AP Physics

AP Physics 1 Lesson 1: Introduction to Physics and Kinematics

Welcome to AP Physics 1!

Hello and welcome to AP Physics 1! This is the first lesson in what I hope will be an exciting journey into the world of physics. Whether you're completely new to physics or have some background, we're going to start from the basics and build up your understanding. Don't worry—everyone starts somewhere, and by the end of this course, you'll be amazed at how much you've learned.

Today, we’re going to cover some foundational concepts that will serve as the building blocks for the rest of the course. Specifically, we’ll talk about what physics is, some basic tools we’ll use in this class, and start diving into Kinematics, which is the study of motion.

What is Physics?

At its core, physics is the study of matter, energy, and the interactions between them. It’s the branch of science that explains how things move, why they move, and what causes them to move in the first place. Physics can describe everything from the motion of a falling apple to the behavior of light, to the forces acting on planets in motion through space.

The goal of physics is not only to understand these phenomena, but to come up with mathematical models and laws that describe how the universe works. We’re interested in understanding how and why things behave the way they do, and physics gives us the tools to do that.

The Scientific Method and Problem Solving

Physics relies on the scientific method, a process for investigating and learning about the world around us. It follows these general steps:

1. Observe the phenomenon.

2. Ask a question based on your observation.

3. Form a hypothesis or a possible explanation.

4. Conduct experiments or gather data to test your hypothesis.

5. Analyze the data to see if the hypothesis holds true.

6. Draw conclusions and refine your hypothesis or theory as needed.

In AP Physics, you’ll be doing a lot of problem-solving. A major part of the scientific method is analyzing data and forming predictions, and we do this using mathematical models. Physics problems often require you to break down complex scenarios into simpler, more manageable parts. We’ll focus a lot on unit analysis, dimensional analysis, and understanding the relationships between variables.

Units and Measurements

Physics is a quantitative science, meaning that we often measure things in numerical terms. But to do so, we need to have consistent units of measurement.

In the United States, we often use the Imperial system (feet, pounds, seconds), but in science, the SI system (International System of Units) is the standard. We’ll primarily work in SI units, which are based on the metric system. Here are some key units you’ll use in AP Physics:

• Length: meter (m)

• Mass: kilogram (kg)

• Time: second (s)

• Velocity: meter per second (m/s)

• Acceleration: meter per second squared (m/s²)

• Force: Newton (N) (1 N = 1 kg·m/s²)

When solving physics problems, it’s essential to keep track of units and make sure they’re consistent. Often, we’ll need to convert between units, so make sure you're comfortable with conversion factors (e.g., 1 km = 1,000 meters).

Kinematics: The Study of Motion

Now, let's start with one of the most important areas of physics: kinematics. Kinematics is the study of motion without worrying about what causes it (that’s dynamics, which we’ll get to later). In kinematics, we focus on describing how objects move: their speed, their direction, their position, and how those things change over time.

To understand motion, we’ll need to understand a few key quantities:

1. Displacement (Δx): Displacement is the change in position of an object. It’s a vector, meaning it has both magnitude (how far) and direction. So if an object moves 5 meters to the east, the displacement is 5 meters east.

2. Velocity (v): Velocity is the rate of change of displacement. It tells you how fast something is moving and in which direction. It’s also a vector. The formula for average velocity is:

   Average velocity=DisplacementTime\text{Average velocity} = \frac{\text{Displacement}}{\text{Time}}Average velocity=TimeDisplacement​

   This tells you how much your position changes per unit of time.

3. Speed: Unlike velocity, speed is a scalar quantity, meaning it only has magnitude (no direction). Speed is how fast something is moving, regardless of direction. It’s just the distance traveled over time.

4. Acceleration (a): Acceleration is the rate of change of velocity. If an object’s velocity is increasing, it’s accelerating; if it’s decreasing, it’s decelerating. If an object’s velocity is constant, its acceleration is zero.

   The formula for average acceleration is:

   Average acceleration=Change in velocityTime interval\text{Average acceleration} = \frac{\text{Change in velocity}}{\text{Time interval}}Average acceleration=Time intervalChange in velocity​ a=vf−vita = \frac{v_f - v_i}{t}a=tvf​−vi​​

   Where:

   • vfv_fvf​ is the final velocity

   • viv_ivi​ is the initial velocity

   • ttt is the time over which the change occurs

Position vs. Time Graphs

A great way to visualize motion is through graphs. Let’s start with a position vs. time graph. In such a graph, the x-axis represents time, and the y-axis represents position.

