krit.club logo

Forces and Motion - Newton’s Laws of Motion

Grade 9IB

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 continue to move at a constant velocity unless acted upon by a non-zero resultant force (∑F≠0\sum F \neq 0).

•

Inertia: The tendency of an object to resist changes in its state of motion. Inertia is directly proportional to the mass (mm) of the object.

•

Newton's Second Law: The acceleration (aa) of an object is directly proportional to the net force (FnetF_{net}) acting on it and inversely proportional to its mass (mm).

•

Newton's Third Law: For every action force, there is an equal and opposite reaction force. If object AA exerts a force F⃗AB\vec{F}_{AB} on object BB, then object BB exerts a force F⃗BA=−F⃗AB\vec{F}_{BA} = -\vec{F}_{AB} on object AA.

•

Weight (WW): The force of gravity acting on an object's mass, calculated using the acceleration due to gravity (g≈9.8 m/s2g \approx 9.8 \text{ m/s}^2 on Earth).

•

Equilibrium: A state where the resultant force is zero (∑F=0\sum F = 0), meaning the object is either at rest or moving with constant velocity (a=0 m/s2a = 0 \text{ m/s}^2).

📐Formulae

F=m⋅aF = m \cdot a

W=m⋅gW = m \cdot g

Fnet=∑F=Fforward−FfrictionF_{net} = \sum F = F_{forward} - F_{friction}

a=Fnetma = \frac{F_{net}}{m}

💡Examples

Problem 1:

A crate with a mass of 50 kg50\text{ kg} is pushed across a floor with a forward force of 200 N200\text{ N}. If the force of friction opposing the motion is 50 N50\text{ N}, calculate the acceleration of the crate.

Solution:

Fnet=200 N−50 N=150 NF_{net} = 200\text{ N} - 50\text{ N} = 150\text{ N}. Using a=Fnetma = \frac{F_{net}}{m}, we get a=150 N50 kg=3 m/s2a = \frac{150\text{ N}}{50\text{ kg}} = 3\text{ m/s}^2.

Explanation:

First, find the resultant (net) force by subtracting the frictional resistance from the applied force. Then, apply Newton's Second Law (F=maF=ma) to solve for acceleration.

Problem 2:

Calculate the weight of an astronaut with a mass of 80 kg80\text{ kg} on the Moon, where the gravitational field strength is gmoon=1.6 m/s2g_{moon} = 1.6\text{ m/s}^2.

Solution:

W=m⋅gmoon=80 kg×1.6 m/s2=128 NW = m \cdot g_{moon} = 80\text{ kg} \times 1.6\text{ m/s}^2 = 128\text{ N}.

Explanation:

Weight is a force and varies depending on the local gravity (gg). The mass remains constant (80 kg80\text{ kg}), but the force exerted by gravity changes based on the environment.

Problem 3:

A book rests on a table. The book exerts a downward force of 10 N10\text{ N} on the table. According to Newton's Third Law, what is the reaction force?

Solution:

The table exerts an upward normal force of 10 N10\text{ N} on the book.

Explanation:

Newton's Third Law pairs involve two different objects. If the 'Action' is Book pushes Table (10 N10\text{ N} down), the 'Reaction' must be Table pushes Book (10 N10\text{ N} up).