Understanding Static and Kinetic Friction

In this video, we're going to focus on static friction and kinetic friction. Now, let's say that this is a carpet floor, and there's a big box. And imagine if you're trying to push the box. So you're trying to apply a force to move it. Now, initially, as you begin to push it, the box doesn't move. But eventually, if you continue... to push it with even more force it will begin to slide and once begin sliding it's easier to continue pushing it but once you stop it it's going to be hard to start it up again the force that prevents the box from sliding at the first place or in the first place is the static frictional force now once it begins to slide the force that impedes the motion is kinetic friction Static means not moving. Kinetic has to do with motion. So the frictional force that opposes motion when the object is sliding against the carpet, that's kinetic friction. The frictional force that prevents you from moving it when you first try to push it, that's static friction. Static friction, let me write the equations on the right. Static friction is less than or equal to mu s times the normal force. Kinetic friction, it's equal to mu k times the normal force. So notice that the static frictional force is represented by an inequality, which means that it's not just a fixed number. It can be a range of numbers up to a maximum point. The kinetic frictional force, however, is not represented by an inequality. So therefore, FK represents a fixed number. Now let's say that this is a five kilogram box. What is the normal force? And calculate these two values with this information. Now this box exerts a downward weight force, and the normal force has to support the weight force. Keep in mind the normal force is a force that the surface exerts on the box, and it's perpendicular to the surface. So in this example, for a horizontal surface, the normal force is equal to the weight force, or mg. So it's going to be 5 kilograms times 9.8 meters per second squared. So the normal force is 49 newtons. Now, I need to give you values for mu s and mu k. Mu s is typically greater than mu k. I haven't seen an example when it's less. So we're going to say that mu s is 0.4. And we're going to say mu k is 0.2. So let's calculate f k and f s. So the kinetic frictional force is going to be 0.2 times 49. which is 9.8. The static frictional force is going to be less than or equal to 0.4 times 49, which is 19.6. So what do these numbers mean? So let me give you an example that's going to illustrate this. let's make a table between the applied force the static frictional force the kinetic frictional force and the net force So, if the applied force is zero, the frictional forces will be zero, and the net force will be zero. If the person doesn't push the box, the box won't move. Nothing's going to happen. But now, let's say if the person applies a force of, let's say, 10 newtons. What's going to happen? Will the box begin to slide? And what are the values of the static and kinetic frictional forces? Now, even though kinetic friction is 9.8, that value only applies if the box slides. Now, what is the static frictional force? Will the box move? Notice that the applied force is less than the maximum static frictional force, so it's not going to move. Which means that there is no kinetic frictional force. The kinetic frictional force only exists If the box is sliding against the carpet, you can't have static friction and kinetic friction present at the same time. If the box is not sliding against the carpet, static friction is present. If it is sliding, kinetic friction is present. So what is static friction in this example? What number should you put here? Now, if you're thinking about putting 19.6, that will not be correct. Because imagine if the person applies a force of 10 newtons to the right to push the box, and if static friction applies a force of 19.6 newtons, that means that there's going to be a net force of 9.6 newtons towards the left. So imagine pushing the box, only to find that the box is pushing you back to the left. That just doesn't happen. So static friction can't be 19.6. It turns out that static friction... It's going to match the applied force until you exceed its maximum value. So if you push it with 10 newtons, it's going to push back on you with 10 newtons. And so the box doesn't move. You're trying to push it, but it doesn't move. So the net force is zero. So let's say if you try to push it with 15 newtons, then it's going to push back on you with 15 newtons. That means you haven't applied enough force to start moving it. So that's what happens when you try to push the box initially. You're pushing hard against it, but it's not moving. Because static friction is matching your applied force until it reaches the maximum value. So this is still going to be 0. So let's say if you apply a force of 19.6 newtons, it's still going to be 19.6. Now, when you exceed 19.6, that's when it begins to slide. And so you no longer have static friction, but you have kinetic friction. So let's say if we go just above 19.6, let's say if we increase it to 20 newtons. Now the box begins to slide, and so there is no more static friction because the surfaces are sliding past each other. But there is kinetic friction, which is always going to be 9.8 once the box begins to slide. So it's 20 minus 9.8, which will