5.6.3d Braking force, energy transfer and deceleration

Braking force and energy transfer

Definition

Braking force

Braking force is the friction force that acts against the motion of a vehicle when its brakes are applied.

When the brakes are applied, friction acts between the brakes and the parts connected to the wheels, and this friction force does work on the vehicle.
This work done reduces the energy in the vehicle's kinetic energy store, so the vehicle slows down.
Energy is not destroyed but transferred: it moves from the vehicle's kinetic energy store to the thermal energy stores of the brakes and the surroundings.
As energy is transferred to the brakes, their temperature increases.

Key Idea

Braking transfers energy: kinetic energy of the vehicle → thermal energy of the brakes and surroundings.

Work done by the braking force

The work done by a braking force is $W = F s$ , where $W$ is the work done in joules ( $J\text{J}$ ), $F$ is the braking force in newtons ( $N\text{N}$ ) and $s$ is the braking distance in metres ( $m\text{m}$ ).
If the vehicle is brought to rest, the work done by the braking force equals the decrease in the vehicle's kinetic energy, provided other energy transfers are negligible.
So the braking force multiplied by the braking distance is approximately the kinetic energy: braking force × braking distance $≈12mv2\approx \tfrac{1}{2}mv^2$ .
A vehicle travelling at a greater speed has more kinetic energy, so more work must be done to stop it.
To stop a faster vehicle within the same distance, the braking force must be greater, because from $W = F s$ a larger work done over the same distance needs a larger force.

Example

A car has $J240\,000\ \text{J}$ of kinetic energy and is brought to rest over a braking distance of $m40\ \text{m}$ . Assuming all of this energy is transferred by a constant braking force, calculate the braking force.

Because the car stops, the work done equals the kinetic energy transferred, so $JW=240\,000\ \text{J}$ .
Rearrange $W = F s$ to make the force the subject: $F=WsF=\frac{W}{s}$ .
Substitute the values: $00040F=\frac{240\,000}{40}$ .
This gives $NF=6000\ \text{N}$ .
If the car were faster with $J360\,000\ \text{J}$ of kinetic energy, stopping in the same $m40\ \text{m}$ would need $NF=\frac{360\,000}{40}=9000\ \text{N}$ , a greater braking force.

Braking force and deceleration

Definition

Deceleration

Deceleration is acceleration in the opposite direction to an object's velocity, so its speed decreases; its magnitude is measured in $m/s2\text{m/s}^2$ .

Resultant force, mass and acceleration are linked by $F = ma$ , where $F$ is in newtons ( $N\text{N}$ ), $m$ in kilograms ( $kg\text{kg}$ ) and $a$ in $m/s2\text{m/s}^2$ .
For a given vehicle the mass is constant, so a greater braking force produces a greater deceleration.
A large braking force can bring a vehicle to rest quickly, but very large decelerations can be dangerous.

Common Mistake

Do not say the vehicle's kinetic energy is "used up" or "lost". Energy is conserved: braking transfers it from the kinetic energy store to thermal energy stores.

Do not confuse deceleration with braking force. Deceleration is measured in $m/s2\text{m/s}^2$ , while braking force is measured in newtons ( $N\text{N}$ ).

Dangers of large decelerations

Injury to occupants: a large deceleration needs a large force to change the passengers' velocity; seat belts and restraints apply this force, but very large forces can still cause injury.
Loss of control: a large braking force can make the tyres lose grip so the vehicle skids, and the driver may lose control of its direction.
Overheating brakes: during heavy braking, kinetic energy is transferred rapidly to the thermal energy store of the brakes, so their temperature can become very high, especially during repeated braking.
Reduced braking effectiveness: overheated brakes may become less effective, making it harder to slow down or stop safely.

Exam technique

In an “explain the dangers” answer, give linked reasoning: a large braking force gives a large deceleration, which can cause skidding and loss of control, and rapid energy transfer that overheats the brakes.
Describe the energy changes using stores: state that kinetic energy is transferred to thermal energy stores rather than being lost.
When explaining dangers, link a large braking force to a large deceleration, then to skidding, loss of control and overheating brakes.
Keep the ideas of deceleration and force separate, and always include the correct units.

Self review

What energy store decreases when a moving vehicle brakes, and where does the energy go?
Why does the temperature of the brakes increase during braking?
Why is a greater braking force needed to stop a faster vehicle in the same distance?
How does increasing the braking force affect the deceleration of the same vehicle?
Explain why a large deceleration can cause injury, brake overheating or loss of control.

5.6.3e Estimating deceleration forces (HT)

Deceleration and resultant force

Definition

Deceleration

Deceleration is acceleration in the direction opposite to an object's velocity; it makes the object's speed decrease and is measured in $m/s2\text{m/s}^2$ .

