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Unit 2.1 | Understanding and Applying Newton’s Law in Depth

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Jason Kuma

Writer | Coach | Builder | Fremont, CA

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This first lesson is important to building the foundation of your Physics journey!

Newton’s Law’s may seem simple, but there is a lot to unpack. You may have to read this lesson a few times to fully understand some of the concepts.

This course article contains videos that can only be accessed once enrolled in the Learn AP Physics from Scratch Course.

Unit 2 Breakdown

You are on Lesson 1 of 8

  1. Unit 2.1 | Understanding and Applying Newton’s Law in Depth [Current Lesson]
  2. Unit 2.2 | Common Linear Forces, Equations, and Misconceptions
  3. Unit 2.3 | Drawing and Understanding Force Body Diagrams
  4. Unit 2.4 | Deriving Equations to Solve Linear Force Problems
  5. Unit 2.5 | Advanced Force Problems – Tension and Elevators
  6. Unit 2.6 | Advanced Force Problems – Ramps (Inclines)
  7. Unit 2.7 | Advanced Force Problems – Pulley System
  8. Unit 2.8 | Advanced Force Problems – Multi-Body System

In this lesson: 

  • You will learn about Newton’s three laws in depth
  • Learn how you can apply the laws to everyday phenomenon
  • Cover fun, counterintuitive examples of Newton’s laws
  • Expand our logic and reasoning skills

The Importance of Newton’s Laws

Chances are you have already covered Newton’s laws in elementary school. 

While these law may seem simple on the surface, they can explain a wide range of everyday Phenomenon. 

This will be the first Unit in which you will make incredible discoveries about the nature of our universe.

Get ready for countless “ah-ha” moments.

Here are some quick warm up question for you to think about. I’ll explain the answers in depth later on.

  1. If you were to drop a 1kg bowling ball and a .2 kg feather at the same time, which would hit the ground first?
  2. While traveling in a car at 50 km/hr, you throw a ball small ball up. Will the ball land back in your hand, in front of you, or behind you?
  3. In a collision between a big truck and a small car, which vehicle hit the other with more force?

Hint: The answer are not what you might think!

LRN Advanced Application of Newton’s Laws

In the video I summarize the three laws and talk about how to apply them to everyday Phenomenon. For more practice, and guidance be sure to throughly read this lesson!

I highly recommend reading the article below before watching this video.

Locked Video Content

Newton’s First Law

All object’s have inertia.

Inertia is an object’s resistance to any changes in its motion. An object at rest will want to stay rest. And an object in motion will stay in motion. 

This only works in the absence of forces. Such as in outer space, where there is no forces.  

Inertia

Inertia is just a fancy word for mass.

So the bigger the object, the more inertia it has. More inertia, means a greater resistance to a change in the object’s motion. 

This makes sense, since a heavier truck (more inertia) is much harder to stop than a bicycle (less inertia).  

Applying the 1st Law

Scenario #1: You’re traveling 100 km/hr in your car, when you suddenly slam on your brakes. Your seatbelt holds you back, but why does your body feels like it is flying forward?

Click to see explanation

Inertia. An object in motion wants to stay in motion. In this case you are the object and you want to keep moving at 100 km/hr. The car can stop because of it massive brakes. But what’s slows you down? Without your seatbelt you would go flying out the window at 100 km/hr! 

Scenario #2: You are traveling 100 km/hr in your car, when your throw a tennis ball directly upwards. You’ll see that the ball will land back perfectly in your hand. Explain why this happens.

Click to see explanation

Inertia! When you throw the ball upwards, it wants to continue moving forward with you at 100 kh/hr. So even while the ball is in the air it is moving forward at the same speed you are. Therefore, it will land back in your hand!

Newton’s Second Law

[katex] F_{net} = ma [/katex]

Simply put, a force ([katex]F[/katex]) acting on an mass ([katex]m[/katex]) will cause an acceleration ([katex]a[/katex]).

Remember, acceleration simply means a change in speed or direction.

This equation is incredibly powerful and versatile. We use it throughout Physics to derive complex equations from scratch starting in Lesson 2.4.

Force vs Net Force

A force is singular. It is ONE particular force.  A net force is the SUM of all forces acting vertically or horizontally. 

For example, think about a force acting on you right now. One force would be your weight directed downwards. Your weight force would imply an acceleration ( F = ma ), however you are not accelerating.

Why? There is a force acting opposite to your weight. We call this the “normal force.” The Normal force cancels out your weight force, thus there is ZERO net force acting on you. And according to Fnet = ma, zero net force means zero acceleration, hence why you are not moving!

FBDs

Look at the force body diagram (FBD) below to see how the two forces acting on you cancel out. We’ll cover FBDs in depth later in Lesson 2.3.

Applying the Second Law

Give the conceptual questions below a shot. They require a bit of critical thinking 

  1. If an object is moving, is it possible for the net force acting on it to be zero?
  2. If the acceleration of an object is zero, are no forces acting on it? Explain.
  3. Only one force acts on an object. Can the object have zero acceleration? Can it have zero velocity? Explain.
  4. Why do you push down harder on the pedals of a bicycle when first starting out than when moving at constant speed? And why do you need to pedal at all when cycling at constant speed?

Newton’s Third Law

This law is one of the most counter-intuitive principles in Physics. But’s it my favorite! You’ll see why in a second. 

The third law states that every action has an equal but opposite reaction. 

[katex] F_{action} = -F_{reaction} [/katex]

Try this!

Push on your desk. Why doesn’t your hand move through the desk?

According to Newton’s third law, your desk is pushing back up at you with the same force!

