4.1 Introduction to force and laws of motion

Achievable AP Physics 1

4. Laws of motion

Our AP Physics 1 course is currently in development and is a work-in-progress.

Introduction to force and laws of motion

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This subchapter explores the core ideas behind forces and motion. You’ll work with three main topics:

Concept of force
Newton’s laws of motion
Applications of laws of motion

Newton’s first law: The law of inertia

Concept and explanation

Newton’s First Law says that an object will stay at rest, or keep moving in a straight line at constant speed, unless a net external force acts on it. This tendency to resist changes in motion is called inertia.

A useful way to state the law is: if the net force ( $F_{net}$ ) is zero, the object’s velocity doesn’t change.

$F_{net} = 0 ⟹ v = constant .$

This is why forces are needed to change motion (speed up, slow down, or turn), not to maintain constant velocity. The law is stated for inertial frames of reference, meaning frames that aren’t accelerating.

Illustrative example

Consider a hockey puck sliding on nearly frictionless ice. With friction close to zero, the puck keeps moving in a straight line at nearly constant speed after it’s pushed. It only changes its motion if another force acts on it (for example, a collision with a barrier).

Hockey puck sliding and colliding with a barrier

Sidenote

Common misconceptions

Misconception: An object must have a force acting on it to remain in motion.
Clarification: Inertia means that no net force is required to maintain constant velocity; a force is only necessary to change that velocity. No force is required to keep an object in motion (constant velocity)

Newton’s second law: The law of acceleration

Concept and explanation

Newton’s Second Law connects net force, mass, and acceleration:

$F_{net} = m a .$

Here,

$F_{net}$ is the vector sum of all forces acting on the object,
$m$ is the mass of the object, and
$a$ is the acceleration of the object.

This law tells you two key things:

For a fixed net force, a larger mass produces a smaller acceleration.
Because force and acceleration are vectors, direction matters just as much as magnitude.

Example problem 1

A $5.0 kg$ block is pulled across a horizontal frictionless surface by a force of $20 N$ acting at an angle of $3 0^{\circ}$ above the horizontal. Determine the acceleration of the block.

Solution:

(spoiler)

Resolve the applied force into components:

$F_{x} = 20 cos 3 0^{\circ} and F_{y} = 20 sin 3 0^{\circ} .$

Using $cos 3 0^{\circ} \approx 0.866$ and $sin 3 0^{\circ} = 0.5$ :

$F_{x} \approx 20 \times 0.866 = 17.32 N, F_{y} = 20 \times 0.5 = 10 N .$
Determine the net force in the horizontal direction:

The surface is frictionless, so there’s no horizontal friction force. The vertical forces balance (the normal force adjusts to account for the upward component $F_{y}$ ), so the only unbalanced force is horizontal:

$F_{net} = F_{x} \approx 17.32 N .$
Apply Newton’s Second Law:

$a = \frac{F _{net}}{m} = \frac{17.32 N}{5.0 kg} \approx 3.46 m/s^{2} .$

The block accelerates at approximately $3.46 m/s^{2}$ in the horizontal direction.

Newton’s third law: Action and reaction

Concept and explanation

Newton’s Third Law says that forces come in pairs: if object A exerts a force on object B, then object B exerts an equal-magnitude force in the opposite direction on object A.

Mathematically, if object A exerts a force ( $F_{AB}$ ) on object B, then object B exerts a force

$F_{BA} = - F_{AB}$

on object A. These two forces are equal in magnitude, opposite in direction, and (most importantly) they act on different objects.

Illustrative example

When you jump off a small boat, your legs push the boat backward. At the same time, the boat pushes you forward with an equal and opposite force. That’s why the boat moves in the opposite direction of your jump.

Action and reaction pair according to 3rd law

Example problem 2

Two ice skaters initially at rest push off from each other. Skater A has a mass of $50 kg$ and accelerates at $2.0 m/s^{2}$ to the right. Skater B has a mass of $70 kg$ . Determine the acceleration of Skater B and verify that the forces between them are equal in magnitude and opposite in direction.

Solution:

(spoiler)

Calculate the force associated with skater A’s acceleration:

By Newton’s Second Law,

$F = m_{A} a_{A} = 50 kg \times 2.0 m/s^{2} = 100 N .$
Apply Newton’s Third Law:

Skater B experiences an equal-magnitude force in the opposite direction, so the force on B is $100 N$ to the left.
Determine the acceleration of skater B:

Using Newton’s Second Law for skater B,

$a_{B} = \frac{F}{m _{B}} = \frac{100 N}{70 kg} \approx 1.43 m/s^{2} .$

Skater B accelerates at approximately $1.43 m/s^{2}$ to the left, and the forces between the skaters form an action-reaction pair of $100 N$ .

Example problem 3

A $3.0 kg$ hockey puck glides on a frictionless ice surface at a constant velocity of $4.0 m/s$ . No external forces act on the puck. Determine the puck’s acceleration and explain why its velocity remains constant.

Solution:

(spoiler)

Apply Newton’s First Law:

Since no net external force is acting on the puck,

$F_{net} = 0 .$

With $F_{net} = 0$ , the acceleration must be zero:

$a = 0 m/s^{2} .$
Explanation:

With zero net force, there’s no change in velocity. So the puck keeps moving at the same constant speed of $4.0 m/s$ .

