Newton's Laws of Motion

Isaac Newton(1643-1727) made many profound contributions to science. He was an important figure in the Scientific Revolution. He worked on optics. He discovered universal gravitation. He also shares credit for inventing calculus.

In his work Philosophiæ Naturalis Principia Mathematica, Newton formulated his three laws of motion that model how objects accelerate. These laws make up the foundation of classical physics.

Hypothesis, Theory and Law

Science doesn't produce unchanging truths. Science is an ongoing process of discovery, and each new explanation has the potential to improve on the previous. Hypotheses, theories and laws could be true, but it is impossible to know anything with 100% certainty.


A phenomenon is an observable event, like the orbits of planets, static electricity, and lactose intolerance.

A hypothesis is a possible explanation for a phenomenon.

A scientific theory is a hypothesis that has been tested many times and never been proven false. Theories are explanations. They answer the question: why does this phenomenon occur? Examples include: quantum field theory, big bang theory, evolution, and germ theory.


A scientific law is also based on the results of many tests, but a law doesn't explain why a phenomenon occurs. Instead, laws are a general description of what has happened and what will happen for a narrow range of conditions.

  • Laws predict the behavior of a natural phenomenon.
  • Laws are only accurate for a limited range of conditions.
  • Laws are typically a mathematical equation, or a rule.

    • Almost all laws are found in the physical sciences. Some examples are: Ohm's Law, Universal Gravitation, Coulomb's law, and Kirchhoff's laws.





    Question: Why are scientific laws so rare in the life sciences?
    answer

    Most biological phenomenon are too complex to be explained by math.





    After observing the motion of the planets, Nicolaus Copernicus published the idea that the planets revolved around the sun. This idea was tested many times and never disproven. Sixty years later, Johannes Kepler discovered an equation that predicted how the planets moved around the sun.

    Question: Identify a phenomenon, hypothesis, theory, and law from the paragraph above.
    answer

    phenomenon: the motion of the planets
    hypothesis: planets revolve around the sun
    theory: planets revolve around the sun
    law: Kepler's equation





    Question: What about a scientific model? What does that word mean?
    answer

    A scientific model is a simplified representation of an aspect of reality. Models help us learn new things, predict how a complex system behaves, or guide future research. Models in physics often take the form of an equation, but they can also be a set of rules, or a physical thing.

    Generally, it is clear that a model doesn't match reality, but if the inaccuracy is small a model still has value.

  • the Ptolemaic model of the solar system
  • an infographic about the water cycle
  • a computer simulation
  • any physics equation

    • Newton's First Law: Inertia

    "A body either remains at rest or continues to move at a constant velocity, unless acted upon by a net external force."

    All matter has the property of Inertia. Matter stays still or keeps moving until a force causes it to accelerate.

    Most people already have a strong intuitive understanding of inertia. You probably know it's difficult to move something very heavy and easy to move something light. That's inertia.

    The Earth has orbited the Sun for billions of years without coming to a stop. It's clear that objects in space follow Newton's first law.

    Question: Objects on Earth seem to not follow Newton's first law. What causes objects on Earth to always come to a stop?
    answer

    The force of friction slows things down. As air particles strike a fast moving object the object slows down and the air speeds up. This spreads out the movement, but the movement is still there.

    m

    Press E to simulate the mass. Then press WASD to apply forces to the mass. Imagine the screen is the floor and you are looking down on the mass as it slides around, like air hockey.

    Simulation: What causes the mass to stop when you aren't pressing a key?
    observation

    Objects only change velocity if there is a force. In this case friction applies a force in the direction opposite the mass's velocity.

    Newton's Second Law: Force

    "The vector sum of the external forces F on an object is equal to the mass m of that object multiplied by the acceleration vector of the object."

