearings, in the gears and between the bit and the wall, so work is done against friction.
That work transfers energy by heating into the thermal energy stores of the drill, the wall and the surrounding air.
The warm casing shows where the wasted joules have gone, and they cannot be recovered to turn the bit.

Exam technique

Explaining dissipation for full marks

Name the store that holds the energy at the start, such as the kinetic energy store of the car.
Name the cause of the unwanted transfer, which is usually friction, air resistance or the squashing of a material.
State that work is done against that force, so energy is transferred by heating.
Name the thermal energy stores that increase, and include the surroundings as well as the object itself.
Finish by saying that the energy has become spread out and is now less useful, which is the mark most answers miss.

Common Mistake

Do not write that energy is lost or destroyed, and write that it is dissipated or transferred to less useful stores instead.
Do not write heat energy, because energy is transferred by heating into a thermal energy store.
Do not claim the total energy falls during dissipation, because the total is unchanged and only its spread has altered.
Do not treat sound as a store, because sound waves are a way of carrying energy away from the system.

Self review

Define dissipation and name the store dissipated energy usually ends in.
Explain what less useful means for dissipated energy.
State what makes a mechanical process wasteful.
Explain why a bouncing ball does not return to its starting height.
Name two forces that cause unwanted energy transfers in a moving vehicle.

8.2.1 Gravitational potential energy

The gravitational potential energy store

Definition

Gravitational potential energy

Gravitational potential energy is energy stored by an object because of its position in a gravitational field.

Definition

Gravitational field strength

Gravitational field strength is the force per unit mass acting on an object placed in a gravitational field, measured in newtons per kilogram (N/kg).

Gravitational potential energy is stored whenever an object is raised in a gravitational field, and the store belongs to the object and the Earth together.
The change in this store depends on the object's mass, the gravitational field strength and the change in vertical height.
A heavier object has a larger weight, so more energy has to be transferred to raise it through the same height.
Raising the same object twice as high transfers twice as much energy into the store.
Gravitational field strength at the Earth's surface is taken as $10\ \text{N/kg}$ unless a question gives a different value.
The value of $g$ on the Moon is much smaller, so the same lift through the same height stores far fewer joules there.
Only changes in this store are calculated, because the height counted as zero is chosen rather than fixed by nature.

The change in GPE equation

The change in the store is calculated from $ΔGPE=m×g×Δh\Delta GPE = m \times g \times \Delta h$ .
Here $ΔGPE\Delta GPE$ is the change in gravitational potential energy in joules, $m$ is the mass in kilograms, $g$ is the gravitational field strength in $N/kg\text{N/kg}$ and $Δh\Delta h$ is the change in vertical height in metres.
The symbol $Δ\Delta$ means change in, so $Δh\Delta h$ is the final height minus the starting height.
Rearranging gives $\dfrac{\Delta GPE}{g \times \Delta h}$ for the mass and $Δh=ΔGPEm×g\Delta h = \dfrac{\Delta GPE}{m \times g}$ for the height.
Convert grams into kilograms and centimetres into metres first, so $kg250\ \text{g} = 0.250\ \text{kg}$ and $m40\ \text{cm} = 0.40\ \text{m}$ .
The height $Δh\Delta h$ is measured straight upwards, so an object pushed up a ramp gains the store set by the vertical rise and not by the length of the slope.
The store increases while the object is being raised and decreases while it falls, so a falling object is emptying this store.

Linking the store to work done

The weight of an object is its mass multiplied by the gravitational field strength, measured in newtons, and this is the force a steady lift has to balance.
The work done in that lift is $\times d = m \times g \times \Delta h$ , which is the same expression as the change in the store.
Lifting a mass therefore transfers the work done straight into the gravitational potential energy store, which is why the two equations agree.
Work done against friction on a ramp is extra to $\times g \times \Delta h$ , so a ramp needs more energy in total than a straight lift to the same height.

Example

Lifting a box onto a shelf

A shelf stacker raises a box of mass $kg25\ \text{kg}$ through a vertical height of $m3.0\ \text{m}$ , where $10\ \text{N/kg}$ .
The equation is $ΔGPE=m×g×Δh\Delta GPE = m \times g \times \Delta h$ .
Substituting gives $ΔGPE=25×10×3.0\Delta GPE = 25 \times 10 \times 3.0$ .
The change in the store is $J\Delta GPE = 750\ \text{J}$ , and the store increases because the box has been raised.

