Definition

Newton's second law

The acceleration of an object is directly proportional to the resultant force acting on it and inversely proportional to its mass.

Definition

Mass

The amount of matter in an object, measured in kilograms.

Newton's second law tells you how much an object accelerates when a resultant force acts on it.
The acceleration is directly proportional to the resultant force when the mass stays the same.
1. Doubling the resultant force on a trolley doubles its acceleration.
The acceleration is inversely proportional to the mass when the resultant force stays the same.
1. Doubling the mass of a trolley halves its acceleration for the same resultant force.
The acceleration always acts in the same direction as the resultant force.
These ideas combine into one equation, $force=mass×acceleration\text{force}=\text{mass}\times\text{acceleration}$ , which you must recall.
In symbols the equation is $F = ma$ .
1. $F$ is the resultant force in newtons, $N\text{N}$ .
2. $m$ is the mass in kilograms, $kg\text{kg}$ .
3. $a$ is the acceleration in metres per second squared, $m/s2\text{m/s}^2$ .

The newton

Definition

Newton

The unit of force, equal to the resultant force that gives a mass of one kilogram an acceleration of one metre per second squared.

The equation $F = ma$ is used to define the unit of force.
One newton is the resultant force that gives a mass of $kg1\ \text{kg}$ an acceleration of $m/s21\ \text{m/s}^2$ .
This means that $m/s21\ \text{N}=1\ \text{kg m/s}^2$ .
Large forces are often given in kilonewtons, where $N1\ \text{kN}=1000\ \text{N}$ .

Second law calculations

The equation can be rearranged to find any one of the three quantities.
1. To find the acceleration, divide both sides by $m$ to give $a=Fma=\dfrac{F}{m}$ .
2. To find the mass, divide both sides by $a$ to give $m=Fam=\dfrac{F}{a}$ .
Always use the resultant force, so work it out first when more than one force acts along the line of motion.
Convert every value into SI units before you substitute.
1. A mass in grams is divided by $1000$ to give kilograms, so $kg250\ \text{g}=0.25\ \text{kg}$ .
2. A force in kilonewtons is multiplied by $1000$ to give newtons, so $N4.5\ \text{kN}=4500\ \text{N}$ .
A negative acceleration from your calculation means the object is slowing down, which happens when the resultant force acts against the motion.

Example

Acceleration of a delivery van

A delivery van with a mass of $kg2500\ \text{kg}$ has a resultant forward force of $N5000\ \text{N}$ .

a=\dfrac{F}{m}

a=\dfrac{5000\ \text{N}}{2500\ \text{kg}}=2.0\ \text{m/s}^2

The van accelerates at $m/s22.0\ \text{m/s}^2$ in the direction of the resultant force.

Example

Acceleration when two forces act

A cyclist and bicycle with a total mass of $kg80\ \text{kg}$ have a forward driving force of $N200\ \text{N}$ and air resistance of $N40\ \text{N}$ .
Find the resultant force first.

F=200\ \text{N}-40\ \text{N}=160\ \text{N}

a=\dfrac{F}{m}=\dfrac{160\ \text{N}}{80\ \text{kg}}=2.0\ \text{m/s}^2

The cyclist accelerates forwards at $m/s22.0\ \text{m/s}^2$ .

Example

Force needed to stop a football

A goalkeeper stops a $g450\ \text{g}$ football, giving it a deceleration of $m/s260\ \text{m/s}^2$ .
Convert the mass into kilograms, so $kgm=0.45\ \text{kg}$ .

F=ma=0.45\ \text{kg}\times60\ \text{m/s}^2=27\ \text{N}

The goalkeeper's gloves exert a force of $N27\ \text{N}$ on the ball, opposite to its motion.