    Newton's second law defined a new term called force. A body accelerates in the direction of the sum of the forces applied to it. The mass of the body determines how much acceleration is felt.

    m F F F

    $$\sum F=ma$$

    \( F \) = force [N, Newtons, kg m/s²] vector
    a push or a pull

    \(m\) = mass [kg, kilogram]
    resistance to acceleration

    \(a\) = acceleration [m/s²] vector

    \( \sum \) = The greek letter Sigma represents a summation of numbers. It means add up all the forces to get a net force.

    $$\sum F=F_{1}+F_{2}+F_{3}+...$$

    If there was a 5 N force to the left and a 20 N force to the right we could get the net force by adding them. Force is a vector, so the left force should be negative.

    $$\sum F= - 5 \, \mathrm{N} + 20 \, \mathrm{N} $$ $$\sum F= 15 \, \mathrm{N}$$
    Example: A ball has a mass of 0.43 kg. Find the acceleration of the ball when it experiences a net force (total force) of 1 N from air friction.
    solution
    $$m = 0.43\,\mathrm{kg}$$ $$\sum F = 1\,\mathrm{N}$$ $$a = \, ?$$
    $$\sum F=ma$$ $$1=(0.43)a$$ $$\frac{1}{0.43}=a$$ $$2.33 \, \mathrm{\tfrac{m}{s^2}}=a$$
    100 kg 2000 N 1900 N Example: A 100 kg boat at rest is being pushed left with a 2000 N force from the wind. The water current is producing a force of 1900 N to the right. How will the boat accelerate? How far will the boat go in 6 s?
    solution $$ \text{Vectors pointed down or left are negative.}$$ $$\sum F=ma$$ $$-2000+1900=(100)(a)$$ $$-100=(100)(a)$$ $$-1 \mathrm{\tfrac{m}{s^{2}}}= a$$ $$ \text{The boat will accelerate to the left.}$$ $$ u = 0 $$ $$ \Delta x = ?$$ $$ \Delta t = 6\mathrm{s}$$ $$ \Delta x = u\Delta t+ \tfrac{1}{2} a \Delta t^{2}$$ $$ \Delta x = (0)(6)+\tfrac{1}{2} (-1)(6)^{2}$$ $$ \Delta x = \tfrac{1}{2} (-1)(6)^{2}$$ $$ \Delta x = -18 \, \mathrm{m}$$ $$ \text{The boat will move 18 meters to the left.}$$

    Play around with the net force simulation to get a feeling for how adding force vectors produces acceleration.

    Forces and Motion: Basics
    Click to Run

    Question: Could the tug of war be moving left, but have a net force to the right?
    answer

    An object could be moving in one direction and have a force in the opposite direction if it was slowing down.

    To test this out, add one person to the left. Click go. Wait a second. Add two on the right.


    Question: On the motion mode, find the mass of the gift.
    answer

    We need to solve a 1-D motion problem with constant accelerate. We can record the initial and final velocities if you check the "values" and "speed" boxes. Get a stopwatch to record the time elapsed.

    After we use a kinematics equation to solve for acceleration, we can use Newton's 2nd law to calculate the mass.

    answer

    I applied a 10N force for 10 seconds and recorded the time and velocities.

    $$u = 0 $$ $$v = 2 \, \mathrm{\tfrac{m}{s}}$$ $$\Delta t = 10 \, \mathrm{s}$$ $$a = \, ?$$
    $$v = u + a \Delta t$$ $$2 \, \mathrm{\tfrac{m}{s}} = 0 + a (10 \, \mathrm{s})$$ $$a = 0.2 \, \mathrm{\tfrac{m}{s^2} }$$
    $$\sum F=ma$$ $$10 \, \mathrm{N} = m (0.2 \, \mathrm{\tfrac{m}{s^2}})$$ $$\frac{10 \, \mathrm{N}}{0.2 \, \mathrm{\tfrac{m}{s^2}}} = m$$ $$50 \, \mathrm{kg} = m$$

    Question: If I apply a force to a mass it accelerates. What happens if I apply a larger force to the same mass?
    answer

    Mass is constant so we can just pretend it is one.

    $$F=ma$$ $$F=a$$ $${\Uparrow \atop F} {\atop =} {\Uparrow \atop a} $$

    Force and acceleration are directly proportional. Increasing F will cause a proportional increase in a.

