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AP Physics 1 review of Forces and Newton's Laws

In this video David quickly explains each concept behind Forces and Newton's Laws and does a sample problem for each concept. Keep an eye on the scroll to the right to see how far along you've made it in the review. Created by David SantoPietro.

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  • blobby green style avatar for user davidsantopietro
    Hi all. Here's a dropbox link to download the Force notes from the video if you wanted them. Good luck on the AP exam!
    https://www.dropbox.com/s/pugma8iq9wdkohu/Forces%20AP%201%20review.pdf?dl=0
    (40 votes)
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  • piceratops seed style avatar for user Thonkham
    The Example question for Kinetic Frictions states that the answers are B and C. Why wouldn't increasing the mass of the car increase the breaking distance required to come to a full stop? Mass is included in the equation.
    (5 votes)
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    • blobby green style avatar for user A4SamuelNelson
      Two ways to think about it, math is below, conceptual is at the end.
      The mass of the car effects the normal force. That does not mean an increase of mass will relate to a change in distance to break. In order for the car to stop it must accelerate in the opposite direction the car is traveling. Acceleration is defined by total force/mass. Frictional force is defined as frictional coefficient x normal force. In this instance normal force is equal and opposite to Gravitational force which is equal to mass x gravity constant. So if you substitute all of this in you are left with two equations, assuming frictional force is equal to the total force.
      a=Ff/m
      Ff=u*m*g (Since Ff=u*Fn and Fn=Fg and Fg=m*g)
      So... substitute Ff
      a=u*m*g/m
      Since mass is both multiplied and divided
      a=u*g
      So in this instance mass does not effect acceleration

      Conceptually, mass does two things we care about right now. The bigger the mass, the more friction. Try dragging an empty bag vs a full one. Mass also makes acceleration harder, pushing an empty light car is way less force than pushing a heavy full car. Since friction is the force 'pushing' the car to a stop, an interesting thing happens. As mass increases the amount of force slowing the car increases, as well as how much force is needed to slow the car. Since mass proportionally increases friction as it decreases acceleration, the net result is the same. It makes it hared to slow, while working harder to stop it, meaning that time and distance used to stop the car does not change as mass increases or decreases.
      Hoped that helped, sorry if that was confusing.
      (3 votes)
  • duskpin ultimate style avatar for user Ishika Agarwal
    At - , David mentions that "internal forces" can be canceled when you're treating systems as a single object. What are some examples of internal forces?
    (2 votes)
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  • blobby green style avatar for user luyasluyanda8
    What happens to acceleration when mass is doubled
    (3 votes)
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  • duskpin ultimate style avatar for user Mo
    At , David said the force of gravity on all objects near on the Earth is always down towards the centre of the Earth, but our teacher told us the gravity is not always towards the centre of the Earth unless we are at the two poles of the Earth...
    (0 votes)
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    • male robot hal style avatar for user Andrew M
      Your teacher's statement is very misleading. There's nothing special about the poles, and there's no reason to worry about the tiny, tiny deviations of the direction or strength of gravity when you are doing AP Physics 1.
      (5 votes)
  • hopper jumping style avatar for user Kailas Moon
    At , the video says the normal force exerted on a box on a slope with angle theta is equal to m*g*cos (theta). Could someone explain the mathematical steps behind this? Thank you in advance and good luck everyone.
    (1 vote)
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  • blobby green style avatar for user han zhang
    At , why X and W are mgsin theta?
    (2 votes)
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  • starky sapling style avatar for user Charm
    In the question for Kinetic Friction, why won't decreasing the mass of the car increase the distance required for the car to stop? Shouldn't lowering the mass cause the normal force, and thus the force of friction, to decrease?
    An explanation would be greatly appreciated. <3
    (2 votes)
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  • blobby green style avatar for user 3, Hendriks Gabe
    A company uses a ramp to slide boxes of parts to a shipping area. Each box has a mass of 26 kilograms. When sliding down the ramp, the boxes accelerate at a rate of 0.1 m/s2. What is the force acting on each box? For this problem, ignore the effects of friction acting on each box.
    (1 vote)
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    • piceratops ultimate style avatar for user Hecretary Bird
      If your question asks about the net force on each box, that is given by Newton's second law. The sum of forces is equal to mass times acceleration. We're given mass and acceleration, so we use them to get the net force:
      F = ma
      F = (26)(.1) = 2.6 N
      (1 vote)
  • old spice man blue style avatar for user mand4796
    For the "Treating Systems as object" example, how could David simply add together all x-direction and y-direction forces - doesn't he have to treat these separately?