• A horizontal line on the graph means the object is not moving (its position is constant).

• A diagonal line means the object is moving. The steeper the line, the faster the object is moving.

• The slope of the line represents the object's velocity. A steeper slope means a greater velocity.

For example, if you have a graph where the line is steeper for the first part and then flattens out, this indicates that the object was moving faster at first and then slowed down.

Velocity vs. Time Graphs

In a velocity vs. time graph, the x-axis is time, and the y-axis is velocity. The key features of these graphs are:

• A horizontal line indicates constant velocity (no acceleration).

• A positive slope means the object is accelerating (speeding up), and a negative slope means the object is decelerating (slowing down).

• The area under the curve represents the object’s displacement over time. If the velocity is constant, the area under the curve is simply the velocity multiplied by time.

Equations of Motion

Now, let’s introduce the key equations of motion that will help us solve kinematics problems. These equations relate displacement, velocity, acceleration, and time. They are derived from the basic definitions of velocity and acceleration, and they assume constant acceleration:

1. First Equation of Motion:

   vf=vi+atv_f = v_i + a tvf​=vi​+at

   Where:

   • vfv_fvf​ = final velocity

   • viv_ivi​ = initial velocity

   • aaa = acceleration

   • ttt = time

2. Second Equation of Motion:

   Δx=vit+12at2\Delta x = v_i t + \frac{1}{2} a t^2Δx=vi​t+21​at2

   Where:

   • Δx\Delta xΔx = displacement

   • viv_ivi​ = initial velocity

   • aaa = acceleration

   • ttt = time

3. Third Equation of Motion:

   vf2=vi2+2aΔxv_f^2 = v_i^2 + 2a \Delta xvf2​=vi2​+2aΔx

   Where:

   • vfv_fvf​ = final velocity

   • viv_ivi​ = initial velocity

   • aaa = acceleration

   • Δx\Delta xΔx = displacement

Example Problem: Free Fall

To see how all of this works, let’s do an example problem. Suppose an object is dropped from a height of 20 meters. We can use the equations of motion to find out how long it will take for the object to hit the ground, as well as its final velocity.

1. We know that the initial velocity viv_ivi​ is 0 m/s (because it’s dropped, not thrown).

2. We know the displacement is Δx=−20\Delta x = -20Δx=−20 meters (it falls downward).

3. The acceleration due to gravity, a=−9.8 m/s2a = -9.8 \, \text{m/s}^2a=−9.8m/s2.

We can use the second equation of motion to find the time it takes to fall:

Δx=vit+12at2\Delta x = v_i t + \frac{1}{2} a t^2Δx=vi​t+21​at2

Substituting the values:

−20=0⋅t+12(−9.8)t2-20 = 0 \cdot t + \frac{1}{2} (-9.8) t^2−20=0⋅t+21​(−9.8)t2

Simplifying:

−20=−4.9t2-20 = -4.9 t^2−20=−4.9t2 t2=204.9≈4.08t^2 = \frac{20}{4.9} \approx 4.08t2=4.920​≈4.08 t≈2.02 secondst \approx 2.02 \, \text{seconds}t≈2.02seconds

Now, we can use the first equation of motion to find the final velocity:

vf=vi+atv_f = v_i + a tvf​=vi​+at vf=0+(−9.8)(2.02)≈−19.8 m/sv_f = 0 + (-9.8)(2.02) \approx -19.8 \, \text{m/s}vf​=0+(−9.8)(2.02)≈−19.8m/s

So, the object takes about 2.02 seconds to fall and hits the ground with a velocity of approximately -19.8 m/s (negative because it’s downward).

Conclusion

This concludes the first lesson in AP Physics 1. Today, we discussed what physics is, the basic units and measurements, and we began our exploration of kinematics. We learned how to describe motion using displacement, velocity, and acceleration, and we worked with some basic equations of motion to solve problems.

In future lessons, we’ll build on these concepts and dive deeper into more advanced topics like forces, momentum, energy, and circular motion.

Remember, physics is all about understanding the world around you through logical thinking and mathematics. Don’t get discouraged if you don’t understand everything right away—it takes time and practice.