A road vehicle decelerates when a resultant force acts opposite to its direction of motion.
During braking, friction between the brakes and the wheels slows their rotation, while friction between the tyres and the road provides the force that slows the vehicle.
The acceleration is found from $a=v−uta=\frac{v-u}{t}$ , where $u$ is the initial velocity, $v$ the final velocity (both in $m/s\text{m/s}$ ) and $t$ the time in seconds ( $s\text{s}$ ).
When a vehicle slows down $v$ is less than $u$ , so the acceleration is negative; the magnitude of the deceleration is $u−vt\frac{u-v}{t}$ .

Estimating the decelerating force

The resultant force on the vehicle is found from $F = ma$ , with $F$ in newtons ( $N\text{N}$ ), $m$ in kilograms ( $kg\text{kg}$ ) and $a$ in $m/s2\text{m/s}^2$ .
A greater mass needs a greater force for the same deceleration, and stopping in a shorter time needs a greater force because the deceleration is larger.
Useful approximate values are: a small or medium car has a mass of about $1000$ to $kg1500\ \text{kg}$ ; $mph30\ \text{mph}$ is roughly $m/s13\ \text{m/s}$ ; and a controlled stop takes a few seconds.
Using these, the resultant decelerating force on a car is typically several thousand newtons, which shows the forces involved in braking are large.
These are estimates: the actual force depends on the vehicle's mass, its initial speed, the stopping time and the road conditions, so answers should stay physically reasonable.

Example

A car of mass $kg1200\ \text{kg}$ travels at about $m/s13\ \text{m/s}$ . The driver brakes and it stops in $s4.0\ \text{s}$ . Estimate the resultant force on the car.

Find the acceleration: $m/s2a=\frac{v-u}{t}=\frac{0-13}{4.0}=-3.25\ \text{m/s}^2$ .
The negative sign shows the acceleration is opposite to the car's motion.
Find the force: $NF=ma=1200\times(-3.25)=-3900\ \text{N}$ .
The magnitude is about $N4000\ \text{N}$ , acting opposite to the car's direction of motion.
If the same car stopped in only $s2.0\ \text{s}$ , the deceleration would be twice as large, so the resultant force would also be about twice as large.

Estimating with energy

The force can also be estimated from energy, because the work done by the braking force equals the vehicle's kinetic energy when it stops: braking force × braking distance $=12mv2=\tfrac{1}{2}mv^2$ .
Rearranging gives $F=12mv2sF=\frac{\tfrac{1}{2}mv^2}{s}$ , where $s$ is the braking distance in metres.
For example, the $kg1200\ \text{kg}$ car at $m/s13\ \text{m/s}$ has kinetic energy $J\tfrac{1}{2}\times1200\times13^2\approx101\,000\ \text{J}$ ; over a braking distance of about $m25\ \text{m}$ this gives $NF\approx4000\ \text{N}$ , agreeing with the $F = ma$ estimate.
Both methods give forces of the same order of magnitude, confirming that typical braking forces are large.

Getting estimates right

Do not substitute the vehicle's speed directly into $F = ma$ ; first calculate the acceleration or deceleration from the change in velocity divided by time.
Convert speeds to $m/s\text{m/s}$ before using them: for a speed in $km/h\text{km/h}$ divide by $3.6$ , and remember $m/s30\ \text{mph}\approx13\ \text{m/s}$ .
$F = ma$ gives the resultant force, not the force from a single source; the negative sign only shows direction, so state the magnitude as a positive value.
For an estimate, show a sensible method, give a realistic order of magnitude with units, and state that the resultant force acts opposite to the vehicle's motion.

Self review

What is deceleration, and in what unit is it measured?
How do you calculate acceleration from the initial and final velocity and the time?
How is the resultant decelerating force calculated using $F = ma$ ?
Why does a shorter stopping time or a greater mass need a larger resultant force?
Roughly what order of magnitude is a typical decelerating force on a car?

5.6.3d Braking force, energy transfer and deceleration

Braking force and energy transfer

Definition

Braking force

Braking force is the friction force that acts against the motion of a vehicle when its brakes are applied.

When the brakes are applied, friction acts between the brakes and the parts connected to the wheels, and this friction force does work on the vehicle.
This work done reduces the energy in the vehicle's kinetic energy store, so the vehicle slows down.
Energy is not destroyed but transferred: it moves from the vehicle's kinetic energy store to the thermal energy stores of the brakes and the surroundings.
As energy is transferred to the brakes, their temperature increases.

Key Idea

Braking transfers energy: kinetic energy of the vehicle → thermal energy of the brakes and surroundings.