Practical

Investigating force and acceleration

Aim: to investigate how the acceleration of a trolley depends on the resultant force acting on it, with the total mass kept constant.
Apparatus: a dynamics trolley, a $m2\ \text{m}$ runway, wooden blocks, a bench pulley, strong thread, a $g100\ \text{g}$ mass hanger with five $g100\ \text{g}$ slotted masses, a double interrupt card, a light gate and a data logger, a metre rule, a top-pan balance and a box of sand or a cushion.
Method:
- Clamp the pulley to the end of the bench and lay the runway on the bench so that the trolley can roll towards the pulley.
- Raise the starting end of the runway with blocks until a trolley given a gentle push rolls down at a steady speed, which shows that friction has been balanced by a small component of the weight.
- Fix the double interrupt card upright on the trolley and position the light gate near the pulley end of the runway.
- Place four $g100\ \text{g}$ masses on the trolley and tie the thread from the trolley over the pulley to the empty hanger, so that the hanger hangs just below the pulley with the box of sand beneath it.
- Each $g100\ \text{g}$ that hangs from the thread has a weight of $N1\ \text{N}$ , so the hanger alone gives an accelerating force of $N1\ \text{N}$ .
- Hold the trolley at a marked start line, then release it without pushing.
- Record the acceleration calculated by the data logger from the speeds of the two halves of the card and the time between them.
- Repeat twice more and calculate the mean acceleration.
- Move one $g100\ \text{g}$ mass from the trolley to the hanger, increasing the force by $N1\ \text{N}$ , and repeat the measurements.
- Continue until all of the masses are on the hanger, which gives five values of force from $N1\ \text{N}$ to $N5\ \text{N}$ .
- Plot a graph of mean acceleration on the vertical axis against force on the horizontal axis.
Variables: the accelerating force is the independent variable, the acceleration is the dependent variable, and the total mass of the trolley, hanger and masses, the runway slope and the start line are control variables.
Results: the acceleration increases in equal steps as the force increases in equal steps, giving a straight line that passes through the origin.
Conclusion: the acceleration is directly proportional to the resultant force when the mass is constant.
Watch out: moving masses from the trolley to the hanger keeps the total moving mass constant, but adding new masses to the hanger would change the mass as well as the force and spoil the test.
Safety: keep feet away from the falling hanger and let it land in the box of sand, and stop the trolley before it hits the pulley.

Common Mistake

Do not use one force from a diagram in $F = ma$ when several forces act, because the equation needs the resultant force.
Do not substitute a mass in grams, because the equation only works with mass in kilograms.
Do not confuse mass and weight, because mass is measured in $kg\text{kg}$ and weight is a force measured in $N\text{N}$ .

Exam technique

Newton's second law calculations

Write $F = ma$ from memory, because it is not usually given to you.
Rearrange the equation before substituting when the unknown is $a$ or $m$ , and show the rearranged form on its own line.
Look out for masses in grams and forces in kilonewtons, as a conversion is often where a mark is lost.
In practical questions, explain that moving masses from the trolley to the hanger keeps the total mass constant.

Self review

State Newton's second law in words.
Write the equation that links resultant force, mass and acceleration, with the unit of each quantity.
Define one newton.
Calculate the acceleration of a $kg1200\ \text{kg}$ car with a resultant force of $N3000\ \text{N}$ .
Why are masses moved from the trolley to the hanger, rather than added to the hanger, in the force and acceleration practical?

Definition

Newton's second law

The acceleration of an object is directly proportional to the resultant force acting on it and inversely proportional to its mass.

Definition

Mass

The amount of matter in an object, measured in kilograms.

Newton's second law tells you how much an object accelerates when a resultant force acts on it.
The acceleration is directly proportional to the resultant force when the mass stays the same.
1. Doubling the resultant force on a trolley doubles its acceleration.
The acceleration is inversely proportional to the mass when the resultant force stays the same.
1. Doubling the mass of a trolley halves its acceleration for the same resultant force.
The acceleration always acts in the same direction as the resultant force.
These ideas combine into one equation, $force=mass×acceleration\text{force}=\text{mass}\times\text{acceleration}$ , which you must recall.
In symbols the equation is $F = ma$ .
1. $F$ is the resultant force in newtons, $N\text{N}$ .
2. $m$ is the mass in kilograms, $kg\text{kg}$ .
3. $a$ is the acceleration in metres per second squared, $m/s2\text{m/s}^2$ .

