5.6.2c Newton's Second Law

Newton’s second law

Definition

Newton’s second law

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

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 $m/s2\text{m/s}^2$ .
The resultant force is the overall force after all the forces on the object are combined, and the object accelerates in its direction.
At constant mass, increasing the resultant force increases the acceleration by the same factor: doubling the force doubles the acceleration.
At constant force, increasing the mass decreases the acceleration, because a larger mass is harder to accelerate: doubling the mass halves the acceleration.
Rearranged, the equation gives $\dfrac{F}{m}$ and $\dfrac{F}{a}$ .

Example

A car has a mass of $kg1200\ \text{kg}$ and accelerates at $m/s22.5\ \text{m/s}^2$ . Calculate the resultant force on the car.

Write the equation: $F = ma$ .
Substitute the values: $1200\ \text{kg} \times 2.5\ \text{m/s}^2$ .
Calculate the force: $3000\ \text{N}$ .

Estimating forces in road transport

You should be able to estimate the speeds, accelerations and forces when road vehicles accelerate rapidly; estimates need sensible values rather than exact ones.
The symbol $≈\approx$ means approximately equal to; for example, $m/s60\ \text{mph} \approx 27\ \text{m/s}$ .
Estimate the acceleration with $\dfrac{\text{change in velocity}}{\text{time}}$ , then find the force with $F = ma$ .

Example

A car of mass $kg1500\ \text{kg}$ accelerates from rest to about $m/s27\ \text{m/s}$ in $s9.0\ \text{s}$ . Estimate its acceleration and the resultant force on it.

Estimate the acceleration: $\dfrac{27 - 0}{9.0} \approx 3.0\ \text{m/s}^2$ .
Find the force: $1500\ \text{kg} \times 3.0\ \text{m/s}^2$ .
Calculate the force: $\approx 4500\ \text{N}$ ; the $≈\approx$ symbol is used because the speed and time are estimates.

Using the equation carefully

Always use the resultant force in $F = ma$ , not just one of the forces acting.
Acceleration does not mean “moving quickly”; it means a change in velocity. An object at high constant velocity has zero acceleration and zero resultant force.
Do not confuse mass (kilograms, $kg\text{kg}$ ) with weight (newtons, $N\text{N}$ ).
Write $F = ma$ before substituting, use mass in $kg\text{kg}$ and give force in $N\text{N}$ ; at Higher tier you may need to select and rearrange the equation yourself.

Practical

Investigation: how force and mass affect the acceleration of a trolley

Aim: investigate how the acceleration of a trolley depends on the force applied to it (at constant mass) and on its mass (at constant force).
Apparatus: a toy car or trolley, a metre ruler, pencil, chalk or masking tape to mark intervals, a bench pulley, string, a small weight stack (for example up to $N1.0\ \text{N}$ in $N0.2\ \text{N}$ steps), a stopwatch, and Blu-Tack to fix masses to the car.
Set up: use the ruler to mark equal intervals along the bench (for example every $cm20\ \text{cm}$ up to $cm100\ \text{cm}$ ); clamp the bench pulley at the end of the bench; tie the string to the car, pass it over the pulley and hang the weight stack on the other end, keeping the string horizontal and in line with the car.
Vary the force (constant mass): attach the full weight stack ( $N1.0\ \text{N}$ ) to the string and hold the car at the start line.
Release and time: release the car and start the stopwatch together, then use lap timing to record the time at each marked interval and the final time at $cm100\ \text{cm}$ .
Change the force: repeat for smaller pulling forces ( $N0.8\ \text{N}$ , $N0.6\ \text{N}$ , $N0.4\ \text{N}$ , $N0.2\ \text{N}$ ); each time you remove a weight from the stack, place it on top of the car so the total mass of the system stays constant.
Find the acceleration for each force from the distance–time data, for example from a velocity–time graph.
Vary the mass (constant force): keep the hanging weight the same so the pulling force is constant, then add a known mass (for example $g200\ \text{g}$ ) to the car.
Release, time and repeat: lap-time the car to $cm100\ \text{cm}$ , find its acceleration, then repeat with more masses added to the car, taking repeat readings and calculating a mean acceleration each time.
Control variables: the same runway, pulley and release point; the total mass is constant when varying the force, and the pulling force is constant when varying the mass.
Result: at constant mass the acceleration is proportional to the force; at constant force the acceleration is inversely proportional to the mass, both consistent with $F = ma$ .