Work done by the braking force

The work done by a braking force is $W = F s$ , where $W$ is the work done in joules ( $J\text{J}$ ), $F$ is the braking force in newtons ( $N\text{N}$ ) and $s$ is the braking distance in metres ( $m\text{m}$ ).
If the vehicle is brought to rest, the work done by the braking force equals the decrease in the vehicle's kinetic energy, provided other energy transfers are negligible.
So the braking force multiplied by the braking distance is approximately the kinetic energy: braking force × braking distance $≈12mv2\approx \tfrac{1}{2}mv^2$ .
A vehicle travelling at a greater speed has more kinetic energy, so more work must be done to stop it.
To stop a faster vehicle within the same distance, the braking force must be greater, because from $W = F s$ a larger work done over the same distance needs a larger force.

Example

Because the car stops, the work done equals the kinetic energy transferred, so $JW=240\,000\ \text{J}$ .
Rearrange $W = F s$ to make the force the subject: $F=WsF=\frac{W}{s}$ .
Substitute the values: $00040F=\frac{240\,000}{40}$ .
This gives $NF=6000\ \text{N}$ .
If the car were faster with $J360\,000\ \text{J}$ of kinetic energy, stopping in the same $m40\ \text{m}$ would need $NF=\frac{360\,000}{40}=9000\ \text{N}$ , a greater braking force.

Braking force and deceleration

Definition

Deceleration

Deceleration is acceleration in the opposite direction to an object's velocity, so its speed decreases; its magnitude is measured in $m/s2\text{m/s}^2$ .

Resultant force, mass and acceleration are linked by $F = ma$ , where $F$ is in newtons ( $N\text{N}$ ), $m$ in kilograms ( $kg\text{kg}$ ) and $a$ in $m/s2\text{m/s}^2$ .
For a given vehicle the mass is constant, so a greater braking force produces a greater deceleration.
A large braking force can bring a vehicle to rest quickly, but very large decelerations can be dangerous.

Common Mistake

Do not say the vehicle's kinetic energy is "used up" or "lost". Energy is conserved: braking transfers it from the kinetic energy store to thermal energy stores.

Do not confuse deceleration with braking force. Deceleration is measured in $m/s2\text{m/s}^2$ , while braking force is measured in newtons ( $N\text{N}$ ).

Dangers of large decelerations

Injury to occupants: a large deceleration needs a large force to change the passengers' velocity; seat belts and restraints apply this force, but very large forces can still cause injury.
Loss of control: a large braking force can make the tyres lose grip so the vehicle skids, and the driver may lose control of its direction.
Overheating brakes: during heavy braking, kinetic energy is transferred rapidly to the thermal energy store of the brakes, so their temperature can become very high, especially during repeated braking.
Reduced braking effectiveness: overheated brakes may become less effective, making it harder to slow down or stop safely.

Exam technique

In an “explain the dangers” answer, give linked reasoning: a large braking force gives a large deceleration, which can cause skidding and loss of control, and rapid energy transfer that overheats the brakes.
Describe the energy changes using stores: state that kinetic energy is transferred to thermal energy stores rather than being lost.
When explaining dangers, link a large braking force to a large deceleration, then to skidding, loss of control and overheating brakes.
Keep the ideas of deceleration and force separate, and always include the correct units.

Self review

What energy store decreases when a moving vehicle brakes, and where does the energy go?
Why does the temperature of the brakes increase during braking?
Why is a greater braking force needed to stop a faster vehicle in the same distance?
How does increasing the braking force affect the deceleration of the same vehicle?
Explain why a large deceleration can cause injury, brake overheating or loss of control.

5.6.3e Estimating deceleration forces (HT)

Deceleration and resultant force

Definition

Deceleration

Deceleration is acceleration in the direction opposite to an object's velocity; it makes the object's speed decrease and is measured in $m/s2\text{m/s}^2$ .

A road vehicle decelerates when a resultant force acts opposite to its direction of motion.
During braking, friction between the brakes and the wheels slows their rotation, while friction between the tyres and the road provides the force that slows the vehicle.
The acceleration is found from $a=v−uta=\frac{v-u}{t}$ , where $u$ is the initial velocity, $v$ the final velocity (both in $m/s\text{m/s}$ ) and $t$ the time in seconds ( $s\text{s}$ ).
When a vehicle slows down $v$ is less than $u$ , so the acceleration is negative; the magnitude of the deceleration is $u−vt\frac{u-v}{t}$ .