give you a net force of 10.2. Now, if you decide to increase the applied force, the static frictional force will still be 0, Fk is going to still be 9.8, and the net force is now 30 minus 9.8, which is 20.2. So hopefully this example helped you to understand the difference between static friction and kinetic friction, and how to calculate it based on the applied force. Let's work on this problem. A 15 kg box rests on a horizontal surface. What is the minimum horizontal force that is required to cause the box to begin to slide if the coefficient of static friction is 0.35? So you can pause the video if you want to work on this problem as well. But let's start with a picture. So this is the 15 kg box. So we wish to apply a horizontal force, which we're going to call capital F, and we know that friction is going to oppose it. Now, if we want the minimum horizontal force that is required to cause the box to begin to slide, we need to use static friction because until the force exceeds static friction, only then can it slide. If it doesn't exceed the static frictional force, then it won't slide. So the threshold is the maximum static frictional force. So we need to set F equal to FS and that's when it begins to slide. when these two have the same magnitude. Now the static frictional force, its maximum value is mu s times the normal force. And in this example, the normal force is going to be mg. So mu s is 0.35, m is 15, and g is 9.8. So the applied force has to be 51.45 Newtons in order for the box to begin to slide. If it's less than this value, the box will not slide. It will not move. Even if it equals this value, the net force will still be 0. It has to be just above. So if it's 51.46, it will move. If it's 51.44, it's not going to move. 51.45. That's the threshold, so it really doesn't move at that point. So technically, the applied force has to be just above 51.45. But for our practical purposes, we're going to say this is the threshold value, so we'll go with that. Now what about part B? What is the acceleration of the system if a person pushes the box with a force of 90? So 90 is greater than 51.45. So the box will begin to slide. So therefore we no longer have static friction present. Because the box is sliding, now we have kinetic friction. Whenever you want to find the acceleration, write an expression for the net force, in this case in the x direction. So this is going to be positive because it's directed towards the positive x-axis, and this is going to be negative since it's directed towards the negative x-axis. The net force, based on Newton's second law, is mass times acceleration. And fk is mu k times normal force, where the normal force is mg. So now we can calculate the acceleration. So the mass is 15, the applied force is 90, mu k is 0.20, m is still 15, and g is 9.8. Let's multiply 0.2 times 15 times 9.8. So that's 29.4. 90 minus 29.4 is 60.6. 60.6, which is F minus FK, that's the net force, by the way, if you needed to find it. So the acceleration is going to be the net force divided by the mass. 60.6 divided by 15. So that will give us an acceleration of 4.04 meters per second squared. And so that's it for this problem. Now let's look at the second example. A force of 65 Newtons is needed to start an 8 kilogram box moving across a horizontal surface. Calculate the coefficient of static friction. So let's draw a picture. So here's the box. It's 8 kilograms in mass and we need to apply a force. And that force is going against static friction. So if this is the minimum force that is necessary to cause its move, then we can say that F is equal to the maximum value of Fs. So the maximum value of F of s is mu s times the normal force. So just like before, it's going to be mu s times mg. The applied force is 65. In this example, we're looking for mu s. m is 8, g is 9.8. so let's multiply 8 times 9.8 and you should get 78.4 so mu s is going to be 65 divided by 78.4 which is pretty high.829 and so that's the coefficient of static friction now let's move on to part b If the box continues to move with an acceleration of 1.4 ms2, what is the coefficient of kinetic friction? So let's replace this with FK. So anytime you're dealing with forces and acceleration, it's helpful to write an expression with the net force. The net force is going to be F minus FK. And the net force is MA. And FK, we know it's mu K times the normal force, which is mg. So our goal is to find mu K, or to find the value of mu K. So M is 8, the acceleration is 1.4, the applied force is still 65. Because once you have a force of 65, it begins to move. And so that force of 65 will continue to apply to the 8 kg box. 8 times 1.4 That's 11.2. And we said 8 times 9.8, that's 78.4. times mu k. So now let's subtract both sides by 65. So 11.2 minus 65, that's negative 53.8, and that's equal to negative 78.4 times mu k. So to calculate mu k, we got to divide both sides by negative 78.4. So negative 53.8 divided by negative 78.4 will give us a mu k value of 0.686. So as you can see, mu s is almost always greater than mu k. I haven't seen an example where mu k is greater than mu s. And so now you know how to calculate it. You can use the same formulas as what we used in the last example. Number 3. A force of 150 Newtons pulls the 30 kg box to the right as shown below. If the coefficient of kinetic friction is 0.25, what is the horizontal acceleration of the box? So