I encourage you to keep practicing problem-solving, review the key concepts, and ask questions if you're stuck. Physics is like learning a new language, and the more you immerse yourself in it, the more fluent you'll become.

Homework:

• Review the definitions of displacement, velocity, and acceleration.

• Work through some practice problems from the textbook on kinematics (look for problems involving free fall, constant velocity, and constant acceleration).

• Try plotting some position vs. time graphs and velocity vs. time graphs for simple motions.

Looking forward to seeing you in the next lesson!

Welcome Back!

I hope you had a chance to review the first lesson on kinematics. Now that we have a solid foundation on basic motion, we’re going to start expanding our understanding to include vectors and scalars, and apply our knowledge of motion to two-dimensional problems. By the end of this lesson, you’ll be comfortable dealing with vectors in multiple directions, and be able to analyze motion in two dimensions using kinematic equations.

Scalars vs. Vectors

Before we dive deeper into two-dimensional motion, let’s take a moment to review the difference between scalars and vectors, which are the two primary types of quantities we deal with in physics.

1. Scalars:

   • A scalar is a quantity that has only magnitude (size or amount) and no direction.

   • Examples of scalar quantities include distance, time, mass, temperature, and speed.

   • You can perform regular arithmetic (addition, subtraction, multiplication, etc.) with scalars.

2. Vectors:

   • A vector is a quantity that has both magnitude and direction.

   • Examples of vector quantities include displacement, velocity, force, and acceleration.

   • To work with vectors, we need to consider both the magnitude (how much) and the direction (which way).

   Vector notation:

   • Vectors are often written in boldface (e.g., v for velocity), or with an arrow above them (e.g., v⃗\vec{v}v).

   • The direction is often indicated by an angle or by using compass directions (e.g., north, east, south, west, etc.).

Adding and Subtracting Vectors

When dealing with vectors, we can add or subtract them, but this isn’t as simple as just adding or subtracting numbers. We need to use vector addition techniques.

1. Tip-to-Tail Method:

   • To add two vectors, place the tail of the second vector at the tip of the first vector. The resultant vector (the vector sum) is the vector that goes from the tail of the first vector to the tip of the second vector.

2. Component Method:

   • A more common and useful approach is to break vectors into their components along the x-axis and y-axis (horizontal and vertical directions). Once the vectors are broken into components, you can add the components separately in the x-direction and y-direction.

Example 1: Adding Vectors Using Components

Suppose you have two vectors:

• Vector A = 5 meters at 30° east of north.

• Vector B = 10 meters at 60° east of north.

To add them, we first need to break them into components. We’ll use trigonometry to do this, converting each vector into horizontal (x) and vertical (y) components.

For Vector A:

• Ax=5⋅cos⁡(30∘)=5⋅0.866=4.33 mA_x = 5 \cdot \cos(30^\circ) = 5 \cdot 0.866 = 4.33 \, \text{m}Ax​=5⋅cos(30∘)=5⋅0.866=4.33m

• Ay=5⋅sin⁡(30∘)=5⋅0.5=2.5 mA_y = 5 \cdot \sin(30^\circ) = 5 \cdot 0.5 = 2.5 \, \text{m}Ay​=5⋅sin(30∘)=5⋅0.5=2.5m

For Vector B:

• Bx=10⋅cos⁡(60∘)=10⋅0.5=5 mB_x = 10 \cdot \cos(60^\circ) = 10 \cdot 0.5 = 5 \, \text{m}Bx​=10⋅cos(60∘)=10⋅0.5=5m

• By=10⋅sin⁡(60∘)=10⋅0.866=8.66 mB_y = 10 \cdot \sin(60^\circ) = 10 \cdot 0.866 = 8.66 \, \text{m}By​=10⋅sin(60∘)=10⋅0.866=8.66m

Now, add the components:

• Total x-component: Ax+Bx=4.33+5=9.33 mA_x + B_x = 4.33 + 5 = 9.33 \, \text{m}Ax​+Bx​=4.33+5=9.33m

• Total y-component: Ay+By=2.5+8.66=11.16 mA_y + B_y = 2.5 + 8.66 = 11.16 \, \text{m}Ay​+By​=2.5+8.66=11.16m

Now, the total vector’s magnitude and direction can be found:

• Magnitude: Magnitude=(9.33)2+(11.16)2=87.07+124.56=211.63≈14.54 m\text{Magnitude} = \sqrt{(9.33)^2 + (11.16)^2} = \sqrt{87.07 + 124.56} = \sqrt{211.63} \approx 14.54 \, \text{m}Magnitude=(9.33)2+(11.16)2​=87.07+124.56​=211.63​≈14.54m

• Direction: θ=tan⁡−1(11.169.33)≈tan⁡−1(1.2)≈50.2∘\theta = \tan^{-1}\left(\frac{11.16}{9.33}\right) \approx \tan^{-1}(1.2) \approx 50.2^\circθ=tan−1(9.3311.16​)≈tan−1(1.2)≈50.2∘

So, the resultant vector has a magnitude of approximately 14.54 meters and is directed at an angle of about 50.2° east of north.

Motion in Two Dimensions

Now that we understand how to add vectors, let’s apply this knowledge to motion in two dimensions. Two-dimensional motion is just motion along both the x-axis and the y-axis at the same time. A good example of two-dimensional motion is projectile motion, such as a ball being kicked in the air.

In two-dimensional motion, the key idea is that the motion in the x-direction and the motion in the y-direction are completely independent of each other, even though they happen simultaneously.

1. Horizontal Motion:

   • If we assume no air resistance, the velocity in the x-direction (horizontal velocity) is constant. The acceleration in the x-direction is zero.

   • The object moves with a constant velocity in the x-direction, so we can apply the same kinematic equations for constant velocity to the horizontal motion.

2. Vertical Motion:

   • The vertical motion is influenced by gravity, which means there’s a constant downward acceleration of g=9.8 m/s2g = 9.8 \, \text{m/s}^2g=9.8m/s2.

   • We apply the kinematic equations for constant acceleration to the vertical motion, where the acceleration is equal to ggg (downward).

Projectile Motion

Projectile motion is a special type of two-dimensional motion where an object moves through the air under the influence of gravity. We break this motion into horizontal and vertical components.

• Horizontal component:

  • The horizontal velocity remains constant (since no horizontal forces act on the object, assuming no air resistance).

  • The horizontal displacement after a time ttt is given by: x=vix⋅tx = v_{i_x} \cdot tx=vix​​⋅t

• Vertical component:

  • The vertical velocity changes due to the acceleration due to gravity.

  • The vertical displacement after a time ttt is given by the kinematic equation: y=viy⋅t+12(−g)t2y = v_{i_y} \cdot t + \frac{1}{2} (-g) t^2y=viy​​⋅t+21​(−g)t2

Example 2: Projectile Motion

Let’s say a ball is kicked at an angle of 30° with an initial velocity of 20 m/s. We want to find how far the ball travels before hitting the ground (its range).

1. Break the velocity into components:

   • Horizontal velocity: vix=20⋅cos⁡(30∘)=20⋅0.866=17.32 m/sv_{i_x} = 20 \cdot \cos(30^\circ) = 20 \cdot 0.866 = 17.32 \, \text{m/s}vix​​=20⋅cos(30∘)=20⋅0.866=17.32m/s

   • Vertical velocity: viy=20⋅sin⁡(30∘)=20⋅0.5=10 m/sv_{i_y} = 20 \cdot \sin(30^\circ) = 20 \cdot 0.5 = 10 \, \text{m/s}viy​​=20⋅sin(30∘)=20⋅0.5=10m/s

2. Find the time the projectile is in the air. To do this, we use the vertical motion equation. The ball will land when its vertical displacement is zero (y=0y = 0y=0):

   y=viy⋅t+12(−g)t2y = v_{i_y} \cdot t + \frac{1}{2} (-g) t^2y=viy​​⋅t+21​(−g)t2

   Substituting the known values:

   0=10⋅t−12(9.8)⋅t20 = 10 \cdot t - \frac{1}{2} (9.8) \cdot t^20=10⋅t−21​(9.8)⋅t2

   Factoring:

   t(10−4.9t)=0t(10 - 4.9 t) = 0t(10−4.9t)=0

   So, t=0t = 0t=0 (the starting point) or t=104.9≈2.04 secondst = \frac{10}{4.9} \approx 2.04 \, \text{seconds}t=4.910​≈2.04seconds.