Example

Pushing a wheelbarrow up a ramp

A wheelbarrow of mass $kg40\ \text{kg}$ is pushed $m5.0\ \text{m}$ up a ramp that rises $m1.2\ \text{m}$ vertically, with $10\ \text{N/kg}$ .
The vertical rise is the value needed, so $m\Delta h = 1.2\ \text{m}$ and the $m5.0\ \text{m}$ measured along the slope is not used.
Substituting gives $ΔGPE=40×10×1.2\Delta GPE = 40 \times 10 \times 1.2$ .
The gravitational potential energy store increases by $J\Delta GPE = 480\ \text{J}$ .

Example

Finding a height from the energy stored

A crane transfers $J36\,000\ \text{J}$ into the gravitational store of a $kg300\ \text{kg}$ girder, with $10\ \text{N/kg}$ .
Rearranging gives $Δh=ΔGPEm×g\Delta h = \dfrac{\Delta GPE}{m \times g}$ .
Substituting gives $000300×10\Delta h = \dfrac{36\,000}{300 \times 10}$ .
The girder has been raised through $m\Delta h = 12\ \text{m}$ .

Exam technique

Getting the GPE values right

Write $ΔGPE=m×g×Δh\Delta GPE = m \times g \times \Delta h$ first, then list the three values you are about to substitute.
Take the value of $g$ from the question, and use $N/kg10\ \text{N/kg}$ only when no value is given.
Read the diagram carefully for the vertical height, because ramp lengths and slope distances are printed there to catch you out.
State whether the store increases or decreases when a description is wanted as well as a number.
Answer in joules, and convert to kilojoules only when the question asks for them.

Common Mistake

Do not put a weight in newtons into the $m$ position, because that position needs a mass in kilograms.
Do not use the distance measured along a slope as $Δh\Delta h$ , because only the vertical rise changes this store.
Do not call $g$ gravity, because the quantity is gravitational field strength measured in $N/kg\text{N/kg}$ .
Do not leave a mass in grams, because the equation needs kilograms.

Self review

State what gravitational potential energy is and which two objects share the store.
Write the equation for the change in gravitational potential energy.
Give the unit of each quantity in that equation.
Explain why the vertical rise is used for a load pushed up a ramp.
Calculate the change in the store when a $kg2.0\ \text{kg}$ book is raised $m1.5\ \text{m}$ with $10\ \text{N/kg}$ .

8.2.2 Kinetic energy

The kinetic energy store

Definition

Kinetic energy

Kinetic energy is energy stored by an object because it is moving.

Kinetic energy is stored by an object because it is moving, so every moving object holds some.
The size of the store depends on the object's mass and on its speed.
Speed matters more than mass, because the store depends on the square of the speed.
A stationary object has a speed of $m/s0\ \text{m/s}$ , so its kinetic energy store holds $J0\ \text{J}$ .
The store fills as an object speeds up and empties as it slows down.
Kinetic energy is measured in joules, $J\text{J}$ , like every other store.

The kinetic energy equation

The store is calculated from $\dfrac{1}{2} \times m \times v^{2}$ .
Here $K E$ is the kinetic energy in joules, $m$ is the mass in kilograms and $v$ is the speed in metres per second, $m/s\text{m/s}$ .
Square the speed before multiplying, so a speed of $m/s6.0\ \text{m/s}$ gives $m2/s2v^{2} = 36\ \text{m}^{2}/\text{s}^{2}$ .
Writing the working as $\times m \times v^{2}$ makes the factor of one half harder to forget.
Rearranging for speed gives $\sqrt{\dfrac{2 \times KE}{m}}$ , and rearranging for mass gives $\dfrac{2 \times KE}{v^{2}}$ .
Doubling the mass at the same speed doubles the store, because the mass is not squared.
Doubling the speed at the same mass makes the store four times larger, since $2^{2} = 4$ .
Tripling the speed makes the store nine times larger, since $3^{2} = 9$ .
Convert grams into kilograms, and convert any speed given in $km/h\text{km/h}$ into $m/s\text{m/s}$ , before substituting.

Kinetic energy in falling and braking

A falling object empties its gravitational potential energy store and fills its kinetic energy store, so $12×m×v2=m×g×Δh\dfrac{1}{2} \times m \times v^{2} = m \times g \times \Delta h$ when resistive forces are small.
Cancelling the mass gives $\sqrt{2 \times g \times \Delta h}$ , which shows that the landing speed does not depend on how heavy the object is.
A vehicle's kinetic energy store has to be emptied before it stops, and the work done by the braking force equals that store, so $\times d = \dfrac{1}{2} \times m \times v^{2}$ .
Because the speed is squared, doubling a car's speed makes its kinetic energy store four times larger and so needs about four times the distance to stop with the same braking force.
Those joules are transferred into the thermal energy stores of the brakes, tyres and road, which is why brake discs become hot.

