Grade 11: Newton's Laws of Motion 🍎💥

Interactive Lesson • $\text{Inertia}$ • $\text{Force} = \text{Mass} \times \text{Acceleration}$ • $\text{Action}$-$\text{Reaction}$

The $\text{Foundation}$ $\text{of}$ $\text{Dynamics}$ 🧱

  $\text{Dynamics}$ is the study of why objects move. $\text{Sir}$ $\text{Isaac}$ $\text{Newton}$ formulated three $\text{Laws}$ of $\text{Motion}$ that describe the relationship between $\text{Force}$ ($\vec{F}$), $\text{Mass}$ ($m$), and $\text{Acceleration}$ ($\vec{a}$).

 
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The $\text{Three}$ $\text{Laws}$ $\text{of}$ $\text{Motion}$ 🌟

 
   

1. $\text{Law}$ $\text{of}$ $\text{Inertia}$

   

An object at $\text{rest}$ stays at $\text{rest}$, and an object in $\text{motion}$ stays in $\text{motion}$ with the $\text{same}$ $\text{speed}$ and in the $\text{same}$ $\text{direction}$ unless acted upon by a $\text{Net}$ $\text{External}$ $\text{Force}$ ($\vec{F}_{\text{net}}$).

   

Key concept: $\text{Inertia}$ (resistance to change in motion).

 
 
   

2. $\vec{F}_{\text{net}} = m\vec{a}$

   

The $\text{acceleration}$ ($\vec{a}$) of an object is $\text{directly}$ $\text{proportional}$ to the $\text{net}$ $\text{force}$ ($\vec{F}_{\text{net}}$) and $\text{inversely}$ $\text{proportional}$ to its $\text{mass}$ ($m$).

    $$\vec{F}_{\text{net}} = m\vec{a}$$    

The unit of $\text{Force}$ is the $\text{Newton}$ ($\text{N}$), where $1 \text{ N} = 1 \text{ kg} \cdot \text{m}/\text{s}^2$.

 
 
   

3. $\text{Action}$-$\text{Reaction}$

   

For every $\text{action}$ $\text{force}$, there is an $\text{equal}$ in $\text{magnitude}$ and $\text{opposite}$ in $\text{direction}$ $\text{reaction}$ $\text{force}$.

    $$\vec{F}_{\text{A on B}} = -\vec{F}_{\text{B on A}}$$    

Note: These forces always act on $\text{different}$ $\text{objects}$.

 

Interactive $\text{Law}$ $\text{II}$ $\text{Calculation}$ ($\vec{F}_{\text{net}} = m\vec{a}$) 📐

Use $\text{Newton's}$ $\text{Second}$ $\text{Law}$ to calculate the $\text{acceleration}$ of an object.

   

Problem:

   

A boy pushes a $\text{cart}$ with a total $\text{mass}$ of $\mathbf{20.0 \text{ kg}}$ with a $\text{net}$ $\text{force}$ of $\mathbf{40.0 \text{ N}}$ ($\text{forward}$). What is the $\text{cart}$'s $\text{acceleration}$?

   

$$\text{Calculation}: \vec{a} = \frac{\vec{F}_{\text{net}}}{m} = \frac{40.0 \text{ N}}{20.0 \text{ kg}} = ? \text{ m}/\text{s}^2$$

   
                $\text{m}/\text{s}^2$            
   

⚡ $\text{Newton's}$ $\text{Law}$ $\text{Application}$ $\text{Check}$!

Identify which of the $\text{Three}$ $\text{Laws}$ best explains the following scenarios.

 
When you jump, the $\text{Earth}$ $\text{pushes}$ $\text{up}$ on $\text{you}$ with the $\text{same}$ $\text{force}$ $\text{you}$ $\text{push}$ $\text{down}$ on $\text{it}$.
 
     
 
 

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