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Newton's Laws of Motion: First and Second Law

Jane Doe

Jane Doe

10 min read

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Study Guide Overview

This study guide covers forces and motion in AP Physics C: Mechanics. Key topics include Newton's Laws, force interactions, inertia, net force, action-reaction, circular motion, free body diagrams (FBDs), and friction (static and kinetic). It emphasizes problem-solving using FBDs, vector components, and calculus applications. Practice questions and exam tips are also provided.

AP Physics C: Mechanics - Forces and Motion Study Guide 🚀

Welcome to your ultimate guide for mastering forces and motion! This guide is designed to be your go-to resource, especially the night before the exam. Let's get started!

Overview

Time to dive into the heart of mechanics: forces! We'll explore how forces cause motion, stop motion, and everything in between. Get ready to understand Newton's laws, which are the foundation of classical mechanics. This unit is crucial for success in the rest of the course, so let's make sure you've got it down!

Big Ideas

  • Force Interactions: Forces define how objects or systems interact. Think of it as the push or pull between objects.
  • Inertia: Why does coffee keep swirling after you stop stirring? That's inertia in action! ☕
  • Net Force: Why doesn't a car move when you push it like a shopping cart? It's all about the net force.
  • Action-Reaction: Why do you push backward to move forward on a skateboard? Newton's third law!
  • Circular Motion: Why does the sun set in nearly the same place each day? Circular motion plays a role.

Exam Impact

This unit accounts for 17-23% of the AP exam! That's a significant chunk, so let's make sure you're well-prepared. You should expect around 24 class periods (45 minutes each) to cover this material. The AP Classroom has 25 MCQs and 1 FRQ for practice.

Newton's Laws of Motion: First and Second Law 🥏

Let's start with Newton's first two laws. These are the foundation for understanding how forces affect motion.

Newton's First Law of Motion: A body at rest will remain at rest unless an outside force acts on it, and a body in motion at a constant velocity will remain in motion in a straight line unless acted upon by a net external force.

Newton's Second Law of Motion: If an unbalanced force acts on a body, that body will experience acceleration (or deceleration).

Key Concept

Newton's First Law introduces inertia, the tendency of an object to resist changes in its motion. Newton's Second Law is all about calculating net force using Free Body Diagrams (FBDs).

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Free Body Diagrams (FBDs)

FBDs are your best friend in this unit! They help you visualize all the forces acting on an object. Remember to always consider direction and angles!

Here's a simple FBD example:

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For the box at rest, we have:

  • Gravitational Force (Weight): FgF_g acting downwards.
  • Normal Force: FnF_n acting upwards.
Exam Tip

When setting up your equations, remember that forces are vectors! Break them into horizontal (x) and vertical (y) components.

Let's say the box has a mass of m=7.3m = 7.3 kg. To find the force, use Newton's Second Law: ΣF=ma\Sigma F = ma.

For the vertical forces (yy-direction):

ΣFy=FnFg\Sigma F_y = F_n - F_g

Since the box is at rest, a=0a = 0 m/s²:

FnFg=ma=(7.3)(0)=0F_n - F_g = ma = (7.3)(0) = 0

Fn=Fg=mg=(7.3)(10)=73F_n = F_g = mg = (7.3)(10) = 73 N

So, the normal force and gravitational force are both 73 N, putting the box in equilibrium.

Friction Force

Friction is a force that opposes motion between surfaces. It's always parallel to the surface and independent of the applied force. There are two types:

Static Friction

Static friction (fsf_s) occurs when surfaces are not sliding relative to each other.

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fsμsFnf_s \le \mu_s F_n

Quick Fact

The static friction force adjusts to match the applied force, up to a maximum value. The maximum static friction is given by fs,max=μsFnf_{s,max} = \mu_s F_n.

Kinetic Friction

Kinetic friction (fkf_k) occurs when surfaces are sliding relative to each other.

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fk=μkFnf_k = \mu_k F_n

Common Mistake

Static friction is generally greater than kinetic friction (μs>μk\mu_s > \mu_k). This means it takes more force to start an object moving than to keep it moving.

Key Differences

  • Static friction: Prevents motion, adjusts to applied force up to a maximum.
  • Kinetic friction: Opposes motion, constant value while sliding.

μ\mu (mu) is the coefficient of friction, a constant for a given surface. It's unit-less and represents the ratio of frictional force to normal force.

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Memory Aid

Remember: Static friction is like a stubborn friend who doesn't want to move, while kinetic friction is like trying to slow down a sled once it's already going.

TIP: The static force increases as the exerted force increases until the object breaks free from the surface.

Need More Help?

Practice is key! Use FBDs, consider angles, and break forces into x and y components. Check out this PhET simulation to visualize forces in one dimension.

https://phet.colorado.edu/sims/cheerpj/forces-1d/latest/forces-1d.html?simulation=forces-1d

Practice Questions

Let's test your understanding with some practice problems. Remember to use FBDs and Newton's laws!

Practice Question

Question 1:

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A wooden box of mass 10 kg is placed on a horizontal surface. A force of 40 N is applied to the box at an angle of 30 degrees above the horizontal, but the box remains at rest.

a) Draw a free body diagram illustrating all forces acting on the box. b) Draw a free body diagram illustrating the net force acting on the box. c) Determine the friction force exerted on the box. d) Determine the coefficient of static friction.