    (1 vote)
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    • male robot hal style avatar for user Andrew M
      No, because there's really only one direction of motion in this problem: motion along the path of the spring. The "horizontal" motion of one object is completely dependent on the "vertical" motion of the other, so it's really one motion, not two (or at least it can be viewed that way)

      You will notice that when you treat the system as two objects, even then, the direction doesn't really come into solving the problem. There's just a pull, and a tension.
      (1 vote)

Video transcript

- [Instructor] What's Newton's first law say? Newton's first law states that objects don't change their velocity unless there's an unbalanced force. So, if there was no force on an object, or the forces are balanced, then the object will continue moving with a constant velocity. Or, if it was at rest, it'll continue sitting at rest. In other words, there doesn't have to be a net force for something to have motion, there only has to be a net force for something to have acceleration. And it's really important to note that Newton's first law does not apply to single objects. It applies to systems of objects as well. In other words, if you consider a system of objects, and look at the center of mass of that system, the center of mass of the system will remain at rest or remain in constant motion as long as there's no external unbalanced forces on the system. So these objects may be exerting forces on each other, but the center of mass will remain at rest, or with constant velocity unless there's an unbalanced external force on this system of particles. So what's an example problem involving Newton's first law look like? Say you were told that a heavy elevator is lifted upward by a cable exerting a force Fc, and the elevator moves up with a constant velocity of five meters per second. We wanna know how the force from the cable compares to the force of gravity. The mistake many people make is they think that since the object was moving upward, the upward force must be larger, but that's not true. Since this is moving upward with constant velocity, the forces actually have to balance. Since Newton's first law states that when the net force is zero, the object maintains a constant velocity. And for the net force to be zero, these forces have to cancel. So, even though it's non-intuitive, this cable force has to equal the force of gravity, so that the elevator can move with constant velocity. What's Newton's second law mean? Newton's second law states that the acceleration of an object is proportional to the net force and inversely proportional to the mass. Which, written in equation form, states that the acceleration of an object is equal to the net force on that object divided by the mass of the object. And this equation works for any single direction as well. In other words, the acceleration in the x direction is equal to the net force in the x direction divided by the mass. And the acceleration in the y direction is equal to the net force in the y direction divided by the mass. So what's an example of Newton's second law look like? Let's say a five kg space rock had the forces acting on it shown in this diagram, and we wanted to determine the acceleration in the horizontal direction. Since horizontal is the x direction, we're only gonna use forces in the x direction to determine the acceleration in the x direction. That means the 15 Newton force, and the five Newton force don't contribute at all to the acceleration in the x direction. The only components that contribute are the horizontal component of the 10 Newton force, which would be 10 cosine of 30, and the 40 Newton force. So the acceleration in the x direction would equal the net force in the x direction, which would be 10 cosine 30. That would be a positive contribution, since it points to the right, minus 40, since that's a negative contribution pointing to the left. And finally we'd divide by five kilograms, which gives us the correct acceleration in the x direction. What's Newton's third law mean? Newton's third law states that if an object A is exerting a force on object B, then object B must be exerting an equal and opposite force back on object A. And this is true even if the objects have different sizes or there's acceleration. In other words, let's say the Earth is pulling on an asteroid. Even though the Earth is much larger than the asteroid, the asteroid's gonna exert an equal and opposite force back on the Earth. And this is true whether the asteroid is moving with constant velocity, or whether it's accelerating. So what's an example problem involving Newton's third law look like? Let's say a metal sphere is sitting on a cardboard box, and we want to determine which of these choices constitute a Newton's third law force pair. The first option says that there's an upward force on the sphere from the box. So to find the third law pair, just reverse the order of the objects, which means the partner to this force would be the force on the box by the sphere, which is not what this says, so it's not option A. Option B refers to an upward force on the box from the table, which we know if we reverse the labels, should have a partner force that would be the force on the table by the box. Which is not what this says, so it's not option B. Option C talks about an upward force on the sphere from the box, which, reversing the labels gives us a partner force of on box by sphere. Which is not what this says, so it's not option C. And D refers to an upward force on the box from the table, which, if we reverse the labels, gives us a partner force on the table by the box, which is what this says, and so the forces in D constitute a Newton's third law force pair. Which means they must always be equal and opposite. Other pairs might be equal and opposite, but no matter what happens, these two forces have to be equal and opposite. How do you find the force of gravity on objects near the Earth? The force of gravity on all objects near the Earth is down toward the center of the Earth, and it's equal to the mass times the acceleration due to gravity. Another word for the force of gravity is the weight of an object. But be careful, the weight is not the mass. Weight is the force of gravity which means weight is m times g not just m. The force of gravity is a vector, and it has units of Newtons. So what's an example problem involving the force of gravity look like? Let's say