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Physics: Forces and Energy - Newton’s Laws of Motion

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Newton’s First Law (Law of Inertia): An object will remain at rest or move at a constant velocity in a straight line unless acted upon by an unbalanced external force. This tendency to resist change in motion is called inertia, which is directly proportional to the mass (mm).

Newton’s Second Law: The acceleration (aa) of an object is directly proportional to the net force (FF) acting on it and inversely proportional to its mass (mm). This is expressed as the vector sum of forces being equal to the product of mass and acceleration.

Newton’s Third Law: For every action, there is an equal and opposite reaction. If object AA exerts a force on object BB (FABF_{AB}), then object BB exerts a force of equal magnitude and opposite direction on object AA (FBA-F_{BA}).

Mass vs. Weight: Mass is the quantity of matter in an object (measured in kgkg), while Weight is the force of gravity acting on that mass (measured in NN). The relationship is defined by the local gravitational acceleration (g9.8 m/s2g \approx 9.8 \text{ m/s}^2 on Earth).

Balanced and Unbalanced Forces: When the net force F=0\sum F = 0, the forces are balanced and the object's velocity remains constant. When F0\sum F \neq 0, the forces are unbalanced, resulting in acceleration.

📐Formulae

F=m×aF = m \times a

W=m×gW = m \times g

Fnet=FappliedFfrictionF_{net} = F_{applied} - F_{friction}

a=vuta = \frac{v - u}{t}

💡Examples

Problem 1:

A toy car with a mass of 0.5 kg0.5 \text{ kg} is pushed with a net force of 2 N2 \text{ N}. Calculate the acceleration of the car.

Solution:

a=Fm=2 N0.5 kg=4 m/s2a = \frac{F}{m} = \frac{2 \text{ N}}{0.5 \text{ kg}} = 4 \text{ m/s}^2

Explanation:

According to Newton's Second Law (F=maF = ma), we rearrange the formula to solve for acceleration by dividing the force by the mass.

Problem 2:

An astronaut has a mass of 70 kg70 \text{ kg}. Calculate their weight on Earth where g=9.8 m/s2g = 9.8 \text{ m/s}^2 and on the Moon where g=1.6 m/s2g = 1.6 \text{ m/s}^2.

Solution:

Weight on Earth: W=70 kg×9.8 m/s2=686 NW = 70 \text{ kg} \times 9.8 \text{ m/s}^2 = 686 \text{ N}. Weight on Moon: W=70 kg×1.6 m/s2=112 NW = 70 \text{ kg} \times 1.6 \text{ m/s}^2 = 112 \text{ N}.

Explanation:

Mass remains constant regardless of location (70 kg70 \text{ kg}), but weight changes because it is a force dependent on the local gravitational field strength (gg).

Problem 3:

A book rests on a table. If the book exerts a downward force of 15 N15 \text{ N} on the table due to gravity, what is the magnitude and direction of the force exerted by the table on the book?

Solution:

The table exerts an upward force (Normal Force) of 15 N15 \text{ N} on the book.

Explanation:

Based on Newton's Third Law, the action-reaction pair consists of the book pushing down on the table and the table pushing up on the book with an equal magnitude of 15 N15 \text{ N}.

Newton’s Laws of Motion - Revision Notes & Key Formulas | IB Grade 8 Science