If the first law tells you what happens when forces balance, the second law tells you what happens when they don't: the object accelerates. Push twice as hard and it speeds up twice as fast; load it with twice the mass and it speeds up half as fast. That whole story fits in one tiny equation, $F = ma$, the engine room of all of mechanics.
You apply the same push to a $1\ \text{kg}$ trolley and then to a $2\ \text{kg}$ trolley.
Before reading: which trolley speeds up faster, and by what factor compared with the other? Write your prediction and one sentence on why.
An unbalanced force makes an object speed up, slow down or turn
When the forces on an object do not cancel, the leftover net force accelerates it, and that acceleration points in the same direction as the net force.
Newton's second law states that the net force equals mass times acceleration:
$$\vec{F}_{net} = m\vec{a}$$
Acceleration is directly proportional to the net force ($a \propto F$) and inversely proportional to mass ($a \propto 1/m$). The unit of force is the newton: $1\ \text{N} = 1\ \text{kg m s}^{-2}$. The empty trolley accelerates more than the loaded one because, for the same force, less mass means more acceleration.
Newton's second law: $\vec{F}_{net} = m\vec{a}$. Acceleration is directly proportional to the net force and inversely proportional to mass, in the direction of the net force. $1\ \text{N} = 1\ \text{kg m s}^{-2}$.
Pause, copy the highlighted law into your book before moving on.
Did you get this? The same net force acts on a $2\ \text{kg}$ object and a $4\ \text{kg}$ object. How does the $4\ \text{kg}$ object's acceleration compare?
Test the claim that a moving object needs a net force in its direction of motion
Use the simulation. Set mass $10\ \text{kg}$ and $\mu = 0.30$, get the crate moving, then find the applied force that holds the velocity constant (the trace goes flat). What is it, and why is it not zero?
Use the simulation. Set $\mu = 0$, push the crate up to speed, then press Release force so the applied force is zero. True or false: the crate slows down and stops.
Find the net force first, then apply the equation
We just saw that net force drives acceleration. That raises a question: how do we put real numbers through $F = ma$ when several forces act? This card answers it → find the net force from the free-body diagram, then rearrange $F = ma$ for the unknown.
The reliable routine is to combine all the forces into a single net force, then apply $F = ma$, rearranging to $a = F/m$ or $F = ma$ as needed.
A $4.0\ \text{kg}$ box is pushed with a $30\ \text{N}$ force while a $6.0\ \text{N}$ friction force opposes it. Find the box's acceleration.
To solve, find the net force from the free-body diagram, then apply $\vec{F}_{net} = m\vec{a}$, rearranging to $a = \dfrac{F_{net}}{m}$ or $F_{net} = ma$ as needed. The acceleration points in the direction of the net force.
Add the highlighted method to your notes before the check below.
Quick check: A net force of $50\ \text{N}$ acts on a $10\ \text{kg}$ object. What is its acceleration?
Pattern B, Plan an experiment
Plan a school experiment to verify $F = ma$ using a dynamics trolley, a string over a pulley with hanging masses, and a motion sensor. In your book: (1) state how you would keep mass constant and vary the force; (2) state what you would plot to test $a \propto F$; (3) state one source of error.
Fill the gaps. Newton’s second law is $F_{net} = $ . For a fixed net force, doubling the mass the acceleration. Weight is the force of gravity, $W = $ .
Dot point: Solve problems involving Newton's second law of motion using F = ma
A special case of the second law with $a = g$
We just learned to apply $F = ma$ with any net force. That raises a question: what is the force of gravity itself in these terms? This card answers it → weight is just $F = ma$ with the acceleration equal to $g$.
An object's weight is the gravitational force on it, and it is simply the second law applied to free fall, where the acceleration is $g$.
Setting $a = g$ in $F = ma$ gives the weight $W = mg$, directed downward. This is why a $2\ \text{kg}$ bag has a weight of about $2 \times 9.8 = 19.6\ \text{N}$. Mass (in kilograms) is the amount of matter and never changes; weight (in newtons) is a force that depends on the local $g$, so the same object weighs less on the Moon. This distinction closes the focus area: every problem you meet now reduces to drawing the forces, finding the net force, and applying $F = ma$.
Weight is gravity as a force: $W = mg$, directed downward, the second law with $a = g$. Mass (kg) is the fixed amount of matter; weight (N) is a force that depends on $g$, so the same mass weighs less where $g$ is smaller.