The newton

Definition

Newton

The unit of force, equal to the resultant force that gives a mass of one kilogram an acceleration of one metre per second squared.

The equation $F = ma$ is used to define the unit of force.
One newton is the resultant force that gives a mass of $kg1\ \text{kg}$ an acceleration of $m/s21\ \text{m/s}^2$ .
This means that $m/s21\ \text{N}=1\ \text{kg m/s}^2$ .
Large forces are often given in kilonewtons, where $N1\ \text{kN}=1000\ \text{N}$ .

Second law calculations

The equation can be rearranged to find any one of the three quantities.
1. To find the acceleration, divide both sides by $m$ to give $a=Fma=\dfrac{F}{m}$ .
2. To find the mass, divide both sides by $a$ to give $m=Fam=\dfrac{F}{a}$ .
Always use the resultant force, so work it out first when more than one force acts along the line of motion.
Convert every value into SI units before you substitute.
1. A mass in grams is divided by $1000$ to give kilograms, so $kg250\ \text{g}=0.25\ \text{kg}$ .
2. A force in kilonewtons is multiplied by $1000$ to give newtons, so $N4.5\ \text{kN}=4500\ \text{N}$ .
A negative acceleration from your calculation means the object is slowing down, which happens when the resultant force acts against the motion.

Example

Acceleration of a delivery van

A delivery van with a mass of $kg2500\ \text{kg}$ has a resultant forward force of $N5000\ \text{N}$ .

a=\dfrac{F}{m}

a=\dfrac{5000\ \text{N}}{2500\ \text{kg}}=2.0\ \text{m/s}^2

The van accelerates at $m/s22.0\ \text{m/s}^2$ in the direction of the resultant force.

Example

Acceleration when two forces act

A cyclist and bicycle with a total mass of $kg80\ \text{kg}$ have a forward driving force of $N200\ \text{N}$ and air resistance of $N40\ \text{N}$ .
Find the resultant force first.

F=200\ \text{N}-40\ \text{N}=160\ \text{N}

a=\dfrac{F}{m}=\dfrac{160\ \text{N}}{80\ \text{kg}}=2.0\ \text{m/s}^2

The cyclist accelerates forwards at $m/s22.0\ \text{m/s}^2$ .

Example

Force needed to stop a football

A goalkeeper stops a $g450\ \text{g}$ football, giving it a deceleration of $m/s260\ \text{m/s}^2$ .
Convert the mass into kilograms, so $kgm=0.45\ \text{kg}$ .

F=ma=0.45\ \text{kg}\times60\ \text{m/s}^2=27\ \text{N}

The goalkeeper's gloves exert a force of $N27\ \text{N}$ on the ball, opposite to its motion.