you knew the mass and weight of a watermelon to be 5 kilograms and 49 Newtons when you measure them on the Earth. What might the values for mass and weight of that watermelon be when it's brought to the moon? The value of the mass isn't gonna change here since it's a measure of the total amount of substance in that object. But the weight of the watermelon on the moon is gonna be less since the gravitational pull is gonna be weaker on the moon. So the only choice consistent with those two conditions is A since the mass stays the same and the weight decreases. What's the normal force? The normal force is the outward force exerted by, and perpendicular to a surface. There's no formula specifically to find the normal force, you simply have to use Newton's second law. Let normal force be one of the unknowns, and then solve for it. Now, if you've just got a mass sitting on a horizontal surface, and there's no extra forces involved, the normal force is just gonna counter the force of gravity, which means the normal force will just be mg. But if there's extra forces, or there's acceleration in the direction of the normal force, then the normal force is not gonna equal mg, and you'd have to use Newton's second law for that direction to solve for it. The word normal in normal force refers to the fact that the force is always perpendicular to the surface exerting that force. And it's good to remember that, for a mass on an incline, that normal force is not gonna be equal to mg. It's gonna be mg times cosine of theta. The normal force is a vector, since it's a force, and it also has units of Newtons. So what's an example problem involving normal force look like? Let's say a person is pushing on a stationary box of mass M against the ceiling with a force Fp, and they do so at an angle theta. We want to know what's the magnitude of the normal force exerted on the box from the ceiling? So we'll draw a force diagram. There's the force Fp from the person, the force mg from gravity. If the box is stationary there'd have to be a force preventing it from sliding across the ceiling, which is most likely static friction. And there's also gonna be a normal force, but that normal force will not point upward. The normal force from the ceiling cannot pull up on the box. The normal force from the ceiling will only push out on the box, which will be downward. Since our normal force is in the vertical direction, we'll analyze the forces in the vertical direction. And we can see that the forces must be balanced vertically, since this box has no motion vertically. In other words, the normal force plus the gravitational force is gonna have to equal the vertical component of the force Fp. Which, since that's the opposite side from this angle, we can write as Fp sine of theta. And now we can solve for the normal force, which gives us Fp sine theta - mg. Note that we did not have to use the force of friction, or the horizontal component since our normal force was in the vertical direction. What's the force of tension mean? The force of tension is any force exerted by a string, rope, cable, cord or any other rope like object. And unlike the normal force that can only push, tension can only pull. In other words, ropes can't push on an object. But, similar to the normal force, there's no formula for tension. To find the tension, you'd insert the tension as an unknown variable into Newton's second law, and then solve for it. Since the force of tension is always pulling on objects, when you draw your force diagram, make sure you always draw those tension forces directed away from the object the string is exerting the tension on. Tension's a vector, since it's a force, and it has units of Newtons. So what's an example problem involving tension look like? Let's say two ropes are holding up a stationary box, and we wanna know how the magnitudes of the tensions in both ropes compare. Drawing our force diagram there'll be a downward force of gravity, a force of tension to the left, and also a diagonal force of tension up and to the right. Since the box is stationary, the forces have to be balanced in every direction. That means the vertical component of T2 has to equal the magnitude of the force of gravity, and the horizontal component of T2 has to equal the magnitude of the force T1. But if a component of T2 equals the entire T1, then the total tension T2 has to be bigger than T1. In other words, if part of T2 is equal to T1, then all of T2 is greater than T1. What's the force of kinetic friction mean? The force of kinetic friction is the force exerted between two surfaces that are sliding across each other. And this force always resists the sliding motion of those two surfaces. The force of kinetic friction is proportional to the normal force between the two surfaces, and it's proportional to the coefficient of kinetic friction between the two surfaces. Note that the force of kinetic friction does not depend on the velocity of the object. In other words, if the normal force and coefficient stay the same, then no matter how fast or slow the object moves, no matter how hard or soft you pull, the force of kinetic friction is gonna maintain the same value. Since kinetic friction is a force, it is a vector and it has units of Newtons. So what's an example problem involving kinetic friction look like? So you've got this question about a car traveling at cruising speed, slamming on the brakes, and skidding to a stop. We want to know what two changes could be made that would increase the distance required for the car to skid to a stop. To get some intuition about what would cause this car to skid farther, we could use a kinematic formula. Since the car skids to a stop, the final velocity would be zero. And if we solve for the distance, we get -v0 squared over two times the acceleration. So in order to get the car to skid further, we could increase the initial speed of the car, or reduce the deceleration. To figure out what reduces the magnitude of the acceleration, we'll use Newton's second law. The force slowing the skidding car is the force of kinetic friction, and since there's no extra vertical forces, the normal force is just m times g. Since the masses cancel, the acceleration won't depend on the mass of the car, but reducing the coefficient of friction will reduce the deceleration, and reducing the deceleration will increase the distance the car skids to a stop. The force of static friction tries to prevent the two surfaces from slipping in the first place, and that