Pause, copy the highlighted relationship into your book before the check below.
Last check: Decide if the statement is true or false.
The weight of a $5\ \text{kg}$ object near Earth's surface is about $49\ \text{N}$ (using $W = mg$ with $g = 9.8\ \text{m s}^{-2}$).
An investigation of $F = ma$: match each variable with its role:
Dot point: Conduct a scientific investigation to illustrate the relationship between force, mass and acceleration
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
ApplyBand 5(4 marks) 1. A $1200\ \text{kg}$ car experiences a forward driving force of $4000\ \text{N}$ and a total resistive force of $1000\ \text{N}$. Calculate the net force and the car's acceleration. Show your working.
1 mark: $F_{net} = 4000 - 1000$ · 1 mark: $F_{net} = 3000\ \text{N}$ · 1 mark: $a = F_{net}/m = 3000/1200$ · 1 mark: $a = 2.5\ \text{m s}^{-2}$
ApplyBand 5(4 marks) 2. A $1200\ \text{kg}$ car accelerates from rest to $25\ \text{m s}^{-1}$ in $10\ \text{s}$. Calculate the acceleration and the net driving force, and state the assumption you have made.
1 mark: $a = 2.5\ \text{m s}^{-2}$ · 1 mark: applies $F_{net} = ma$ · 1 mark: $F_{net} = 3000\ \text{N}$ · 1 mark: assumes uniform acceleration, and that this is the NET force
Dot point: Solve problems involving Newton's second law of motion using F = ma
AnalyseBand 6(3 marks) 3. In an investigation of the relationship between force, mass and acceleration, a student plots acceleration against net force for a trolley of fixed mass. Describe the expected graph and explain how its gradient gives the mass.
1 mark: a straight line through the origin · 1 mark: acceleration is directly proportional to net force · 1 mark: gradient $= 1/m$, so mass is the reciprocal of the gradient
Dot point: Conduct a scientific investigation to illustrate the relationship between force, mass and acceleration
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (4 marks): First find the net force along the direction of motion. The driving force is forward and the resistive force is backward, so they subtract: $F_{net} = 4000 - 1000 = 3000\ \text{N}$ forward. Then apply Newton's second law, $\vec{F}_{net} = m\vec{a}$, rearranged for acceleration: $a = \dfrac{F_{net}}{m} = \dfrac{3000}{1200} = 2.5\ \text{m s}^{-2}$. So the car accelerates at $2.5\ \text{m s}^{-2}$ in the forward direction.
Q2 (4 marks): The acceleration is $$a = \frac{v-u}{t} = \frac{25 - 0}{10} = 2.5\ \text{m s}^{-2}$$ Applying Newton’s second law, $$F_{net} = ma = 1200 \times 2.5 = 3000\ \text{N}$$ Two assumptions are involved. First, that the acceleration is uniform over the $10\ \text{s}$, which is what allows $a$ to be found from the overall change in velocity. Second, that $3000\ \text{N}$ is the NET force, not the force the engine produces: the engine must supply more than this to overcome friction and air resistance as well, so the actual driving force is larger.
Q3 (3 marks): Rearranging Newton’s second law as $a = F_{net}/m$ and comparing with $y = mx$ shows that plotting $a$ on the vertical axis against $F_{net}$ on the horizontal gives a straight line passing through the origin, because zero net force must produce zero acceleration. The line being straight confirms that acceleration is directly proportional to net force when the mass is held constant. Its gradient is $1/m$, so the mass of the trolley system is found by taking the reciprocal of the gradient. Any systematic friction that has not been compensated shows up as a positive intercept on the force axis, since some force is then needed before the trolley moves at all.
You can now apply $\vec{F}_{net} = m\vec{a}$ to find force, mass or acceleration, design an investigation to verify it, and relate weight to mass. With vectors, kinematics and Newton's three laws established, the focus area continues into friction, connected bodies and inclined planes.
↩ Back to focus-area hubAt the start you compared a $1\ \text{kg}$ and a $2\ \text{kg}$ trolley under the same push. The $1\ \text{kg}$ trolley accelerates twice as fast, because $a = F/m$ and halving the mass doubles the acceleration. The loaded shopping trolley barely moves for exactly this reason: more mass, less acceleration for the same force. Revisit Card 1 if the inverse relationship still surprises you.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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