Estimating the decelerating force

The resultant force on the vehicle is found from $F = ma$ , with $F$ in newtons ( $N\text{N}$ ), $m$ in kilograms ( $kg\text{kg}$ ) and $a$ in $m/s2\text{m/s}^2$ .
A greater mass needs a greater force for the same deceleration, and stopping in a shorter time needs a greater force because the deceleration is larger.
Useful approximate values are: a small or medium car has a mass of about $1000$ to $kg1500\ \text{kg}$ ; $mph30\ \text{mph}$ is roughly $m/s13\ \text{m/s}$ ; and a controlled stop takes a few seconds.
Using these, the resultant decelerating force on a car is typically several thousand newtons, which shows the forces involved in braking are large.
These are estimates: the actual force depends on the vehicle's mass, its initial speed, the stopping time and the road conditions, so answers should stay physically reasonable.

Example

A car of mass $kg1200\ \text{kg}$ travels at about $m/s13\ \text{m/s}$ . The driver brakes and it stops in $s4.0\ \text{s}$ . Estimate the resultant force on the car.

Find the acceleration: $m/s2a=\frac{v-u}{t}=\frac{0-13}{4.0}=-3.25\ \text{m/s}^2$ .
The negative sign shows the acceleration is opposite to the car's motion.
Find the force: $NF=ma=1200\times(-3.25)=-3900\ \text{N}$ .
The magnitude is about $N4000\ \text{N}$ , acting opposite to the car's direction of motion.
If the same car stopped in only $s2.0\ \text{s}$ , the deceleration would be twice as large, so the resultant force would also be about twice as large.

Estimating with energy

The force can also be estimated from energy, because the work done by the braking force equals the vehicle's kinetic energy when it stops: braking force × braking distance $=12mv2=\tfrac{1}{2}mv^2$ .
Rearranging gives $F=12mv2sF=\frac{\tfrac{1}{2}mv^2}{s}$ , where $s$ is the braking distance in metres.
For example, the $kg1200\ \text{kg}$ car at $m/s13\ \text{m/s}$ has kinetic energy $J\tfrac{1}{2}\times1200\times13^2\approx101\,000\ \text{J}$ ; over a braking distance of about $m25\ \text{m}$ this gives $NF\approx4000\ \text{N}$ , agreeing with the $F = ma$ estimate.
Both methods give forces of the same order of magnitude, confirming that typical braking forces are large.

Getting estimates right

Do not substitute the vehicle's speed directly into $F = ma$ ; first calculate the acceleration or deceleration from the change in velocity divided by time.
Convert speeds to $m/s\text{m/s}$ before using them: for a speed in $km/h\text{km/h}$ divide by $3.6$ , and remember $m/s30\ \text{mph}\approx13\ \text{m/s}$ .
$F = ma$ gives the resultant force, not the force from a single source; the negative sign only shows direction, so state the magnitude as a positive value.
For an estimate, show a sensible method, give a realistic order of magnitude with units, and state that the resultant force acts opposite to the vehicle's motion.

Self review

What is deceleration, and in what unit is it measured?
How do you calculate acceleration from the initial and final velocity and the time?
How is the resultant decelerating force calculated using $F = ma$ ?
Why does a shorter stopping time or a greater mass need a larger resultant force?
Roughly what order of magnitude is a typical decelerating force on a car?

5.6.3d Factors affecting braking distance 2

5.6.3d Braking force, energy transfer and deceleration

Braking force and energy transfer

Work done by the braking force

Braking force and deceleration

Dangers of large decelerations

5.6.3e Estimating deceleration forces (HT)

Deceleration and resultant force

Estimating the decelerating force

Estimating with energy

Getting estimates right

Review 5.6.3d Factors affecting braking distance 2 by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.6.3d Factors affecting braking distance 2

5.6.3d Braking force, energy transfer and deceleration

Braking force and energy transfer

Work done by the braking force

Braking force and deceleration

Dangers of large decelerations

5.6.3e Estimating deceleration forces (HT)

Deceleration and resultant force

Estimating the decelerating force

Estimating with energy

Getting estimates right

Review 5.6.3d Factors affecting braking distance 2 by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.6.3d Factors affecting braking distance 2

5.6.3d Braking force, energy transfer and deceleration

Braking force and energy transfer

Work done by the braking force

Braking force and deceleration

Dangers of large decelerations

5.6.3e Estimating deceleration forces (HT)

Deceleration and resultant force

Estimating the decelerating force

Estimating with energy

Getting estimates right

Review 5.6.3d Factors affecting braking distance 2 by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.6.3d Factors affecting braking distance 2 Revision Guide

5.6.3d Factors affecting braking distance 2

5.6.3d Braking force, energy transfer and deceleration

Braking force and energy transfer

Work done by the braking force

Braking force and deceleration

Dangers of large decelerations

5.6.3e Estimating deceleration forces (HT)

Deceleration and resultant force

Estimating the decelerating force

Estimating with energy

Getting estimates right

Review 5.6.3d Factors affecting braking distance 2 by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.6.3d Factors affecting braking distance 2 Revision Guide