3. Find the horizontal displacement (range): Since horizontal velocity is constant, we can use the formula:

   x=vix⋅t=17.32⋅2.04≈35.3 metersx = v_{i_x} \cdot t = 17.32 \cdot 2.04 \approx 35.3 \, \text{meters}x=vix​​⋅t=17.32⋅2.04≈35.3meters

So, the ball will travel approximately 35.3 meters before hitting the ground.

Conclusion

In this lesson, we expanded on the concept of motion by introducing vectors and how to add them. We learned that vectors have both magnitude and direction, and we saw how to add them using components. We then applied this to two-dimensional motion, specifically projectile motion. We broke the motion into horizontal and vertical components, each of which follows its own set of rules, and we solved problems involving projectiles.

In the next lesson, we will dive deeper into forces and explore the role they play in causing motion. This will lay the groundwork for understanding Newton’s laws, which are essential for understanding how objects move in the real world.

Homework:

1. Practice solving problems with vectors and components. For example:

   • Add two vectors at different angles.

   • Break a vector into components and find its resultant.

2. Work on some projectile motion problems. You can try varying the initial velocity or angle and see how the range and time of flight change.

Welcome to Lesson 3!

Now that we’ve covered the basics of kinematics and two-dimensional motion, it’s time to introduce one of the most important concepts in physics: forces. Specifically, we’re going to focus on Newton's Laws of Motion, which describe the relationship between the motion of an object and the forces acting on it. These laws form the foundation for understanding how objects move (or don't move) under the influence of forces.

By the end of this lesson, you’ll understand how Newton’s three laws describe motion, how to identify forces acting on an object, and how to apply these laws to solve problems involving forces and acceleration.

Newton’s First Law of Motion: The Law of Inertia

Newton’s First Law states:

This law describes an object’s inertia, which is its resistance to changes in its state of motion. Inertia is the property of an object that makes it want to keep doing what it’s doing—either staying at rest or moving with a constant velocity.

Key Concepts:

• Rest: If an object is not moving, it will stay at rest unless a force causes it to move.

• Motion: If an object is moving at a constant velocity, it will continue to move at that velocity unless something changes it (like a force).

• External Force: A force that comes from outside the object. If there’s no external force, the object won’t change its motion.

For example:

• A book sitting on a table stays at rest because there’s no external force acting on it to move it.

• A car driving at a constant speed on a flat road will continue moving at that speed unless something—like friction, air resistance, or the driver—causes it to slow down or stop.

In this context, force is the reason why things change their motion. Without a force, objects don’t accelerate; they simply keep doing what they were doing.

Newton’s Second Law of Motion: Force and Acceleration

Newton’s Second Law gives us a relationship between force, mass, and acceleration. It states:

In equation form:

F⃗=m⋅a⃗\vec{F} = m \cdot \vec{a}F=m⋅a

Where:

• F⃗\vec{F}F is the net force acting on the object (in Newtons, N),

• mmm is the mass of the object (in kilograms, kg),

• a⃗\vec{a}a is the acceleration of the object (in meters per second squared, m/s²).

Key Concepts:

• Acceleration: The rate at which an object changes its velocity. If there’s a net force acting on an object, it will accelerate in the direction of that force.

• Mass: A measure of an object’s resistance to acceleration (inertia). The larger the mass, the less it accelerates when a given force is applied.

• Net Force: The sum of all the forces acting on an object. If multiple forces act on an object in different directions, you need to combine them to find the net force.

Let’s break it down:

• If you push a car and a bike with the same force, the bike will accelerate much more because it has less mass. The car, with more mass, will accelerate less for the same applied force.

• Conversely, if you apply the same force to both objects, the more massive car will have a smaller acceleration than the less massive bike.

Example 1: Applying Newton’s Second Law

Suppose you apply a force of 50 N to a 10 kg object. To find the acceleration, we can use Newton's second law:

F⃗=m⋅a⃗\vec{F} = m \cdot \vec{a}F=m⋅a

Rearranging the formula to solve for acceleration:

a⃗=F⃗m\vec{a} = \frac{\vec{F}}{m}a=mF​

Substitute the given values:

a⃗=50 N10 kg=5 m/s2\vec{a} = \frac{50 \, \text{N}}{10 \, \text{kg}} = 5 \, \text{m/s}^2a=10kg50N​=5m/s2

So, the acceleration of the object is 5 meters per second squared.

Newton’s Third Law of Motion: Action and Reaction

Newton’s Third Law states:

This means that if object A exerts a force on object B, object B exerts a force of equal magnitude but in the opposite direction on object A.