go ahead and try this problem. So now, based on this problem, we can tell that the box is moving. So, there's kinetic friction. Plus, the question asked us to look for the coefficient of kinetic friction. So, we have to assume that the box is in motion. Now, what do we need to do in order to find the horizontal acceleration? Well, let's write an expression for the sum of all forces in the x direction. F is not directly in the x direction, but a component of F, which we'll call F of x, is. So the sum of all forces in the x direction is going to be this value minus that one. This value is going in the positive x direction, so it's going to be positive f of x. And this one is in the negative x direction, so negative fk. Now we know this force, based on Newton's second law, is equal to mass times acceleration. f of x is f cosine theta. Now what about FK? FK is mu K times normal force. Now what is the normal force in this problem? In this example, the normal force does not equal mg. Make sure you understand why. Now to understand this, let's go over a few things. So let's say if we have a 5 kilogram box. The weight force of this box is going to be 5 times 9.8, which is 49. Now in order for the box to rest on a horizontal surface, the net force in the y-direction has to be 0, which means the normal force has to be equal to the weight force. So whenever you have a box on a horizontal surface, the normal force is equal to mg. Now, what happens if you take the same box and if you apply a downward force of 10 newtons? What's going to happen? Now, we still have a weight force of 49 newtons. But what's the normal force now? Before, the surface only needed to support the weight of the object, which is 49 N. But now, the surface not only has to support the downward weight force of the object, but it must also support the downward force that you have to support. apply as you press down on the object. So the normal force increases anytime you press the block against the surface. So now the normal force is 59 N. Now what about if we take a rope and we pull, if we try to lift up the box with a force that's less than the weight force. So let's say the weight force is still 49, but the upward tension force is 20 N. What's the normal force now? In this case, the normal force is going to be less than 49 because it doesn't have to fully support the weight of the object on its own. The tension force supports 20 newtons out of the 49 newtons of weight that the object has. So the normal force has to support the other 29. So basically, the sum of the upward forces must equate to the sum of the downward forces. So here's what you want to take from this. Anytime you press down on an object, you increase the normal force. When you try to lift it up... the normal force decreases. And in this example, this block is being lifted up by the y component of the force. And so it is this y component that changes the normal force, it decreases it. So that's why the normal force doesn't equal mg, as it usually does in other examples. But anytime you have a force that partially lifts up the block, it decreases the normal force. So we have to come up with an expression to calculate Fm. We have to take this into account. Now let's draw some other forces that are on this box. So we have the downward weight force, and we have an upward normal force, plus the upward y component of the applied force. So if we write an expression dealing with the sum of all forces in the y direction, it's going to be Fn, it's upward so it's positive, plus Fy, and the weight force is downward. Now, the net force in the y direction is going to be 0 if this force does not exceed the weight force. So we can do a quick test. The weight force is 30 times 9.8, which is 294. Fy is going to be 150 times sine 30, which is 75. So Fy doesn't exceed the weight. If it did, this object would be lifted above the ground. So it remains in contact with the surface. So if it's not being lifted up, we could say the sum of all forces in the y direction will be equal to 0. There's no acceleration in the y direction. So this is 0. And solving for Fn, we need to subtract both sides by Fy. And we need to add W to both sides. So therefore, the normal force for this particular problem is going to be the weight force, which is mg, but minus this force, fy. As we said, any time you try to lift up the object, the normal force is going to decrease, and that's why we have the minus sign. If we apply a downward force, this will be a plus sign, because the normal force would increase. so let's calculate the normal force first that's going to be the mass of 30 times 9.8 minus FY which is F that's 150 times sine 30 so we know 30 times 9.8 we said that's 294 and 150 times sine 30 that's 75 So 294 minus 75, that's 219. So the normal force in this example is 219 newtons. Now let's use that to calculate the horizontal acceleration. So m is 30, f is still 150, times cosine 30, that's going to give us fx, and then minus mu k, which is 0.25, times the normal force of 219. Let me just separate these two parts of the problem. So 150 cosine 30, that's 129.9. And 0.25 times 219, that's 54.75. So if we subtract those two numbers... This will give you 75.15, and so that's equal to 30 times the acceleration. So the acceleration in the x direction is 75.15 divided by 30, which is 2.505 meters per second squared. So this is the answer.