force of static friction will match whatever force is trying to budge the object until that budging force matches the maximum possible static frictional force, which is proportional to the normal force and the coefficient of static friction. So if the maximum value of the static frictional force is 100 Newtons, and you try to budge the object with 80 Newtons, the static frictional force will just oppose you with 80 Newtons, preventing the object from slipping. If you exert 90 Newtons, the static frictional force will increase to 90 Newtons, preventing the object from slipping. But if you exert 110 Newtons, since this exceeds the maximum possible static frictional force, the object will budge and there will only be a kinetic frictional force now that the object is sliding. So what's an example problem involving the force of static friction look like? Let's say you push on a refrigerator that's 180 kilograms, and the coefficient of static friction between the floor and the fridge is .8. If you exert 50 Newtons on the refrigerator, what's the magnitude of the static friction force exerted on the refrigerator? We'll first find the maximum possible static frictional force using mew s times Fn. Since there's no extra vertical forces, the normal force will just be mg. Plugging in values, we get a maximum possible static frictional force of 1,411 Newtons, but this will not be the value of the static frictional force. This is just the maximum value of the static frictional force. So if we exert 50 Newtons to the right, since that does not exceed this maximum possible static frictional force, static friction will just oppose us with an equal 50 Newtons to the left. And it will continue to match whatever force we exert, until we exceed the maximum possible static frictional force. How do you deal with inclines? Inclines are just angled surfaces that objects can slide up or down, and since the object can't move into the incline, or off of the incline, the motion will only be taking place parallel to the surface of the incline. There will be no acceleration perpendicular to the surface of the incline. So instead of breaking our forces into x and y, we break them into forces perpendicular to the surface and parallel to the surface. The component of gravity that's parallel to the surface is gonna equal mg sine theta, where theta is the angle between the horizontal floor, and the inclined surface. And the component of gravity perpendicular to the surface is gonna be mg cosine theta, where again, theta is the angle measured between the horizontal floor and the angled surface. Since there's no acceleration perpendicular to the surface, the net force in the perpendicular direction has to be zero. And that means this perpendicular component of gravity has to be exactly canceled by the normal force, which is why the value for the normal force is the same as the perpendicular component of gravity. And since those perpendicular components cancel, the total net force on an object on an incline is just gonna equal the component of the net force that's parallel to the surface of the incline. Which, if there's no friction, would simply be mg sine theta, and if there was friction, it would be mg sine theta minus the force of friction. But be careful when you're finding the force of friction on an incline, the normal force will not be m times g, the normal force is gonna be mg cosine theta. So what's an example problem involving inclines look like? Let's say a box started with a huge speed at the bottom of a frictionless ramp, and then it slides up the ramp and through these points shown w, x, y and z. We want to rank the magnitudes of the net force on the box for these points that are indicated. Well, when the box is flying through the air, we know the net force is simply the force of gravity straight down which is m times g. So the net force at y and z are equal. And, on an incline, the net force is the forced component that's parallel to the surface of the incline, which is gonna be mg sine theta. Note that the net force on the incline points down the incline, even though the mass moves up the incline. That just means the mass is slowing down, but since mg is greater than mg sine theta, z and y are greater than the net force at x and w. What does treating systems as as single object mean? This is a trick you can use when two or more objects are required to move with the same speed and acceleration, which will allow us to avoid having to use multiple equations to find the acceleration, and instead use one equation to get the acceleration. When you treat a system of objects as a single object, you get to ignore internal forces, since the internal forces will always cancel. This means you can find the acceleration of the system by looking at only the external forces on that system, and then dividing by the total mass of that system. So what's an example of treating systems as a single object look like? Let's say a mass m1 is pulled across a rough horizontal table by a rope connected to mass m2. If the coefficient of kinetic friction between m1 and the table is mew k, then what's an expression for the magnitude of the acceleration of the masses? So instead of analyzing the forces on each mass individually, which would give us multiple equations and multiple unknowns, we'll use one equation of Newton's second law but we'll treat this system as if it were a single object. Which means we're basically just gonna ask what external forces are gonna make this system go? And what external forces are gonna make this system stop? The external force makes this system go is the force of gravity on m2. It's an external force, since it's exerted by the Earth, which is not part of our system, and it's making the system go, so we'll call that force positive. And we'll call forces that try to make our system stop negative, like this force of kinetic friction on m1, which is also external because the table is not part of the mass of our system. But we will not include the force of tension, since this is an internal force, and these forces will cancel. Now, since we're treating this system as a single mass, we'll divide by the total mass of our system. And then if we write the force of kinetic friction in terms of the coefficient, we get mew k times the normal force, and the normal force on m1 is gonna be m1 times g. Which, with a single equation, gives us an expression for the acceleration of our system, without having to solve multiple equations with multiple unknowns.