Key Concepts:

• Action and Reaction Forces: These are two forces that are equal in magnitude and opposite in direction, and they act on different objects. Even though they are equal in size, they don’t cancel each other out because they act on different objects.

For example:

• When you push against a wall, the wall pushes back on you with the same force, but in the opposite direction.

• A rocket in space expels gas out the back (action), and the rocket moves forward (reaction).

• When you jump off a boat, the boat moves backward (this is why boats appear to "drift" when you jump off them).

Even though the forces are equal and opposite, the motion they cause can be very different depending on the masses involved. For example, when you push on a small object (like a ball), it accelerates much more than a massive object like a car, even though the forces are equal.

Forces and Free-Body Diagrams

In problems involving Newton’s Laws, we often draw free-body diagrams to help visualize all the forces acting on an object. A free-body diagram shows an object as a simple point or box, with arrows representing the forces acting on it. The length of each arrow represents the magnitude of the force, and the direction of the arrow represents the direction of the force.

Types of Forces:

• Gravitational Force (Weight): The force due to gravity pulling an object downward. Fg=m⋅gF_g = m \cdot gFg​=m⋅g, where ggg is the acceleration due to gravity (approximately 9.8 m/s²).

• Normal Force: The force exerted by a surface to support the weight of an object resting on it. It’s perpendicular to the surface.

• Frictional Force: The force that resists motion between two surfaces in contact. It acts opposite to the direction of motion.

• Tension Force: The force transmitted through a string, rope, or cable when it is pulled tight.

• Applied Force: Any force that is applied to an object by another object.

Example 2: Free-Body Diagram and Solving for Acceleration

Let’s consider a box of mass 5 kg5 \, \text{kg}5kg resting on a horizontal surface. You apply a horizontal force of 20 N20 \, \text{N}20N to the box. We’re going to solve for the acceleration of the box.

Step 1: Draw a Free-Body Diagram

1. Gravitational Force: The box experiences a downward gravitational force: Fg=m⋅g=5 kg⋅9.8 m/s2=49 NF_g = m \cdot g = 5 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 49 \, \text{N}Fg​=m⋅g=5kg⋅9.8m/s2=49N downward.

2. Normal Force: The surface exerts an upward normal force equal in magnitude to the gravitational force: FN=49 NF_N = 49 \, \text{N}FN​=49N upward.

3. Applied Force: You apply a horizontal force of 20 N20 \, \text{N}20N to the box to the right.

4. Frictional Force: The frictional force opposes the motion. Let’s assume it’s 5 N5 \, \text{N}5N to the left.

Step 2: Apply Newton’s Second Law

In the horizontal direction, the net force is:

Fnet, x=Fapplied−Ffriction=20 N−5 N=15 NF_{\text{net, x}} = F_{\text{applied}} - F_{\text{friction}} = 20 \, \text{N} - 5 \, \text{N} = 15 \, \text{N}Fnet, x​=Fapplied​−Ffriction​=20N−5N=15N

Using Newton’s Second Law:

Fnet=m⋅aF_{\text{net}} = m \cdot aFnet​=m⋅a

Solving for acceleration:

a=Fnetm=15 N5 kg=3 m/s2a = \frac{F_{\text{net}}}{m} = \frac{15 \, \text{N}}{5 \, \text{kg}} = 3 \, \text{m/s}^2a=mFnet​​=5kg15N​=3m/s2

So, the box accelerates at 3 m/s² to the right.

Conclusion

In today’s lesson, we explored Newton's Three Laws of Motion:

1. First Law (Inertia): Objects don’t change their motion unless acted on by a net force.

2. Second Law (Force and Acceleration): The acceleration of an object depends on the net force acting on it and its mass.

3. Third Law (Action and Reaction): For every action, there is an equal and opposite reaction.

We also learned how to draw free-body diagrams to visualize the forces acting on an object and used Newton’s Second Law to solve for acceleration. Forces are at the core of understanding how objects move, and by applying Newton’s Laws, we can predict the motion of objects in a variety of scenarios.

In the next lesson, we will explore friction, circular motion, and how forces act in specific situations like inclined planes.

Homework:

1. Draw free-body diagrams for various objects and solve for acceleration using Newton's Second Law.

2. Solve problems involving forces on an object, including problems with friction, tension, and gravitational force.