Linking force and acceleration

Definition

Directly proportional

A relationship between two quantities in which multiplying one by a factor multiplies the other by the same factor.

Definition

Inversely proportional

A relationship between two quantities in which multiplying one by a factor divides the other by the same factor.

Force, mass, velocity and acceleration are tied together by two equations that you must recall.
1. $F = ma$ links the resultant force to the mass and the acceleration.
2. $a=v−uta=\dfrac{v-u}{t}$ links the acceleration to the change in velocity and the time taken.
For a fixed mass, the acceleration is directly proportional to the resultant force.
1. If the resultant force on a shopping trolley is tripled, its acceleration is tripled.
For a fixed resultant force, the acceleration is inversely proportional to the mass.
1. If the same push is given to a trolley with three times the mass, its acceleration is one third as large.
Proportional reasoning lets you predict changes without working out every value from the beginning.
1. If the force is doubled and the mass is also doubled, the acceleration does not change, because $2F2m=Fm\dfrac{2F}{2m}=\dfrac{F}{m}$ .
2. If the force is halved and the mass is doubled, the acceleration falls to one quarter, because $F/22m=14×Fm\dfrac{F/2}{2m}=\dfrac{1}{4}\times\dfrac{F}{m}$ .

Force and change in velocity

Definition

Initial velocity

The velocity of an object at the start of the time interval being considered.

Definition

Final velocity

The velocity of an object at the end of the time interval being considered.

Substituting $a=v−uta=\dfrac{v-u}{t}$ into $F = ma$ gives $F=m(v−u)tF=\dfrac{m(v-u)}{t}$ .
This equation shows how the resultant force depends on the mass, the change in velocity and the time over which the change happens.
1. A larger mass needs a larger force to produce the same change in velocity in the same time.
2. A larger change in velocity in the same time needs a larger force.
3. The same change in velocity spread over a longer time needs a smaller force.
The last of these ideas explains why stopping something gently needs a smaller force than stopping it suddenly.
1. A fielder who pulls their hands back while catching a cricket ball makes the stopping time longer, so the force on their hands is smaller.
2. Bending your knees when you land from a jump makes the time to stop longer, so the force on your legs is smaller.
If the time is not known, the acceleration can be found from the velocities and the distance with $v^2-u^2=2as$ , which you need to select and apply rather than recall.
Combining $v^2-u^2=2as$ with $F = ma$ shows that, for a given mass and change of speed, a larger force changes the velocity over a shorter distance.

Solving multi-step problems

Many questions need two equations, with the answer from the first used in the second.
Work out which quantity links the two equations, which is usually the acceleration.
Find the acceleration from the motion data first, then use it in $F = ma$ , or find the acceleration from $F = ma$ first, then use it to find a velocity, a time or a distance.
Keep unrounded values in your calculator between steps and round only the final answer.
Check the sign of the acceleration, because a negative value means the resultant force acts against the motion and the object is slowing down.

Example

Force to accelerate a car

A car of mass $kg1200\ \text{kg}$ speeds up from $m/s8.0\ \text{m/s}$ to $m/s20\ \text{m/s}$ in $s6.0\ \text{s}$ .
Find the acceleration first.

a=\dfrac{v-u}{t}=\dfrac{20\ \text{m/s}-8.0\ \text{m/s}}{6.0\ \text{s}}=2.0\ \text{m/s}^2

Then use Newton's second law.

F=ma=1200\ \text{kg}\times2.0\ \text{m/s}^2=2400\ \text{N}

The resultant force on the car is $N2400\ \text{N}$ forwards.

Example

Time for a force to change a velocity

A resultant force of $N150\ \text{N}$ acts on a $kg60\ \text{kg}$ cyclist who starts from rest.

a=\dfrac{F}{m}=\dfrac{150\ \text{N}}{60\ \text{kg}}=2.5\ \text{m/s}^2

Rearrange $a=v−uta=\dfrac{v-u}{t}$ to find the time to reach $m/s10\ \text{m/s}$ .

t=\dfrac{v-u}{a}=\dfrac{10\ \text{m/s}-0}{2.5\ \text{m/s}^2}=4.0\ \text{s}

The cyclist reaches $m/s10\ \text{m/s}$ after $s4.0\ \text{s}$ .

Example

Braking force from a stopping distance

A $kg900\ \text{kg}$ car travelling at $m/s20\ \text{m/s}$ brakes and stops in a distance of $m25\ \text{m}$ .
No time is given, so select the equation that links $u$ , $v$ , $a$ and $s$ .

a=\dfrac{v^2-u^2}{2s}=\dfrac{0^2-20^2}{2\times25}=\dfrac{-400}{50}=-8.0\ \text{m/s}^2

F=ma=900\ \text{kg}\times(-8.0\ \text{m/s}^2)=-7200\ \text{N}

The braking force is $N7200\ \text{N}$ , and the negative sign shows that it acts opposite to the motion.