    Question: If I apply a 100 N force to a mass it accelerates at 2 m/s². What happens if I double the mass while keeping the force the same?
    answer

    Force is constant so we can just pretend it is one.

    $$F=ma$$ $$1=ma$$ $$1=(2m)\left(\tfrac{1}{2} a \right)$$

    Mass and acceleration are inversely proportional. Doubling m will cause a to be halved.

    Newton's Third Law:
    Equal and Opposite Force Pairs

    "When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body."


    F 1 F 2 $$F_1 = -F_2$$

    If you push on something it will push back with the same force, but in the opposite direction. Forces always come in equal, but opposite pairs.

    Most of the time force pairs come from objects in contact, but even non-contact forces, like gravity, still obey this law.

    Question: You push down on the ground with a 1000 N force. Describe the magnitude and direction of the force the ground pushes on you.
    answer

    The force the ground applies is equal and opposite to the force you apply on the ground. The magnitude is 1000 N. The direction is up.

    Question: A boxer's glove applies a 140 N force to the face of another boxer. Describe the magnitude of the force the face applies to the glove.
    answer

    When a fist is punching a face, the face is also punching the fist. That's deep...

    Anyways, the force is equal and opposite. The magnitude is 140 N.

    Example: You (100 kg) are standing on a frictionless skateboard. You throw your 2 kg bottle of water to the right. During the throw, the water bottle briefly accelerates at 40 m/s². How much acceleration do you feel at that time?
    strategy

    Calculate the force the water bottle feels when it accelerates at 40 m/s². Plug the negative value of that force in as the force you feel.

    solution

    The force the water bottle feels is equal and opposite to the force you feel.

    $$ \text{water bottle}$$ $$F = ma $$ $$F=(2\, \mathrm{kg})(40\,\mathrm{\tfrac{m}{s^2}} )$$ $$F=\color{#80f}80 \, \mathrm{N}$$
    $$ \text{you}$$ $$F = ma $$ $$- {\color{#80f}80 \, \mathrm{N}} = (100 \, \mathrm{kg}) a$$ $$-0.8 \, \mathrm{\tfrac{m}{s^2}}=a$$
    practice problems (16)

    In case you wanted more practice I used AI to make some more problems. The rest of the site I made by hand, but generating endless problems seemed safe. I did find mistakes in the AI generated problems, and there are probably some I didn't find. Let me know if something could be fixed. I also added a practice problem on each page with no solution. That's intentional. Have fun!

    Keep left and down negative unless the problem gives a different direction.

    printout.pdf

    Example: An 8.0 kg crate of glassware is pushed to the right with a 30 N force. Friction pushes left on the crate with a 6.0 N force. What is the crate's acceleration?
    solution $$m = 8.0\,\mathrm{kg}$$ $$F_{\mathrm{push}} = 30\,\mathrm{N}$$ $$F_{\mathrm{friction}} = -6.0\,\mathrm{N}$$ $$a = \,?$$
    $$\sum F = ma$$ $$30 - 6.0 = 8.0a$$ $$24 = 8.0a$$ $$a = \frac{24}{8.0}$$ $$a = 3.0\,\mathrm{\tfrac{m}{s^{2}}}$$

    The answer is positive, so the crate accelerates to the right.

    Example: A 4.0 kg cart starts from rest on a 2.0 m level track. A net force of 10 N pushes it to the right for 1.2 s. What is its acceleration, and how far does it move?
    solution $$m = 4.0\,\mathrm{kg}$$ $$\sum F = 10\,\mathrm{N}$$ $$a = \,?$$
    $$\sum F = ma$$ $$10 = 4.0a$$ $$a = \frac{10}{4.0}$$ $$a = 2.5\,\mathrm{\tfrac{m}{s^{2}}}$$

    Now use the acceleration in an equation of motion from the kinematics page.

    $$u = 0$$ $$a = 2.5\,\mathrm{\tfrac{m}{s^{2}}}$$ $$\Delta t = 1.2\,\mathrm{s}$$ $$\Delta x = \,?$$
    $$\Delta x = u\Delta t + \tfrac{1}{2}a\Delta t^{2}$$ $$\Delta x = (0)(1.2) + \tfrac{1}{2}(2.5)(1.2)^{2}$$ $$\Delta x = 1.8\,\mathrm{m}$$

    The cart stays on the 2.0 m track.