Example

Kinetic energy of a cyclist

A cyclist and bicycle have a combined mass of $kg80\ \text{kg}$ and travel at $m/s6.0\ \text{m/s}$ .
The equation is $\dfrac{1}{2} \times m \times v^{2}$ .
Squaring the speed gives $v^{2} = 6.0^{2} = 36$ .
Substituting gives $\times 80 \times 36$ .
The kinetic energy store holds $1440\ \text{J}$ .

Example

A mass given in grams

A cricket ball of mass $g160\ \text{g}$ is thrown at $m/s18\ \text{m/s}$ .
Converting the mass gives $kg160\ \text{g} = 0.160\ \text{kg}$ .
Squaring the speed gives $v^{2} = 18^{2} = 324$ .
Substituting gives $\times 0.160 \times 324$ .
The kinetic energy store holds $25.9\ \text{J}$ to three significant figures.

Example

Speed of a falling stone

A stone is dropped from a bridge $m20\ \text{m}$ above a river, with $10\ \text{N/kg}$ and air resistance small enough to ignore.
The gravitational store emptied equals the kinetic store filled, so $12×m×v2=m×g×Δh\dfrac{1}{2} \times m \times v^{2} = m \times g \times \Delta h$ .
Cancelling the mass and rearranging gives $\sqrt{2 \times g \times \Delta h}$ .
Substituting gives $\sqrt{2 \times 10 \times 20} = \sqrt{400}$ .
The stone reaches the water at $20\ \text{m/s}$ .

Example

Finding a speed from the store

A trolley of mass $kg2.5\ \text{kg}$ holds $J45\ \text{J}$ in its kinetic energy store.
Rearranging gives $\sqrt{\dfrac{2 \times KE}{m}}$ .
Substituting gives $\sqrt{\dfrac{2 \times 45}{2.5}} = \sqrt{36}$ .
The trolley is moving at $6.0\ \text{m/s}$ .

Exam technique

Working with the squared speed

Write $\dfrac{1}{2} \times m \times v^{2}$ and square the speed on its own line so the working can be followed.
Square only the speed, never the mass and never the one half.
Keep the full value in the calculator when a rearranged answer is wanted, then take the square root last.
Compare two kinetic energy stores by comparing the masses and the squares of the speeds rather than the speeds themselves.
Round to the same number of significant figures as the data in the question, which is usually two or three.

Common Mistake

Do not multiply by the speed once instead of squaring it, because that badly underestimates the store.
Do not leave the mass in grams, because using grams in place of kilograms makes the answer a thousand times too large.
Do not forget the factor of $12\dfrac{1}{2}$ , which is the most commonly dropped mark in this calculation.
Do not say that a heavier object always holds more kinetic energy, because a light object moving quickly can hold more.

Self review

State what kinetic energy is and give its unit.
Write the equation for kinetic energy and name each quantity in it.
Explain what happens to the kinetic energy store when the speed triples.
Calculate the kinetic energy of a $kg1200\ \text{kg}$ car travelling at $m/s15\ \text{m/s}$ .
Explain why the landing speed of a dropped object does not depend on its mass.

8.2.3 Dissipation of energy

Dissipation of energy

Definition

Dissipation

Dissipation is the spreading of transferred energy into the thermal energy stores of the surroundings, which makes the energy less useful for further transfers.

Every real change in a system transfers some energy into stores that were not wanted, and that energy is said to be dissipated.
Dissipated energy almost always ends up in the thermal energy stores of the object and its surroundings.
Conservation of energy still holds, so all of the joules are still there and none have been destroyed.
Dissipated energy is described as less useful because it has been shared out between huge numbers of particles in the surroundings.
Spread that thinly, the energy raises the temperature of the surroundings by only a tiny amount and cannot be gathered back to drive the change again.
Whether a transfer counts as useful depends on the job the device is meant to do, so heating is the useful transfer in a kettle and an unwanted one in a drill.
Sound waves also carry energy away from a system, and those joules finish in the thermal energy store of the surroundings as well.

Why mechanical processes waste energy

A mechanical process is one in which forces do work, such as a motor turning a shaft or a wheel rolling along a road.
Whenever two surfaces slide or rub across each other, friction acts between them and work is done against it.
That work transfers energy into the thermal energy stores of both surfaces, so their temperature rises.
A mechanical process becomes wasteful when it causes this rise in temperature, because the energy is then dissipated into the surroundings by heating.
Moving through air or water means doing work against air resistance or drag, which warms both the fluid and the moving object.
The rise in temperature is the signature of the waste, so a machine that runs hot is dissipating a large share of the energy supplied to it.
The unwanted transfer also depends on how long the process runs, because a longer run dissipates more joules.