Example

Predicting a change with proportion

A loaded van has twice the mass of the same van when empty, and the same resultant force acts on it in both cases.
The acceleration is inversely proportional to the mass, so the loaded van has half the acceleration of the empty van.
With half the acceleration, the loaded van takes twice as long to reach the same speed from rest.

Common Mistake

Do not use the time in $F = ma$ directly, because you must find the acceleration first.
Do not drop the minus sign from a deceleration partway through a calculation, because the sign tells you the direction of the force.
Do not assume that a larger force always gives a larger acceleration, because a larger mass can reduce the acceleration more than the force increases it.

Exam technique

Multi-step force and motion problems

Write the two equations you plan to use at the start, so the examiner can follow your method.
Label each intermediate answer, such as "acceleration $m/s2=2.0\ \text{m/s}^2$ ", as this often carries its own mark.
For a "how would the acceleration change" question, use proportion and give a factor, such as "halves" or "doubles", rather than only "decreases".
When you explain why a longer stopping time helps, link it to a smaller deceleration and so to a smaller resultant force.

Self review

How does the acceleration of an object change if the resultant force doubles and the mass stays the same?
Write the equation that combines $F = ma$ and $a=v−uta=\dfrac{v-u}{t}$ .
Explain why bending your knees on landing reduces the force on your legs.
Calculate the resultant force on a $kg50\ \text{kg}$ skater who speeds up from $m/s2.0\ \text{m/s}$ to $m/s5.0\ \text{m/s}$ in $s1.5\ \text{s}$ .
What happens to the acceleration if both the force and the mass are tripled?

Linking force and acceleration

Definition

Directly proportional

A relationship between two quantities in which multiplying one by a factor multiplies the other by the same factor.

Definition

Inversely proportional

A relationship between two quantities in which multiplying one by a factor divides the other by the same factor.

Force, mass, velocity and acceleration are tied together by two equations that you must recall.
1. $F = ma$ links the resultant force to the mass and the acceleration.
2. $a=v−uta=\dfrac{v-u}{t}$ links the acceleration to the change in velocity and the time taken.
For a fixed mass, the acceleration is directly proportional to the resultant force.
1. If the resultant force on a shopping trolley is tripled, its acceleration is tripled.
For a fixed resultant force, the acceleration is inversely proportional to the mass.
1. If the same push is given to a trolley with three times the mass, its acceleration is one third as large.
Proportional reasoning lets you predict changes without working out every value from the beginning.
1. If the force is doubled and the mass is also doubled, the acceleration does not change, because $2F2m=Fm\dfrac{2F}{2m}=\dfrac{F}{m}$ .
2. If the force is halved and the mass is doubled, the acceleration falls to one quarter, because $F/22m=14×Fm\dfrac{F/2}{2m}=\dfrac{1}{4}\times\dfrac{F}{m}$ .

Force and change in velocity

Definition

Initial velocity

The velocity of an object at the start of the time interval being considered.

Definition

Final velocity

The velocity of an object at the end of the time interval being considered.

Substituting $a=v−uta=\dfrac{v-u}{t}$ into $F = ma$ gives $F=m(v−u)tF=\dfrac{m(v-u)}{t}$ .
This equation shows how the resultant force depends on the mass, the change in velocity and the time over which the change happens.
1. A larger mass needs a larger force to produce the same change in velocity in the same time.
2. A larger change in velocity in the same time needs a larger force.
3. The same change in velocity spread over a longer time needs a smaller force.
The last of these ideas explains why stopping something gently needs a smaller force than stopping it suddenly.
1. A fielder who pulls their hands back while catching a cricket ball makes the stopping time longer, so the force on their hands is smaller.
2. Bending your knees when you land from a jump makes the time to stop longer, so the force on your legs is smaller.
If the time is not known, the acceleration can be found from the velocities and the distance with $v^2-u^2=2as$ , which you need to select and apply rather than recall.
Combining $v^2-u^2=2as$ with $F = ma$ shows that, for a given mass and change of speed, a larger force changes the velocity over a shorter distance.

Solving multi-step problems

Many questions need two equations, with the answer from the first used in the second.
Work out which quantity links the two equations, which is usually the acceleration.
Find the acceleration from the motion data first, then use it in $F = ma$ , or find the acceleration from $F = ma$ first, then use it to find a velocity, a time or a distance.
Keep unrounded values in your calculator between steps and round only the final answer.
Check the sign of the acceleration, because a negative value means the resultant force acts against the motion and the object is slowing down.

Example

Force to accelerate a car

A car of mass $kg1200\ \text{kg}$ speeds up from $m/s8.0\ \text{m/s}$ to $m/s20\ \text{m/s}$ in $s6.0\ \text{s}$ .
Find the acceleration first.

a=\dfrac{v-u}{t}=\dfrac{20\ \text{m/s}-8.0\ \text{m/s}}{6.0\ \text{s}}=2.0\ \text{m/s}^2

Then use Newton's second law.