Answer:

a) FBD with labeled forces: Applied force (F_app) at 30 degrees, weight (F_g) down, normal force (F_n) up, and static friction (f_s) opposite the direction of motion.

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b) Since the box is at rest, the net force is zero. FBD shows all forces canceling out.

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c) ΣFx=0\Sigma F_x = 0, so fs=Fappcos(30)=40cos(30)=34.6f_s = F_{app} \cos(30) = 40 \cos(30) = 34.6 N

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d) ΣFy=0\Sigma F_y = 0, so Fn+Fappsin(30)Fg=0F_n + F_{app} \sin(30) - F_g = 0. Fn=FgFappsin(30)=10040sin(30)=80F_n = F_g - F_{app} \sin(30) = 100 - 40 \sin(30) = 80 N. fsμsFnf_s \le \mu_s F_n, so μsfs/Fn=34.6/80=0.43\mu_s \ge f_s / F_n = 34.6 / 80 = 0.43

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Quick Fact

Coefficients of friction are unit-less, representing the ratio of frictional force to normal force.

Practice Question

Question 2:

A 2 kg object's position is given by x(t)=5t2+2tx(t) = 5t^2 + 2t and y(t)=3t3y(t) = 3t^3. Find the magnitude of the net force acting on the object at t=2t = 2 seconds.

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Answer:

Use calculus! Find the acceleration by taking the second derivative of position:

vx(t)=dxdt=10t+2v_x(t) = \frac{dx}{dt} = 10t + 2, ax(t)=dvxdt=10a_x(t) = \frac{dv_x}{dt} = 10

vy(t)=dydt=9t2v_y(t) = \frac{dy}{dt} = 9t^2, ay(t)=dvydt=18ta_y(t) = \frac{dv_y}{dt} = 18t

At t=2t=2 s: ax(2)=10a_x(2) = 10 m/s², ay(2)=36a_y(2) = 36 m/s²

Resultant acceleration: a=102+362=37.36a = \sqrt{10^2 + 36^2} = 37.36 m/s²

Net force: F=ma=237.36=74.72F = ma = 2 * 37.36 = 74.72 N

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Practice Question

Question 3:

Multiple Choice Question (Based on College Board Sample Questions)

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Answer:

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Practice Question

Question 4:

Free Response Question (Based on College Board Sample Questions)

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Answer:

Side note: When creating a graph given the values of acceleration and two blocks of mass, it is usually created to make a graph that would demonstrate a slope of g, the acceleration due to gravity.

Practice Question

Question 5:

Two teams of nine members each engage in a tug of war. Each of the first team's members has an average mass of 68 kg and exerts an average force of 1350 N horizontally. Each of the second team's members has an average mass of 73 kg and exerts an average force of 1365 N horizontally. (Taken from Lumen Learning)

a) What is the magnitude of the acceleration of the two teams? b) What is the tension in the section of rope between the teams?

Answer:

a) 0.11 m/s²

Draw FBDs for each team. Include: Applied Force, Force of Gravity, Normal Force, and Tension.

Set up equations for the forces in the x-direction:

Fnet1=Fa1Ft=m1aF_{net1} = F_{a1} - F_t = m_1 a

Fnet2=FtFa2=m2aF_{net2} = F_t - F_{a2} = m_2 a

Fa1=91350=12150F_{a1} = 9 * 1350 = 12150 N, m1=968=612m_1 = 9 * 68 = 612 kg

Fa2=91365=12285F_{a2} = 9 * 1365 = 12285 N, m2=973=657m_2 = 9 * 73 = 657 kg

12150Ft=612a12150 - F_t = 612a

Ft12285=657aF_t - 12285 = 657a

Solve for a using substitution: 12150(657a+12285)=612a12150 - (657a + 12285) = 612a, a=0.106a = -0.106 m/s²

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b) 1.2 * 10^4 N

Substitute the acceleration into one of the equations to find tension:

Ft=657(0.106)+12285=12215F_t = 657 * (-0.106) + 12285 = 12215 N

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Final Exam Focus

Alright, you're almost there! Here's a quick rundown of the most important things to focus on:

  • Newton's Laws: Know them inside and out! Apply them to various scenarios.
  • Free Body Diagrams: Master FBDs! They are essential for solving force problems.
  • Friction: Understand static and kinetic friction, and how they affect motion.
  • Net Force: Be able to calculate net force in both x and y directions.
  • Combined Concepts: AP questions often combine multiple concepts. Be prepared to apply your knowledge in different contexts.

Last-Minute Tips

  • Time Management: Don't spend too long on one question. Move on and come back if you have time.
  • Common Pitfalls: Watch out for angles and direction! Make sure to break forces into components.
  • Challenging Questions: If you're stuck, draw an FBD and write down what you know. Start with the basics and build from there.
Exam Tip

Always start with a free body diagram! It's the key to unlocking most force problems. Also, remember to check your units and make sure your answers make sense.

You've got this! Stay calm, trust your preparation, and go ace that exam! 💪

Question 1 of 12

A hockey puck slides on frictionless ice with constant velocity. According to Newton's First Law, what will happen to the puck's motion?

It will gradually slow down and stop

It will continue to move at the same constant velocity

It will accelerate

It will suddenly reverse direction