[katex] F_{push} = -F_{desk-pushing} [/katex]

If you kick a soccer ball. What is the equal but opposite force? It’s the soccer ball kicking you back!

[katex] F_{kick} = -F_{ball-kicking-you} [/katex]

These are what we call “action and reaction pairs.” 

Action and Reaction Pairs

To find the action and reaction pair, simply then switch the two things involved in the action. Look at the example below.

You push a box → Now swap the two → The box pushes you.

It’s as simple as that!

Give it a try

State the reaction to each action:

A baseball player hits a ball with a bat.

Click to see the reaction force

The bat hits the balls.

An archer pulls back the string of a bow

Click to see the reaction force

The string pulls back on the archer

A big truck collides with the small car.

Click to see the reaction force

The small car collides with the big truck. (Yes, the force is equal and opposite!)

How do things move?

After solving the last problem you might wonder:

“How can the big truck and the small car exert the same force on each other? And if they did wouldn’t the forces cancel out and there would be no moment?”

How would anything, like the soccer ball, move if the force you apply is exactly the same force applied back?

Let’s dive in.

Kicking a Ball

If you kick a soccer ball with a force of 10 Newtons, then the soccer ball will kick you back with a force of 10 Newtons. The two kicking forces will cancel out (zero net force).

Zero net force means zero acceleration.

But wait a minute. When you kick a soccer ball, we know the ball will move. So what’s happening? 

Although the forces are the same the masses of the ball and person kicking are very different. This means the lighter object (the ball in this case) will accelerate further.

Kicking a Ball (With Numbers)

Let’s say the mass of the person is 70kg, while the mass of the ball of is .5kg. Since the force on BOTH is 10 Newtons lets find the acceleration of each applying F = ma.

The acceleration of the person → 10 N = (70kg) x apersonaperson = .14 m/s2.

The acceleration of the ball → 10 N = (.5kg) x aballaball = 20 m/s2.

Therefore, what we are really seeing is not a greater force on the ball, but a greater acceleration of ball due to it’s smaller mass.

More Practice

Let’s apply Newton’s 3 Laws to some everyday situations. 

If you can answer all these correctly you are ready to move on to the next lesson!

  1. According to Newton’s third law, each team in a tug of war pulls with equal force on the other team. What, then, determines which team will win?
  2. When a golf ball is dropped to the pavement, it bounces back up. Is a force needed to make it bounce back up? If so, what exerts the force?
  3. How does a rocket continue moving upwards when it is not close to the surface of the earth?
  4. When you stand still on the ground, how large a force does the ground exert on you? Why doesn’t this force make you rise up into the air?
  5. Whiplash sometimes results from an automobile accident when the victim’s car is struck violently from the rear. Explain why the head of the victim seems to be thrown backward in this situation. Is it really?
  6. Why is the stopping distance of a truck much shorter than for a train going the same speed?
  7. If you walk along a log floating on a lake, why does the log move in the opposite direction?

Lesson 2.1 Recap

In this lessons you learned about Newton’s 3 laws in depth.

The first law covers inertia. The second law covers net forces. The third law covers action and reaction forces.

You figured out how to apply these laws to everyday situations, then applied them to various problems yourself.

Everything in this section was conceptual. Over the next few lessons we will begin to apply numbers and equations.

Lesson 2.2 Preview

In the next lesson you will cover common types of forces and their equations. This will help you to draw accurate force diagrams (covered in 2.3) and derive equations (covered in 2.4)

Picture of Jason Kuma
Jason Kuma

Writer | Coach | Builder | Fremont, CA

Units in AP Physics 1

Unit 1 – Linear Kinematics

Unit 2 – Linear Forces

Unit 3 – Circular Motion

Unit 4 – Energy 

Unit 5 – Momentum 

Unit 6 – Torque 

Unit 7 – Oscillations 

Unit 8 – Fluids

Reading Key

LRN
RE
PS
PQ
Black
White
Blue
Orange

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KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque and Rotations
\(p = mv\)\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)\(I = \sum mr^2\)
\(p_i = p_f\)\(L = I \cdot \omega\)
Simple Harmonic MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[katex]g[/katex]Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]Mass of the Sun
VariableSI Unit
[katex]s[/katex] (Displacement)[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)[katex]\text{kilograms (kg)}[/katex]
VariableDerived SI Unit
[katex]F[/katex] (Force)[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)[katex]\text{hertz (Hz)}[/katex]

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

  2. Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]

  3. Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]

  4. Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

[katex]10^{-12}[/katex]

Nano-

n

[katex]10^{-9}[/katex]

Micro-

µ

[katex]10^{-6}[/katex]

Milli-

m

[katex]10^{-3}[/katex]

Centi-

c

[katex]10^{-2}[/katex]

Deci-

d

[katex]10^{-1}[/katex]

(Base unit)

[katex]10^{0}[/katex]

Deca- or Deka-

da

[katex]10^{1}[/katex]

Hecto-

h

[katex]10^{2}[/katex]

Kilo-

k

[katex]10^{3}[/katex]

Mega-

M

[katex]10^{6}[/katex]

Giga-

G

[katex]10^{9}[/katex]

Tera-

T

[katex]10^{12}[/katex]

  1. 1. Some answers may vary by 1% due to rounding.
  2. Gravity values may differ: \(9.81 \, \text{m/s}^2\) or \(10 \, \text{m/s}^2\).
  3. Variables can be written differently. For example, initial velocity (\(v_i\)) may be \(u\), and displacement (\(\Delta x\)) may be \(s\).
  4. Bookmark questions you can’t solve to revisit them later
  5. 5. Seek help if you’re stuck. The sooner you understand, the better your chances on tests.

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