F=ma=1200\ \text{kg}\times2.0\ \text{m/s}^2=2400\ \text{N}

The resultant force on the car is $N2400\ \text{N}$ forwards.

Example

Time for a force to change a velocity

A resultant force of $N150\ \text{N}$ acts on a $kg60\ \text{kg}$ cyclist who starts from rest.

a=\dfrac{F}{m}=\dfrac{150\ \text{N}}{60\ \text{kg}}=2.5\ \text{m/s}^2

Rearrange $a=v−uta=\dfrac{v-u}{t}$ to find the time to reach $m/s10\ \text{m/s}$ .

t=\dfrac{v-u}{a}=\dfrac{10\ \text{m/s}-0}{2.5\ \text{m/s}^2}=4.0\ \text{s}

The cyclist reaches $m/s10\ \text{m/s}$ after $s4.0\ \text{s}$ .

Example

Braking force from a stopping distance

A $kg900\ \text{kg}$ car travelling at $m/s20\ \text{m/s}$ brakes and stops in a distance of $m25\ \text{m}$ .
No time is given, so select the equation that links $u$ , $v$ , $a$ and $s$ .

a=\dfrac{v^2-u^2}{2s}=\dfrac{0^2-20^2}{2\times25}=\dfrac{-400}{50}=-8.0\ \text{m/s}^2

F=ma=900\ \text{kg}\times(-8.0\ \text{m/s}^2)=-7200\ \text{N}

The braking force is $N7200\ \text{N}$ , and the negative sign shows that it acts opposite to the motion.

Example

Predicting a change with proportion

A loaded van has twice the mass of the same van when empty, and the same resultant force acts on it in both cases.
The acceleration is inversely proportional to the mass, so the loaded van has half the acceleration of the empty van.
With half the acceleration, the loaded van takes twice as long to reach the same speed from rest.

Common Mistake

Do not use the time in $F = ma$ directly, because you must find the acceleration first.
Do not drop the minus sign from a deceleration partway through a calculation, because the sign tells you the direction of the force.
Do not assume that a larger force always gives a larger acceleration, because a larger mass can reduce the acceleration more than the force increases it.

Exam technique

Multi-step force and motion problems

Write the two equations you plan to use at the start, so the examiner can follow your method.
Label each intermediate answer, such as "acceleration $m/s2=2.0\ \text{m/s}^2$ ", as this often carries its own mark.
For a "how would the acceleration change" question, use proportion and give a factor, such as "halves" or "doubles", rather than only "decreases".
When you explain why a longer stopping time helps, link it to a smaller deceleration and so to a smaller resultant force.

Self review

How does the acceleration of an object change if the resultant force doubles and the mass stays the same?
Write the equation that combines $F = ma$ and $a=v−uta=\dfrac{v-u}{t}$ .
Explain why bending your knees on landing reduces the force on your legs.
Calculate the resultant force on a $kg50\ \text{kg}$ skater who speeds up from $m/s2.0\ \text{m/s}$ to $m/s5.0\ \text{m/s}$ in $s1.5\ \text{s}$ .
What happens to the acceleration if both the force and the mass are tripled?

2.2.12 Relating force, mass, velocity and acceleration

Linking force and acceleration

Force and change in velocity

Solving multi-step problems

Force to accelerate a car

Time for a force to change a velocity

Braking force from a stopping distance

Predicting a change with proportion

Multi-step force and motion problems

Review 2.2.12 Relating force, mass, velocity and acceleration by teaching Genie

2.2.12 Relating force, mass, velocity and acceleration

Linking force and acceleration

Force and change in velocity

Solving multi-step problems

Force to accelerate a car

Time for a force to change a velocity

Braking force from a stopping distance

Predicting a change with proportion

Multi-step force and motion problems

Review 2.2.12 Relating force, mass, velocity and acceleration by teaching Genie

2.2.12 Relating force, mass, velocity and acceleration

Linking force and acceleration

Force and change in velocity

Solving multi-step problems

Force to accelerate a car

Time for a force to change a velocity

Braking force from a stopping distance

Predicting a change with proportion

Multi-step force and motion problems

Review 2.2.12 Relating force, mass, velocity and acceleration by teaching Genie

2.2.12 Relating force, mass, velocity and acceleration Revision Guide

2.2.12 Relating force, mass, velocity and acceleration

Linking force and acceleration

Force and change in velocity

Solving multi-step problems

Force to accelerate a car

Time for a force to change a velocity

Braking force from a stopping distance

Predicting a change with proportion

Multi-step force and motion problems

Review 2.2.12 Relating force, mass, velocity and acceleration by teaching Genie

2.2.12 Relating force, mass, velocity and acceleration Revision Guide