But eventually, if you continue... to push it with even more force it will begin to slide and once begin sliding it's easier to continue pushing it but once you stop it it's going to be hard to start it up again the force that prevents the box from sliding at the first place or in the first place is the static frictional force now once it begins to slide the force that impedes the motion is kinetic friction Static means not moving. Kinetic has to do with motion.

So the frictional force that opposes motion when the object is sliding against the carpet, that's kinetic friction. The frictional force that prevents you from moving it when you first try to push it, that's static friction. Static friction, let me write the equations on the right.

Static friction is less than or equal to mu s times the normal force. Kinetic friction, it's equal to mu k times the normal force. So notice that the static frictional force is represented by an inequality, which means that it's not just a fixed number.

It can be a range of numbers up to a maximum point. The kinetic frictional force, however, is not represented by an inequality. So therefore, FK represents a fixed number. Now let's say that this is a five kilogram box. What is the normal force?

And calculate these two values with this information. Now this box exerts a downward weight force, and the normal force has to support the weight force. Keep in mind the normal force is a force that the surface exerts on the box, and it's perpendicular to the surface.

So in this example, for a horizontal surface, the normal force is equal to the weight force, or mg. So it's going to be 5 kilograms times 9.8 meters per second squared. So the normal force is 49 newtons. Now, I need to give you values for mu s and mu k.

Mu s is typically greater than mu k. I haven't seen an example when it's less. So we're going to say that mu s is 0.4.

And we're going to say mu k is 0.2. So let's calculate f k and f s. So the kinetic frictional force is going to be 0.2 times 49. which is 9.8. The static frictional force is going to be less than or equal to 0.4 times 49, which is 19.6. So what do these numbers mean?

So let me give you an example that's going to illustrate this. let's make a table between the applied force the static frictional force the kinetic frictional force and the net force So, if the applied force is zero, the frictional forces will be zero, and the net force will be zero. If the person doesn't push the box, the box won't move. Nothing's going to happen.

But now, let's say if the person applies a force of, let's say, 10 newtons. What's going to happen? Will the box begin to slide? And what are the values of the static and kinetic frictional forces?

Now, even though kinetic friction is 9.8, that value only applies if the box slides. Now, what is the static frictional force? Will the box move?

Notice that the applied force is less than the maximum static frictional force, so it's not going to move. Which means that there is no kinetic frictional force. The kinetic frictional force only exists If the box is sliding against the carpet, you can't have static friction and kinetic friction present at the same time. If the box is not sliding against the carpet, static friction is present.

If it is sliding, kinetic friction is present. So what is static friction in this example? What number should you put here?

Now, if you're thinking about putting 19.6, that will not be correct. Because imagine if the person applies a force of 10 newtons to the right to push the box, and if static friction applies a force of 19.6 newtons, that means that there's going to be a net force of 9.6 newtons towards the left. So imagine pushing the box, only to find that the box is pushing you back to the left.

That just doesn't happen. So static friction can't be 19.6. It turns out that static friction...

It's going to match the applied force until you exceed its maximum value. So if you push it with 10 newtons, it's going to push back on you with 10 newtons. And so the box doesn't move. You're trying to push it, but it doesn't move. So the net force is zero.

So let's say if you try to push it with 15 newtons, then it's going to push back on you with 15 newtons. That means you haven't applied enough force to start moving it. So that's what happens when you try to push the box initially. You're pushing hard against it, but it's not moving.

Because static friction is matching your applied force until it reaches the maximum value. So this is still going to be 0. So let's say if you apply a force of 19.6 newtons, it's still going to be 19.6. Now, when you exceed 19.6, that's when it begins to slide. And so you no longer have static friction, but you have kinetic friction.

So let's say if we go just above 19.6, let's say if we increase it to 20 newtons. Now the box begins to slide, and so there is no more static friction because the surfaces are sliding past each other. But there is kinetic friction, which is always going to be 9.8 once the box begins to slide.

So it's 20 minus 9.8, which will give you a net force of 10.2. Now, if you decide to increase the applied force, the static frictional force will still be 0, Fk is going to still be 9.8, and the net force is now 30 minus 9.8, which is 20.2. So hopefully this example helped you to understand the difference between static friction and kinetic friction, and how to calculate it based on the applied force. Let's work on this problem. A 15 kg box rests on a horizontal surface.