    Example: Two students pull on a 15 kg cart. One pulls left with 40 N and the other pulls right with 40 N. What is the cart's acceleration?
    solution $$m = 15\,\mathrm{kg}$$ $$\sum F = -40 + 40$$ $$\sum F = 0$$
    $$\sum F = ma$$ $$0 = 15a$$ $$a = 0\,\mathrm{\tfrac{m}{s^{2}}}$$

    Balanced forces don't cause acceleration. If the cart was already rolling, it keeps rolling at the same velocity.

    Example: A student pushes a loaded wagon with a net force of 72 N, and it accelerates at 3.0 m/s². What is the mass of the wagon and its load?
    solution $$\sum F = 72\,\mathrm{N}$$ $$a = 3.0\,\mathrm{\tfrac{m}{s^{2}}}$$ $$m = \,?$$
    $$\sum F = ma$$ $$72 = m(3.0)$$ $$m = \frac{72}{3.0}$$ $$m = 24\,\mathrm{kg}$$
    Example: A 12 kg box of books is pushed across a carpeted floor with a 40 N horizontal force. What is the box's acceleration?
    solution

    This cannot be solved from the information given. Newton's second law uses the net force, not just the push. Carpet produces friction, and we don't know how large the friction force is. Without it we can't find the net force.

    Example: A 12 kg sled accelerates to the right at 2.0 m/s². Friction pushes left on the sled with 5.0 N. What applied force to the right is needed?
    solution $$m = 12\,\mathrm{kg}$$ $$a = 2.0\,\mathrm{\tfrac{m}{s^{2}}}$$ $$F_{\mathrm{friction}} = -5.0\,\mathrm{N}$$ $$F_{\mathrm{applied}} = \,?$$
    $$\sum F = ma$$ $$F_{\mathrm{applied}} - 5.0 = (12)(2.0)$$ $$F_{\mathrm{applied}} - 5.0 = 24$$ $$F_{\mathrm{applied}} = 29\,\mathrm{N}$$

    24 N of the push goes into accelerating the sled, and the other 5 N cancels friction.

    Example: A 4.5 m canoe and its paddler have a total mass of 90 kg. While the paddler rests, the wind pushes the canoe left with 60 N and the current pushes it right with 42 N. What is the canoe's acceleration?
    solution $$m = 90\,\mathrm{kg}$$ $$F_{\mathrm{wind}} = -60\,\mathrm{N}$$ $$F_{\mathrm{current}} = 42\,\mathrm{N}$$ $$a = \,?$$
    $$\sum F = ma$$ $$-60 + 42 = 90a$$ $$-18 = 90a$$ $$a = \frac{-18}{90}$$ $$a = -0.20\,\mathrm{\tfrac{m}{s^{2}}}$$

    The negative sign means the canoe accelerates to the left, the direction of the larger force.

    Example: A 3.0 kg delivery drone has a 30 N force on it to the right, but it accelerates to the left at 4.0 m/s². What leftward force must also be acting on it?
    solution

    Let right be positive, so the acceleration is negative.

    $$m = 3.0\,\mathrm{kg}$$ $$a = -4.0\,\mathrm{\tfrac{m}{s^{2}}}$$ $$F_{\mathrm{right}} = 30\,\mathrm{N}$$ $$F_{\mathrm{left}} = \,?$$
    $$\sum F = ma$$ $$30 + F_{\mathrm{left}} = (3.0)(-4.0)$$ $$30 + F_{\mathrm{left}} = -12$$ $$F_{\mathrm{left}} = -42\,\mathrm{N}$$

    The leftward force must be 42 N. It cancels the 30 N force and still leaves a 12 N net force to the left.