Practical

Investigating force and acceleration

Aim: to investigate how the acceleration of a trolley depends on the resultant force acting on it, with the total mass kept constant.
Apparatus: a dynamics trolley, a $m2\ \text{m}$ runway, wooden blocks, a bench pulley, strong thread, a $g100\ \text{g}$ mass hanger with five $g100\ \text{g}$ slotted masses, a double interrupt card, a light gate and a data logger, a metre rule, a top-pan balance and a box of sand or a cushion.
Method:
- Clamp the pulley to the end of the bench and lay the runway on the bench so that the trolley can roll towards the pulley.
- Raise the starting end of the runway with blocks until a trolley given a gentle push rolls down at a steady speed, which shows that friction has been balanced by a small component of the weight.
- Fix the double interrupt card upright on the trolley and position the light gate near the pulley end of the runway.
- Place four $g100\ \text{g}$ masses on the trolley and tie the thread from the trolley over the pulley to the empty hanger, so that the hanger hangs just below the pulley with the box of sand beneath it.
- Each $g100\ \text{g}$ that hangs from the thread has a weight of $N1\ \text{N}$ , so the hanger alone gives an accelerating force of $N1\ \text{N}$ .
- Hold the trolley at a marked start line, then release it without pushing.
- Record the acceleration calculated by the data logger from the speeds of the two halves of the card and the time between them.
- Repeat twice more and calculate the mean acceleration.
- Move one $g100\ \text{g}$ mass from the trolley to the hanger, increasing the force by $N1\ \text{N}$ , and repeat the measurements.
- Continue until all of the masses are on the hanger, which gives five values of force from $N1\ \text{N}$ to $N5\ \text{N}$ .
- Plot a graph of mean acceleration on the vertical axis against force on the horizontal axis.
Variables: the accelerating force is the independent variable, the acceleration is the dependent variable, and the total mass of the trolley, hanger and masses, the runway slope and the start line are control variables.
Results: the acceleration increases in equal steps as the force increases in equal steps, giving a straight line that passes through the origin.
Conclusion: the acceleration is directly proportional to the resultant force when the mass is constant.
Watch out: moving masses from the trolley to the hanger keeps the total moving mass constant, but adding new masses to the hanger would change the mass as well as the force and spoil the test.
Safety: keep feet away from the falling hanger and let it land in the box of sand, and stop the trolley before it hits the pulley.

Common Mistake

Do not use one force from a diagram in $F = ma$ when several forces act, because the equation needs the resultant force.
Do not substitute a mass in grams, because the equation only works with mass in kilograms.
Do not confuse mass and weight, because mass is measured in $kg\text{kg}$ and weight is a force measured in $N\text{N}$ .

Exam technique

Newton's second law calculations

Write $F = ma$ from memory, because it is not usually given to you.
Rearrange the equation before substituting when the unknown is $a$ or $m$ , and show the rearranged form on its own line.
Look out for masses in grams and forces in kilonewtons, as a conversion is often where a mark is lost.
In practical questions, explain that moving masses from the trolley to the hanger keeps the total mass constant.

Self review

State Newton's second law in words.
Write the equation that links resultant force, mass and acceleration, with the unit of each quantity.
Define one newton.
Calculate the acceleration of a $kg1200\ \text{kg}$ car with a resultant force of $N3000\ \text{N}$ .
Why are masses moved from the trolley to the hanger, rather than added to the hanger, in the force and acceleration practical?

2.2.9 Newton's second law

The newton

Second law calculations

Acceleration of a delivery van

Acceleration when two forces act

Force needed to stop a football

Investigating force and acceleration

Newton's second law calculations

Review 2.2.9 Newton's second law by teaching Genie

Remember key concepts with flashcards

2.2.9 Newton's second law

The newton

Second law calculations

Acceleration of a delivery van

Acceleration when two forces act

Force needed to stop a football

Investigating force and acceleration

Newton's second law calculations

Review 2.2.9 Newton's second law by teaching Genie

Remember key concepts with flashcards

2.2.9 Newton's second law

The newton

Second law calculations

Acceleration of a delivery van

Acceleration when two forces act

Force needed to stop a football

Investigating force and acceleration

Newton's second law calculations

Review 2.2.9 Newton's second law by teaching Genie

Remember key concepts with flashcards

2.2.9 Newton's second law Revision Guide

2.2.9 Newton's second law

The newton

Second law calculations

Acceleration of a delivery van

Acceleration when two forces act

Force needed to stop a football

Investigating force and acceleration

Newton's second law calculations

Review 2.2.9 Newton's second law by teaching Genie

Remember key concepts with flashcards

2.2.9 Newton's second law Revision Guide