Tracking dissipation in real systems

A bouncing ball rebounds to a lower height each time, because energy is dissipated at every bounce.
During a bounce the ball squashes and then recovers, and the work done inside the rubber warms both the ball and the ground.
The bounce also makes a sound, so a further share of the energy leaves the ball as sound waves.
Less energy is therefore returned to the gravitational potential energy store each time, so each bounce is lower than the one before.
A car left in neutral rolls to a stop because friction in the bearings and air resistance move its kinetic energy store into thermal energy stores.
A filament lamp transfers most of the energy supplied to it by heating the surroundings rather than by light, so most of that transfer is unwanted.
Brakes are the case that proves the rule, because they are designed to raise a thermal energy store, yet once those joules are spread through the brakes and air they can no longer move the car.

Example

Bounce heights of a dropped ball

A tennis ball is dropped from $m1.0\ \text{m}$ onto a hard playground surface and rebounds to about $m0.55\ \text{m}$ .
Falling empties the ball's gravitational potential energy store and fills its kinetic energy store.
At the moment of impact the ball squashes, so work is done inside the rubber and against the ground.
That work transfers energy into the thermal energy stores of the ball, the ground and the air, and some energy leaves as sound waves.
Only the energy still in the ball's kinetic store after the bounce can refill its gravitational store, so the ball rises to a lower height.
Every later bounce dissipates more energy, so the heights keep falling until the ball stays on the ground.

Example

Dissipation in an electric drill

A drill is used to make a hole in a wall, and its casing feels warm afterwards.
Energy is transferred electrically from the supply and increases the kinetic energy store of the rotating bit, which is the useful transfer.
Friction acts in the bearings, in the gears and between the bit and the wall, so work is done against friction.
That work transfers energy by heating into the thermal energy stores of the drill, the wall and the surrounding air.
The warm casing shows where the wasted joules have gone, and they cannot be recovered to turn the bit.

Exam technique

Explaining dissipation for full marks

Name the store that holds the energy at the start, such as the kinetic energy store of the car.
Name the cause of the unwanted transfer, which is usually friction, air resistance or the squashing of a material.
State that work is done against that force, so energy is transferred by heating.
Name the thermal energy stores that increase, and include the surroundings as well as the object itself.
Finish by saying that the energy has become spread out and is now less useful, which is the mark most answers miss.

Common Mistake

Do not write that energy is lost or destroyed, and write that it is dissipated or transferred to less useful stores instead.
Do not write heat energy, because energy is transferred by heating into a thermal energy store.
Do not claim the total energy falls during dissipation, because the total is unchanged and only its spread has altered.
Do not treat sound as a store, because sound waves are a way of carrying energy away from the system.

Self review

Define dissipation and name the store dissipated energy usually ends in.
Explain what less useful means for dissipated energy.
State what makes a mechanical process wasteful.
Explain why a bouncing ball does not return to its starting height.
Name two forces that cause unwanted energy transfers in a moving vehicle.

8.2 Energy calculations

8.2.1 Gravitational potential energy

The gravitational potential energy store

The change in GPE equation

Linking the store to work done

Lifting a box onto a shelf

Pushing a wheelbarrow up a ramp

Finding a height from the energy stored

Getting the GPE values right

8.2.2 Kinetic energy

The kinetic energy store

The kinetic energy equation

Kinetic energy in falling and braking

Kinetic energy of a cyclist

A mass given in grams

Speed of a falling stone

Finding a speed from the store

Working with the squared speed

8.2.3 Dissipation of energy

Dissipation of energy

Why mechanical processes waste energy

Tracking dissipation in real systems

Bounce heights of a dropped ball

Dissipation in an electric drill

Explaining dissipation for full marks

Review 8.2 Energy calculations by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

8.2 Energy calculations Revision Guide

8.2 Energy calculations

8.2.1 Gravitational potential energy

The gravitational potential energy store

The change in GPE equation

Linking the store to work done

Lifting a box onto a shelf

Pushing a wheelbarrow up a ramp

Finding a height from the energy stored

Getting the GPE values right

8.2.2 Kinetic energy

The kinetic energy store

The kinetic energy equation

Kinetic energy in falling and braking

Kinetic energy of a cyclist

A mass given in grams

Speed of a falling stone

Finding a speed from the store

Working with the squared speed

8.2.3 Dissipation of energy

Dissipation of energy

Why mechanical processes waste energy

Tracking dissipation in real systems

Bounce heights of a dropped ball

Dissipation in an electric drill

Explaining dissipation for full marks

Review 8.2 Energy calculations by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

8.2 Energy calculations Revision Guide