Definition

Newton's second law

Newton's second law states that the resultant force on an object equals its mass multiplied by its acceleration.

The relationship is $F = ma$ , where $F$ is resultant force in newtons, $m$ is mass in kilograms and $a$ is acceleration in metres per second squared.
The acceleration is in the same direction as the resultant force. A larger resultant force causes a larger acceleration in that direction.
For constant mass, acceleration is directly proportional to resultant force: $a∝Fa\propto F$ . Doubling $F$ doubles $a$ .
For constant resultant force, acceleration is inversely proportional to mass: $a∝1ma\propto\dfrac{1}{m}$ . Doubling $m$ halves $a$ .
One newton is the resultant force that gives a mass of $kg1\ \text{kg}$ an acceleration of $s−21\ \text{m s}^{-2}$ , so $s−21\ \text{N}=1\ \text{kg m s}^{-2}$ .

Using the force equation

Use the equation $F = ma$ . Its rearranged forms are $a=Fma=\dfrac{F}{m}$ and $m=Fam=\dfrac{F}{a}$ .
The force in this equation is the resultant force. Subtract opposing forces before substitution.
Convert mass to kilograms before using the equation. For example, $kg750\ \text{g}=0.750\ \text{kg}$ .
If velocity data are given, first calculate acceleration using $a=v−uta=\dfrac{v-u}{t}$ , then use $F = ma$ .

Example

Calculating acceleration

A van of mass $kg1500\ \text{kg}$ has a driving force of $N4200\ \text{N}$ and resistive forces of $N1200\ \text{N}$ .
The resultant force is $NF=4200-1200=3000\ \text{N}$ .
Rearranging $F = ma$ gives $a=Fma=\dfrac{F}{m}$ .
Substitution gives $s−2a=\dfrac{3000}{1500}=2.0\ \text{m s}^{-2}$ .
The acceleration is $s−22.0\ \text{m s}^{-2}$ forwards.

Example

Finding resultant force

A $kg900\ \text{kg}$ car increases its velocity from $s−18.0\ \text{m s}^{-1}$ to $s−120\ \text{m s}^{-1}$ in $s6.0\ \text{s}$ .
First calculate acceleration: $s−2a=\dfrac{v-u}{t}=\dfrac{20-8.0}{6.0}=2.0\ \text{m s}^{-2}$ .
Then use $F = ma$ : $NF=900\times2.0=1800\ \text{N}$ .
The resultant force is $N1800\ \text{N}$ in the direction of the acceleration.

Practical

Force, mass and acceleration

Aim: to investigate how acceleration depends on resultant force and total moving mass using a trolley system.
Apparatus: dynamics trolley, level or gently inclined runway, pulley and clamp, light string, mass hanger, slotted masses, additional trolley masses, two light gates with a data logger or a motion sensor, interrupt card, balance, metre rule and stop block.
Variables, mass investigation: total moving mass is the independent variable, acceleration is the dependent variable, and the hanging mass, pulling force, release position, runway angle, light-gate positions and pulley arrangement are controlled.
Method, set-up:
- Measure the mass of the trolley, hanger, string and every slotted mass that will move. Record the total moving mass in kilograms.
- Attach the trolley to the hanger using a light string over the pulley. Keep the string parallel to the runway and secure the pulley firmly.
- Fit the interrupt card, align and zero the light gates or motion sensor, mark a fixed release position and place a stop block before the pulley.
- Check whether the runway is level. If friction compensation is used, raise one end slightly until the trolley moves at nearly constant velocity after a gentle push, then leave the angle unchanged.
Method, changing mass at constant pulling force:
- Place a fixed mass on the hanger so its weight provides an approximately constant pulling force, $F=mhgF=m_{\mathrm{h}}g$ .
- Release the trolley from the marked point without pushing it. Record the velocity-time data and determine the acceleration.
- Repeat at least three times and calculate the mean acceleration. Investigate anomalous readings rather than removing them automatically.
- Add a known mass to the trolley while leaving the hanging mass unchanged. Recalculate the total moving mass and repeat for at least five masses.
Method, changing force at constant total mass:
- Place several slotted masses on the trolley and keep the total moving mass fixed throughout.
- Transfer one mass at a time from the trolley to the hanger. Each transfer increases the pulling force without changing the total mass being accelerated.
- For each pulling force, release the trolley from the same point, record at least three accelerations and calculate a mean.
Results: increasing total mass decreases acceleration when the pulling force is constant. Increasing resultant force increases acceleration when total mass is constant.
Maths: calculate acceleration using $a=v−uta=\dfrac{v-u}{t}$ where required. Plot $a$ against $m$ for the mass investigation and $a$ against $F$ for the force investigation. A plot of $a$ against $1m\dfrac{1}{m}$ should be linear for constant force, while $a$ against $F$ should be a straight line through the origin when mass is constant.
Watch out: use the mass of the whole moving system, including the hanger. Friction and pulley resistance mean the hanging weight is not exactly the resultant force. Slack string, a sloping runway, inconsistent release and misaligned light gates can also distort the results.
Safety: secure the runway and pulley, use a stop block, keep feet clear of the falling hanger and use modest masses so the trolley remains controlled.