Interpreting the results

When the force is varied at constant mass, a graph of acceleration against force is a straight line through the origin: acceleration is proportional to force.
When the mass is varied at constant force, the acceleration decreases as the mass increases, because acceleration is inversely proportional to mass.
For the practical, clearly state what is changed, what is measured and what is kept constant; when varying the force, the key point is that the total mass stays constant.
Take repeat readings and use mean accelerations to reduce random error.

Self review

State Newton’s second law.
Give the equation and the units of each quantity.
What happens to the acceleration if the force doubles at constant mass?
What happens if the mass doubles at constant force?
In the investigation, why must the total mass stay constant when the force is changed?

5.6.2d Inertial mass (HT)

What inertial mass means

Definition

Inertial mass

Inertial mass is a measure of how difficult it is to change the velocity of an object.

Changing an object's velocity means accelerating it, which can involve changing its speed, its direction, or both.
An object with a large inertial mass is harder to accelerate than one with a small inertial mass, because a greater resultant force is needed to produce the same acceleration.
For example, with the same resultant force an empty shopping trolley has the smaller inertial mass, so it gains the greater acceleration.
A fully loaded trolley has the larger inertial mass, so with the same force it gains the smaller acceleration.

Key Idea

For a fixed resultant force, a larger inertial mass gives a smaller acceleration; for a fixed acceleration, a larger inertial mass needs a greater resultant force.

Defining inertial mass with force and acceleration

Inertial mass is defined as the ratio of the resultant force acting on an object to the acceleration produced: $\frac{F}{a}$ .
In this equation $m$ is the inertial mass in kilograms, $kg\text{kg}$ ; $F$ is the resultant force in newtons, $N\text{N}$ ; and $a$ is the acceleration in $m/s2\text{m/s}^2$ .
The word resultant matters: if several forces act, $F$ is the overall force once their sizes and directions have been combined.
The ratio $Fa\frac{F}{a}$ shows how strongly an object resists a change in velocity; a larger ratio means a larger inertial mass.
The relationship can also be written as $F = ma$ , so for a particular object doubling the resultant force doubles its acceleration.
If two objects feel the same resultant force, the one with the greater inertial mass has the smaller acceleration.

Example

A resultant force of $N18\ \text{N}$ makes an object accelerate at $m/s23.0\ \text{m/s}^2$ . Calculate its inertial mass.

Write the equation: $\frac{F}{a}$ .
Substitute the values: $\frac{18\ \text{N}}{3.0\ \text{m/s}^2}$ .
The inertial mass is $6.0\ \text{kg}$ .

Inertial mass is not weight

Inertial mass measures how difficult it is to change an object's velocity and is measured in $kg\text{kg}$ .
Weight is a force and is measured in $N\text{N}$ , so it is not the same as inertial mass.
An object with a large inertial mass can still accelerate; it simply needs a greater resultant force to reach a particular acceleration.

Common Mistake

Do not confuse inertial mass (in $kg\text{kg}$ ) with weight (a force, in $N\text{N}$ ). Also, do not say that an object with a large inertial mass cannot accelerate; it can, but only with a greater resultant force.

Exam technique

To explain inertial mass, state that it measures how difficult it is to change an object's velocity.
Then state that it is defined by $\frac{F}{a}$ .
In a comparison, link all three quantities, for example: the object has a larger inertial mass, so the same resultant force produces a smaller acceleration.

Self review

What does inertial mass measure?
Write the equation that defines inertial mass and give the unit of each quantity.
Which force is used in the equation for inertial mass?
For the same resultant force, how does a larger inertial mass affect the acceleration?
For the same acceleration, how does a larger inertial mass affect the resultant force needed?