    Example: A 1500 kg car cruises down a straight highway at a constant 25 m/s. The engine pushes the car forward with 600 N. What is the net force on the car, and how large are the air and road resistance forces?
    solution

    Constant velocity means the acceleration is zero.

    $$m = 1500\,\mathrm{kg}$$ $$a = 0$$ $$F_{\mathrm{engine}} = 600\,\mathrm{N}$$ $$F_{\mathrm{resistance}} = \,?$$
    $$\sum F = ma$$ $$\sum F = (1500)(0)$$ $$\sum F = 0$$
    $$600 + F_{\mathrm{resistance}} = 0$$ $$F_{\mathrm{resistance}} = -600\,\mathrm{N}$$

    The car is moving fast, but the net force is zero. By Newton's first law, a moving object doesn't need a net force to keep moving. The 600 N of air and road resistance points backward.

    Example: A student throws a 500 g ball, and the ball accelerates to the right at 30 m/s² while it's in the student's hand. What force does the student apply to the ball?
    solution $$500\,\textcolor{DeepPink}{\mathrm{g}}\left(\frac{1\,\mathrm{kg}}{1000\,\textcolor{DeepPink}{\mathrm{g}}}\right)$$ $$0.50\,\mathrm{kg}$$
    $$m = 0.50\,\mathrm{kg}$$ $$a = 30\,\mathrm{\tfrac{m}{s^{2}}}$$ $$\sum F = \,?$$
    $$\sum F = ma$$ $$\sum F = (0.50)(30)$$ $$\sum F = 15\,\mathrm{N}$$

    The student pushes the ball to the right with 15 N. By Newton's third law, the ball pushes back on the student's hand with 15 N to the left.

    Question: A mosquito hits the windshield of a truck on the highway. Which feels a larger force, the mosquito or the truck? Which has a larger acceleration?
    answer

    The forces are the same size. By Newton's third law, the mosquito pushes on the truck with exactly as much force as the truck pushes on the mosquito, just in the opposite direction.

    The accelerations are very different. The mosquito has a tiny mass, so the force gives it a huge acceleration. The truck's mass is enormous, so the same force barely changes its velocity.

    Example: Two ice skaters stand facing each other and push off. Skater A has a mass of 60 kg and skater B has a mass of 45 kg. During the push, each skater feels a 90 N force. Skater A moves left and skater B moves right. What is each skater's acceleration?
    solution

    By Newton's third law, the forces are equal in size and opposite in direction. Let right be positive.

    $$\text{skater A}$$ $$\sum F = ma$$ $$-90 = 60a$$ $$a = \frac{-90}{60}$$ $$a = -1.5\,\mathrm{\tfrac{m}{s^{2}}}$$
    $$\text{skater B}$$ $$\sum F = ma$$ $$90 = 45a$$ $$a = \frac{90}{45}$$ $$a = 2.0\,\mathrm{\tfrac{m}{s^{2}}}$$

    The forces are the same, but the lighter skater gets the larger acceleration.

    Example: A 12 kg backpack falls off a ledge. Ignoring air friction, it accelerates downward at 9.8 m/s², like everything in free fall on Earth. What net force causes that acceleration?
    solution

    Let up be positive. The acceleration comes from the kinematics page.

    $$m = 12\,\mathrm{kg}$$ $$a = -9.8\,\mathrm{\tfrac{m}{s^{2}}}$$ $$\sum F = \,?$$
    $$\sum F = ma$$ $$\sum F = (12)(-9.8)$$ $$\sum F = -118\,\mathrm{N}$$

    The only force on the falling backpack is gravity, so the force of gravity on it is 118 N down. The next page calls this force weight.

