Newton’s Laws of Motion: High School Physics Study Guide | Tutolia Study Library

Master Newton’s three laws of motion with clear explanations, formulas, real-world examples, and practice problems.

Subject: Physics | Grade Level: Grade 10-12 | Curriculum: CAIE / IB / National

This study guide explains the fundamental principles of classical mechanics: Newton's Three Laws of Motion. Understanding these laws is essential for solving physics problems involving forces, acceleration, and mass.

1. Newton's First Law (Law of Inertia)

"An object at rest remains at rest, and an object in motion continues in motion with a constant velocity, unless acted upon by a net external force."

• Inertia is the resistance of any physical object to any change in its velocity.
• Example: When a bus suddenly brakes, passengers lurch forward because their bodies want to maintain their forward motion.

2. Net Force Equation (\(F = ma\))

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

The equation is:

\[F = ma\]

Where:

• \(F\) is the net force acting on the object (measured in Newtons, N).
• \(m\) is the mass of the object (measured in kilograms, kg).
• \(a\) is the acceleration of the object (measured in \(m/s^2\)).

Sample Problem

A box with a mass of \(5 ext{ kg}\) is pushed with a net force of \(20 ext{ N}\). What is its acceleration?

Solution:

\[a = rac{F}{m} = rac{20 ext{ N}}{5 ext{ kg}} = 4 ext{ m/s}^2\]

3. Newton's Third Law (Action-Reaction)

"For every action, there is an equal and opposite reaction."

• This means forces always occur in pairs. If Object A exerts a force on Object B, Object B exerts an equal and opposite force on Object A.
• Example: When you jump off a small boat, you push the boat backward (action), and the boat pushes you forward into the water (reaction).

4. Free-Body Diagrams (FBDs) & Resolving Forces

In Cambridge IGCSE/A-Level and IB Physics Paper 2, almost every mechanics problem begins with a Free-Body Diagram (FBD). An FBD isolates a single object as a point mass and represents all external forces acting on that object as vectors pointing away from the center:

• Weight ($W = mg$): Always acts vertically downward toward the center of the Earth ($g approx 9.81 ext{ m/s}^2$).
• Normal Contact Force ($F_N$ or $R$): Acts perpendicular ($90^circ$) to the surface of contact. On a horizontal surface with no vertical acceleration, $F_N = mg$. On an inclined plane at angle $ heta$, $F_N = mgcos heta$.
• Friction ($f$): Acts parallel to the surface, opposing the direction of motion (kinetic friction, $f_k = mu_k F_N$) or opposing the tendency to slip (static friction, $f_s le mu_s F_N$).
• Tension ($T$): Pulls along a taut rope, string, or cable away from the object.

5. Multi-Step Exam Worked Problems

Worked Problem 1: Block on an Inclined Plane with Friction

A $10 ext{ kg}$ crate slides down a rough ramp inclined at $30^circ$ to the horizontal. The coefficient of kinetic friction between the crate and the ramp is $mu_k = 0.20$. Take $g = 9.8 ext{ m/s}^2$. Calculate:

1. The normal reaction force $F_N$.
2. The kinetic friction force $f_k$.
3. The acceleration $a$ of the crate down the ramp.

Step-by-Step Solution:

1. Resolve weight perpendicular to the ramp:

[F_N = mgcos(30^circ) = 10 imes 9.8 imes 0.866 = 84.87 ext{ N}]

1. Calculate kinetic friction:

[f_k = mu_k F_N = 0.20 imes 84.87 ext{ N} = 16.97 ext{ N}]

1. Apply Newton's Second Law parallel to the ramp:

The component of weight pulling the crate down the slope is $mgsin(30^circ) = 10 imes 9.8 imes 0.5 = 49.0 ext{ N}$.

[F_{ ext{net}} = mgsin(30^circ) - f_k = 49.0 - 16.97 = 32.03 ext{ N}]

[a = rac{F_{ ext{net}}}{m} = rac{32.03 ext{ N}}{10 ext{ kg}} = 3.20 ext{ m/s}^2]

Worked Problem 2: Connected Masses (Atwood Machine / Pulley)

Two masses, $m_1 = 3 ext{ kg}$ and $m_2 = 5 ext{ kg}$, hang vertically on either side of a frictionless, massless pulley connected by a light inextensible string. Find the acceleration of the system and the tension $T$ in the string ($g = 9.8 ext{ m/s}^2$).

Step-by-Step Solution:

• For the heavier mass ($m_2 = 5 ext{ kg}$, accelerating downward): $m_2 g - T = m_2 a$
• For the lighter mass ($m_1 = 3 ext{ kg}$, accelerating upward): $T - m_1 g = m_1 a$
• Add the two equations to eliminate $T$:

[(m_2 - m_1)g = (m_1 + m_2)a implies (5 - 3)(9.8) = (3 + 5)a]

[19.6 = 8a implies a = 2.45 ext{ m/s}^2]

• Substitute $a = 2.45 ext{ m/s}^2$ back into the second equation to find Tension:

[T = m_1(g + a) = 3(9.8 + 2.45) = 36.75 ext{ N}]

6. Common Examiner Pitfalls to Avoid

1. Confusing Mass and Weight: Mass ($ ext{kg}$) is a scalar property of matter that remains constant everywhere in the universe. Weight ($ ext{N}$) is a vector gravitational force that changes with $g$.
2. Misidentifying Newton's Third Law Pairs: The Normal force ($F_N$) and Weight ($mg$) acting on a book resting on a table are NOT an action-reaction pair! They both act on the same object (the book) and balance each other by Newton's First Law. A true Newton's Third Law pair always acts on two different objects and involves the exact same type of force (e.g., Earth's gravity pulling the book down $leftrightarrow$ the book's gravity pulling the Earth up).
3. Assuming Zero Velocity Means Zero Acceleration: When a ball is thrown straight up, at its highest peak its instantaneous velocity is $v = 0 ext{ m/s}$, but the net force is still $mg$ downward, so its acceleration is still $9.81 ext{ m/s}^2$ downward.