Example

Using a graph gradient

An acceleration against force graph passes through $s−2)(0.20\ \text{N},0.25\ \text{m s}^{-2})$ and $s−2)(1.00\ \text{N},1.25\ \text{m s}^{-2})$ .
The gradient is $N−1\dfrac{1.25-0.25}{1.00-0.20}=1.25\ \text{m s}^{-2}\text{ N}^{-1}$ .
Since $a=Fma=\dfrac{F}{m}$ , the gradient equals $1m\dfrac{1}{m}$ .
Therefore $kgm=\dfrac{1}{1.25}=0.80\ \text{kg}$ .

Exam technique

Force calculation method

Write $F = ma$ before substituting, then show the resultant-force calculation separately.
Use the unit $N\text{N}$ for force, $kg\text{kg}$ for mass and $s−2\text{m s}^{-2}$ for acceleration.
For a practical improvement, name the change and link it to its effect, such as using an automatic release so the trolley is not given an extra push.
Use directly proportional only when the graph is a straight line through the origin.

Common Mistake

Common force errors

Do not use the driving or pulling force as $F$ without considering opposing forces.
Do not put weight in newtons into the mass position. Calculate mass in kilograms first.
Do not change force and mass at the same time unless total moving mass is deliberately kept constant by transferring masses.

Self review

State Newton's second law and give the unit of each quantity in $F = ma$ .
Calculate the acceleration of a $kg1200\ \text{kg}$ car acted on by a resultant force of $N3600\ \text{N}$ .
Explain how acceleration changes when mass doubles at constant resultant force.
Describe how the practical changes force while keeping total moving mass constant.
Explain why repeats, a mean and friction control improve the investigation.

Definition

Newton's second law

Newton's second law states that the resultant force on an object equals its mass multiplied by its acceleration.

The relationship is $F = ma$ , where $F$ is resultant force in newtons, $m$ is mass in kilograms and $a$ is acceleration in metres per second squared.
The acceleration is in the same direction as the resultant force. A larger resultant force causes a larger acceleration in that direction.
For constant mass, acceleration is directly proportional to resultant force: $a∝Fa\propto F$ . Doubling $F$ doubles $a$ .
For constant resultant force, acceleration is inversely proportional to mass: $a∝1ma\propto\dfrac{1}{m}$ . Doubling $m$ halves $a$ .
One newton is the resultant force that gives a mass of $kg1\ \text{kg}$ an acceleration of $s−21\ \text{m s}^{-2}$ , so $s−21\ \text{N}=1\ \text{kg m s}^{-2}$ .

Using the force equation

Use the equation $F = ma$ . Its rearranged forms are $a=Fma=\dfrac{F}{m}$ and $m=Fam=\dfrac{F}{a}$ .
The force in this equation is the resultant force. Subtract opposing forces before substitution.
Convert mass to kilograms before using the equation. For example, $kg750\ \text{g}=0.750\ \text{kg}$ .
If velocity data are given, first calculate acceleration using $a=v−uta=\dfrac{v-u}{t}$ , then use $F = ma$ .

Example

Calculating acceleration

A van of mass $kg1500\ \text{kg}$ has a driving force of $N4200\ \text{N}$ and resistive forces of $N1200\ \text{N}$ .
The resultant force is $NF=4200-1200=3000\ \text{N}$ .
Rearranging $F = ma$ gives $a=Fma=\dfrac{F}{m}$ .
Substitution gives $s−2a=\dfrac{3000}{1500}=2.0\ \text{m s}^{-2}$ .
The acceleration is $s−22.0\ \text{m s}^{-2}$ forwards.

Example

Finding resultant force

A $kg900\ \text{kg}$ car increases its velocity from $s−18.0\ \text{m s}^{-1}$ to $s−120\ \text{m s}^{-1}$ in $s6.0\ \text{s}$ .
First calculate acceleration: $s−2a=\dfrac{v-u}{t}=\dfrac{20-8.0}{6.0}=2.0\ \text{m s}^{-2}$ .
Then use $F = ma$ : $NF=900\times2.0=1800\ \text{N}$ .
The resultant force is $N1800\ \text{N}$ in the direction of the acceleration.