What is the minimum horizontal force that is required to cause the box to begin to slide if the coefficient of static friction is 0.35? So you can pause the video if you want to work on this problem as well. But let's start with a picture. So this is the 15 kg box.

So we wish to apply a horizontal force, which we're going to call capital F, and we know that friction is going to oppose it. Now, if we want the minimum horizontal force that is required to cause the box to begin to slide, we need to use static friction because until the force exceeds static friction, only then can it slide. If it doesn't exceed the static frictional force, then it won't slide. So the threshold is the maximum static frictional force. So we need to set F equal to FS and that's when it begins to slide.

when these two have the same magnitude. Now the static frictional force, its maximum value is mu s times the normal force. And in this example, the normal force is going to be mg.

So mu s is 0.35, m is 15, and g is 9.8. So the applied force has to be 51.45 Newtons in order for the box to begin to slide. If it's less than this value, the box will not slide. It will not move.

Even if it equals this value, the net force will still be 0. It has to be just above. So if it's 51.46, it will move. If it's 51.44, it's not going to move. 51.45. That's the threshold, so it really doesn't move at that point.

So technically, the applied force has to be just above 51.45. But for our practical purposes, we're going to say this is the threshold value, so we'll go with that. Now what about part B? What is the acceleration of the system if a person pushes the box with a force of 90? So 90 is greater than 51.45.

So the box will begin to slide. So therefore we no longer have static friction present. Because the box is sliding, now we have kinetic friction. Whenever you want to find the acceleration, write an expression for the net force, in this case in the x direction. So this is going to be positive because it's directed towards the positive x-axis, and this is going to be negative since it's directed towards the negative x-axis.

The net force, based on Newton's second law, is mass times acceleration. And fk is mu k times normal force, where the normal force is mg. So now we can calculate the acceleration. So the mass is 15, the applied force is 90, mu k is 0.20, m is still 15, and g is 9.8.

Let's multiply 0.2 times 15 times 9.8. So that's 29.4. 90 minus 29.4 is 60.6.

60.6, which is F minus FK, that's the net force, by the way, if you needed to find it. So the acceleration is going to be the net force divided by the mass. 60.6 divided by 15. So that will give us an acceleration of 4.04 meters per second squared. And so that's it for this problem.

Now let's look at the second example. A force of 65 Newtons is needed to start an 8 kilogram box moving across a horizontal surface. Calculate the coefficient of static friction.

So let's draw a picture. So here's the box. It's 8 kilograms in mass and we need to apply a force. And that force is going against static friction.

So if this is the minimum force that is necessary to cause its move, then we can say that F is equal to the maximum value of Fs. So the maximum value of F of s is mu s times the normal force. So just like before, it's going to be mu s times mg. The applied force is 65. In this example, we're looking for mu s.

m is 8, g is 9.8. so let's multiply 8 times 9.8 and you should get 78.4 so mu s is going to be 65 divided by 78.4 which is pretty high.829 and so that's the coefficient of static friction now let's move on to part b If the box continues to move with an acceleration of 1.4 ms2, what is the coefficient of kinetic friction? So let's replace this with FK.

So anytime you're dealing with forces and acceleration, it's helpful to write an expression with the net force. The net force is going to be F minus FK. And the net force is MA. And FK, we know it's mu K times the normal force, which is mg.

So our goal is to find mu K, or to find the value of mu K. So M is 8, the acceleration is 1.4, the applied force is still 65. Because once you have a force of 65, it begins to move. And so that force of 65 will continue to apply to the 8 kg box.

8 times 1.4 That's 11.2. And we said 8 times 9.8, that's 78.4. times mu k. So now let's subtract both sides by 65. So 11.2 minus 65, that's negative 53.8, and that's equal to negative 78.4 times mu k. So to calculate mu k, we got to divide both sides by negative 78.4.

So negative 53.8 divided by negative 78.4 will give us a mu k value of 0.686. So as you can see, mu s is almost always greater than mu k. I haven't seen an example where mu k is greater than mu s.

And so now you know how to calculate it. You can use the same formulas as what we used in the last example. Number 3. A force of 150 Newtons pulls the 30 kg box to the right as shown below. If the coefficient of kinetic friction is 0.25, what is the horizontal acceleration of the box?

So go ahead and try this problem. So now, based on this problem, we can tell that the box is moving. So, there's kinetic friction.