    Example: A 900 kg car has a 3.2 kN engine force forward and 0.80 kN of air and road resistance backward. What is the car's acceleration?
    solution $$3.2\,\mathrm{kN} = 3.2(1000)\,\mathrm{N}$$ $$3.2\,\mathrm{kN} = 3200\,\mathrm{N}$$ $$0.80\,\mathrm{kN} = 0.80(1000)\,\mathrm{N}$$ $$0.80\,\mathrm{kN} = 800\,\mathrm{N}$$
    $$m = 900\,\mathrm{kg}$$ $$\sum F = 3200 - 800$$ $$\sum F = 2400\,\mathrm{N}$$ $$a = \,?$$
    $$\sum F = ma$$ $$2400 = 900a$$ $$a = \frac{2400}{900}$$ $$a = 2.67\,\mathrm{\tfrac{m}{s^{2}}}$$
    Example: A 1200 kg car goes from rest to 100 km/h in 9.3 s. What average net force acts on the car?
    solution $$100\,\mathrm{\tfrac{\textcolor{DeepPink}{km}}{\textcolor{DodgerBlue}{h}}}\left(\frac{1000\,\mathrm{m}}{1\,\textcolor{DeepPink}{\mathrm{km}}}\right)\left(\frac{1\,\textcolor{DodgerBlue}{\mathrm{h}}}{3600\,\mathrm{s}}\right)$$ $$27.8\,\mathrm{\tfrac{m}{s}}$$

    First use kinematics to find the acceleration.

    $$u = 0$$ $$v = 27.8\,\mathrm{\tfrac{m}{s}}$$ $$\Delta t = 9.3\,\mathrm{s}$$ $$a = \,?$$
    $$v = u + a\Delta t$$ $$27.8 = 0 + a(9.3)$$ $$a = \frac{27.8}{9.3}$$ $$a = 2.99\,\mathrm{\tfrac{m}{s^{2}}}$$
    $$\sum F = ma$$ $$\sum F = (1200)(2.99)$$ $$\sum F = 3600\,\mathrm{N}$$

    This is the net force. The engine has to push even harder, because air and road resistance push backward.

    Example: An empty 10 kg shopping cart accelerates at 2.0 m/s² when you push it. You load it with 15 kg of groceries and push with the same force. What is the new acceleration?
    solution

    Find the force from the empty cart.

    $$\sum F = ma$$ $$\sum F = (10)(2.0)$$ $$\sum F = 20\,\mathrm{N}$$

    The loaded cart has a total mass of 25 kg.

    $$\sum F = ma$$ $$20 = 25a$$ $$a = \frac{20}{25}$$ $$a = 0.80\,\mathrm{\tfrac{m}{s^{2}}}$$

    The mass went up by a factor of 2.5, so the acceleration went down by a factor of 2.5. With the same force, mass and acceleration are inversely proportional.

    Reading (10 minutes): Read The Greatest Physics Demo of All Time Happened on the Moon by Rhett Allain from WIRED. Then answer these questions.

    What did Aristotle predict about a heavy object and a light object falling from the same height, and what did the Apollo 15 demonstration show instead?
    answer

    Aristotle predicted that the heavier object would fall faster. On the airless Moon, the hammer and feather landed together because they had the same gravitational acceleration.


    A heavier object has a larger gravitational force on it. Why does that not make its free-fall acceleration larger in the simple model?
    answer

    The larger force is paired with a proportionally larger mass. In F = ma, using weight mg for the force leaves a = g, so the mass cancels.


    Why can a rock and feather usually land at different times on Earth without disproving the Moon experiment?
    answer

    Air drag adds an upward force, and it matters much more for a light, wide feather than for a compact rock. The Moon has essentially no atmosphere, so that extra force is absent.

    Reading (10 minutes): Read Cargo Cult Science by Richard Feynman from Caltech. Then answer these questions.

    What does Feynman mean by "cargo cult science"?
    answer

    It is work that copies the visible form of science without producing reliable results. The key missing part is careful testing that can reveal when an idea is wrong.


    What kind of scientific integrity does Feynman ask researchers to practice?
    answer

    He asks them to report possible mistakes, alternative explanations, and details that could weaken their conclusion. This gives other people a fair chance to check the work.


    Why are repeated experiments important even when the original researcher believes their result is correct?
    answer

    Independent repetition can reveal an unnoticed error or show that the effect is real. Scientific confidence comes from nature continuing to agree with the evidence, not from one person's certainty.