Practical

Force, mass and acceleration

Aim: to investigate how acceleration depends on resultant force and total moving mass using a trolley system.
Apparatus: dynamics trolley, level or gently inclined runway, pulley and clamp, light string, mass hanger, slotted masses, additional trolley masses, two light gates with a data logger or a motion sensor, interrupt card, balance, metre rule and stop block.
Variables, mass investigation: total moving mass is the independent variable, acceleration is the dependent variable, and the hanging mass, pulling force, release position, runway angle, light-gate positions and pulley arrangement are controlled.
Method, set-up:
- Measure the mass of the trolley, hanger, string and every slotted mass that will move. Record the total moving mass in kilograms.
- Attach the trolley to the hanger using a light string over the pulley. Keep the string parallel to the runway and secure the pulley firmly.
- Fit the interrupt card, align and zero the light gates or motion sensor, mark a fixed release position and place a stop block before the pulley.
- Check whether the runway is level. If friction compensation is used, raise one end slightly until the trolley moves at nearly constant velocity after a gentle push, then leave the angle unchanged.
Method, changing mass at constant pulling force:
- Place a fixed mass on the hanger so its weight provides an approximately constant pulling force, $F=mhgF=m_{\mathrm{h}}g$ .
- Release the trolley from the marked point without pushing it. Record the velocity-time data and determine the acceleration.
- Repeat at least three times and calculate the mean acceleration. Investigate anomalous readings rather than removing them automatically.
- Add a known mass to the trolley while leaving the hanging mass unchanged. Recalculate the total moving mass and repeat for at least five masses.
Method, changing force at constant total mass:
- Place several slotted masses on the trolley and keep the total moving mass fixed throughout.
- Transfer one mass at a time from the trolley to the hanger. Each transfer increases the pulling force without changing the total mass being accelerated.
- For each pulling force, release the trolley from the same point, record at least three accelerations and calculate a mean.
Results: increasing total mass decreases acceleration when the pulling force is constant. Increasing resultant force increases acceleration when total mass is constant.
Maths: calculate acceleration using $a=v−uta=\dfrac{v-u}{t}$ where required. Plot $a$ against $m$ for the mass investigation and $a$ against $F$ for the force investigation. A plot of $a$ against $1m\dfrac{1}{m}$ should be linear for constant force, while $a$ against $F$ should be a straight line through the origin when mass is constant.
Watch out: use the mass of the whole moving system, including the hanger. Friction and pulley resistance mean the hanging weight is not exactly the resultant force. Slack string, a sloping runway, inconsistent release and misaligned light gates can also distort the results.
Safety: secure the runway and pulley, use a stop block, keep feet clear of the falling hanger and use modest masses so the trolley remains controlled.

Example

Using a graph gradient

An acceleration against force graph passes through $s−2)(0.20\ \text{N},0.25\ \text{m s}^{-2})$ and $s−2)(1.00\ \text{N},1.25\ \text{m s}^{-2})$ .
The gradient is $N−1\dfrac{1.25-0.25}{1.00-0.20}=1.25\ \text{m s}^{-2}\text{ N}^{-1}$ .
Since $a=Fma=\dfrac{F}{m}$ , the gradient equals $1m\dfrac{1}{m}$ .
Therefore $kgm=\dfrac{1}{1.25}=0.80\ \text{kg}$ .

Exam technique

Force calculation method

Write $F = ma$ before substituting, then show the resultant-force calculation separately.
Use the unit $N\text{N}$ for force, $kg\text{kg}$ for mass and $s−2\text{m s}^{-2}$ for acceleration.
For a practical improvement, name the change and link it to its effect, such as using an automatic release so the trolley is not given an extra push.
Use directly proportional only when the graph is a straight line through the origin.

Common Mistake

Common force errors

Do not use the driving or pulling force as $F$ without considering opposing forces.
Do not put weight in newtons into the mass position. Calculate mass in kilograms first.
Do not change force and mass at the same time unless total moving mass is deliberately kept constant by transferring masses.

Self review

State Newton's second law and give the unit of each quantity in $F = ma$ .
Calculate the acceleration of a $kg1200\ \text{kg}$ car acted on by a resultant force of $N3600\ \text{N}$ .
Explain how acceleration changes when mass doubles at constant resultant force.
Describe how the practical changes force while keeping total moving mass constant.
Explain why repeats, a mean and friction control improve the investigation.

2.3.2 Newton's second law

Using the force equation

Calculating acceleration

Finding resultant force

Force, mass and acceleration

Using a graph gradient

Force calculation method

Common force errors

Review 2.3.2 Newton's second law by teaching Genie

Remember key concepts with flashcards

2.3.2 Newton's second law

Using the force equation

Calculating acceleration

Finding resultant force

Force, mass and acceleration

Using a graph gradient

Force calculation method

Common force errors

Review 2.3.2 Newton's second law by teaching Genie

Remember key concepts with flashcards

2.3.2 Newton's second law

Using the force equation

Calculating acceleration

Finding resultant force

Force, mass and acceleration

Using a graph gradient

Force calculation method

Common force errors

Review 2.3.2 Newton's second law by teaching Genie

Remember key concepts with flashcards

2.3.2 Newton's second law Revision Guide

2.3.2 Newton's second law

Using the force equation

Calculating acceleration

Finding resultant force

Force, mass and acceleration

Using a graph gradient

Force calculation method

Common force errors

Review 2.3.2 Newton's second law by teaching Genie

Remember key concepts with flashcards

2.3.2 Newton's second law Revision Guide