So the puck’s acceleration is $0 m/s^{2}$ ; it maintains a constant velocity due to the absence of any net external force.

Concept of force\

Force: push or pull; causes changes in motion
Net force ( $F_{net}$ ): vector sum of all forces on an object
Inertia: resistance to changes in motion

Newton’s first law: The law of inertia\

Object remains at rest or constant velocity if $F_{net} = 0$
Inertia: no force needed to maintain constant velocity
Applies only in inertial (non-accelerating) frames

Newton’s second law: The law of acceleration\

$F_{net} = m a$ (force = mass × acceleration)
Acceleration direction matches net force direction
Larger mass → smaller acceleration for same force

Newton’s third law: Action and reaction\

Forces always occur in equal and opposite pairs: $F_{AB} = - F_{BA}$
Action-reaction forces act on different objects
Explains motion from interactions (e.g., jumping off a boat)

Applications of laws of motion\

Use component analysis for forces at angles
Zero net force → zero acceleration (constant velocity)
Identify which law applies and calculate net force for problem-solving

Next | 4.2 Free body diagrams and equilibrium

Introduction to force and laws of motion

This subchapter explores the core ideas behind forces and motion. You’ll work with three main topics:

Concept of force
Newton’s laws of motion
Applications of laws of motion

Newton’s first law: The law of inertia

Concept and explanation

A useful way to state the law is: if the net force ( $F_{net}$ ) is zero, the object’s velocity doesn’t change.

$F_{net} = 0 ⟹ v = constant .$

Illustrative example

Sidenote

Common misconceptions

Misconception: An object must have a force acting on it to remain in motion.
Clarification: Inertia means that no net force is required to maintain constant velocity; a force is only necessary to change that velocity. No force is required to keep an object in motion (constant velocity)

Newton’s second law: The law of acceleration

Concept and explanation

Newton’s Second Law connects net force, mass, and acceleration:

$F_{net} = m a .$

Here,

$F_{net}$ is the vector sum of all forces acting on the object,
$m$ is the mass of the object, and
$a$ is the acceleration of the object.

This law tells you two key things:

For a fixed net force, a larger mass produces a smaller acceleration.
Because force and acceleration are vectors, direction matters just as much as magnitude.

Example problem 1

A $5.0 kg$ block is pulled across a horizontal frictionless surface by a force of $20 N$ acting at an angle of $3 0^{\circ}$ above the horizontal. Determine the acceleration of the block.

Solution:

(spoiler)

Resolve the applied force into components:

$F_{x} = 20 cos 3 0^{\circ} and F_{y} = 20 sin 3 0^{\circ} .$

Using $cos 3 0^{\circ} \approx 0.866$ and $sin 3 0^{\circ} = 0.5$ :

$F_{x} \approx 20 \times 0.866 = 17.32 N, F_{y} = 20 \times 0.5 = 10 N .$
Determine the net force in the horizontal direction:

The surface is frictionless, so there’s no horizontal friction force. The vertical forces balance (the normal force adjusts to account for the upward component $F_{y}$ ), so the only unbalanced force is horizontal:

$F_{net} = F_{x} \approx 17.32 N .$
Apply Newton’s Second Law:

$a = \frac{F _{net}}{m} = \frac{17.32 N}{5.0 kg} \approx 3.46 m/s^{2} .$

The block accelerates at approximately $3.46 m/s^{2}$ in the horizontal direction.

Newton’s third law: Action and reaction

Concept and explanation

Newton’s Third Law says that forces come in pairs: if object A exerts a force on object B, then object B exerts an equal-magnitude force in the opposite direction on object A.

Mathematically, if object A exerts a force ( $F_{AB}$ ) on object B, then object B exerts a force

$F_{BA} = - F_{AB}$

on object A. These two forces are equal in magnitude, opposite in direction, and (most importantly) they act on different objects.

Illustrative example

Example problem 2

Solution:

(spoiler)

Calculate the force associated with skater A’s acceleration:

By Newton’s Second Law,

$F = m_{A} a_{A} = 50 kg \times 2.0 m/s^{2} = 100 N .$
Apply Newton’s Third Law:

Skater B experiences an equal-magnitude force in the opposite direction, so the force on B is $100 N$ to the left.
Determine the acceleration of skater B:

Using Newton’s Second Law for skater B,

$a_{B} = \frac{F}{m _{B}} = \frac{100 N}{70 kg} \approx 1.43 m/s^{2} .$

Skater B accelerates at approximately $1.43 m/s^{2}$ to the left, and the forces between the skaters form an action-reaction pair of $100 N$ .

Example problem 3

Solution:

(spoiler)

Apply Newton’s First Law:

Since no net external force is acting on the puck,

$F_{net} = 0 .$

With $F_{net} = 0$ , the acceleration must be zero:

$a = 0 m/s^{2} .$
Explanation:

With zero net force, there’s no change in velocity. So the puck keeps moving at the same constant speed of $4.0 m/s$ .

So the puck’s acceleration is $0 m/s^{2}$ ; it maintains a constant velocity due to the absence of any net external force.

Achievable AP Physics 1

4. Laws of motion

Our AP Physics 1 course is currently in development and is a work-in-progress.