Plus, the question asked us to look for the coefficient of kinetic friction. So, we have to assume that the box is in motion. Now, what do we need to do in order to find the horizontal acceleration? Well, let's write an expression for the sum of all forces in the x direction.

F is not directly in the x direction, but a component of F, which we'll call F of x, is. So the sum of all forces in the x direction is going to be this value minus that one. This value is going in the positive x direction, so it's going to be positive f of x. And this one is in the negative x direction, so negative fk. Now we know this force, based on Newton's second law, is equal to mass times acceleration.

f of x is f cosine theta. Now what about FK? FK is mu K times normal force.

Now what is the normal force in this problem? In this example, the normal force does not equal mg. Make sure you understand why. Now to understand this, let's go over a few things.

So let's say if we have a 5 kilogram box. The weight force of this box is going to be 5 times 9.8, which is 49. Now in order for the box to rest on a horizontal surface, the net force in the y-direction has to be 0, which means the normal force has to be equal to the weight force. So whenever you have a box on a horizontal surface, the normal force is equal to mg. Now, what happens if you take the same box and if you apply a downward force of 10 newtons?

What's going to happen? Now, we still have a weight force of 49 newtons. But what's the normal force now?

Before, the surface only needed to support the weight of the object, which is 49 N. But now, the surface not only has to support the downward weight force of the object, but it must also support the downward force that you have to support. apply as you press down on the object. So the normal force increases anytime you press the block against the surface. So now the normal force is 59 N.

Now what about if we take a rope and we pull, if we try to lift up the box with a force that's less than the weight force. So let's say the weight force is still 49, but the upward tension force is 20 N. What's the normal force now? In this case, the normal force is going to be less than 49 because it doesn't have to fully support the weight of the object on its own.

The tension force supports 20 newtons out of the 49 newtons of weight that the object has. So the normal force has to support the other 29. So basically, the sum of the upward forces must equate to the sum of the downward forces. So here's what you want to take from this.

Anytime you press down on an object, you increase the normal force. When you try to lift it up... the normal force decreases. And in this example, this block is being lifted up by the y component of the force.

And so it is this y component that changes the normal force, it decreases it. So that's why the normal force doesn't equal mg, as it usually does in other examples. But anytime you have a force that partially lifts up the block, it decreases the normal force.

So we have to come up with an expression to calculate Fm. We have to take this into account. Now let's draw some other forces that are on this box. So we have the downward weight force, and we have an upward normal force, plus the upward y component of the applied force.

So if we write an expression dealing with the sum of all forces in the y direction, it's going to be Fn, it's upward so it's positive, plus Fy, and the weight force is downward. Now, the net force in the y direction is going to be 0 if this force does not exceed the weight force. So we can do a quick test.

The weight force is 30 times 9.8, which is 294. Fy is going to be 150 times sine 30, which is 75. So Fy doesn't exceed the weight. If it did, this object would be lifted above the ground. So it remains in contact with the surface.

So if it's not being lifted up, we could say the sum of all forces in the y direction will be equal to 0. There's no acceleration in the y direction. So this is 0. And solving for Fn, we need to subtract both sides by Fy. And we need to add W to both sides. So therefore, the normal force for this particular problem is going to be the weight force, which is mg, but minus this force, fy. As we said, any time you try to lift up the object, the normal force is going to decrease, and that's why we have the minus sign.

If we apply a downward force, this will be a plus sign, because the normal force would increase. so let's calculate the normal force first that's going to be the mass of 30 times 9.8 minus FY which is F that's 150 times sine 30 so we know 30 times 9.8 we said that's 294 and 150 times sine 30 that's 75 So 294 minus 75, that's 219. So the normal force in this example is 219 newtons. Now let's use that to calculate the horizontal acceleration. So m is 30, f is still 150, times cosine 30, that's going to give us fx, and then minus mu k, which is 0.25, times the normal force of 219. Let me just separate these two parts of the problem. So 150 cosine 30, that's 129.9.

And 0.25 times 219, that's 54.75. So if we subtract those two numbers... This will give you 75.15, and so that's equal to 30 times the acceleration.

So the acceleration in the x direction is 75.15 divided by 30, which is 2.505 meters per second squared. So this is the answer.

Transcript for:Understanding Static and Kinetic Friction

Transcript for:
Understanding Static and Kinetic Friction