Motion as a function of time: Friction example

0:00- [Instructor] A block of mass M slides across a flat table. The coefficient of kinetic friction between the block and the table surface is mu K, and at T equals zero, the block is moving to the right at speed V zero. Find the equations for the block's acceleration, velocity, and the position as a function of time until it comes to rest.

0:25So we are given a scenario, and our goal is to build equations for the acceleration, velocities, and the positions, so how do we do that? Well, the first and the big question for us is, what exactly is this mu K, the coefficient of kinetic friction? Well, we probably know that, whenever something is sliding, or it's attempted to slide on a surface, there is a force that tends to oppose that sliding, or attempted sliding.

0:51We call this force the force of friction, and that's one of the reasons why, if this block was sliding, then it will eventually come to rest. But what does the magnitude of this frictional force depend on? Well it turns, out it depends on, it's actually proportional to the normal force.

1:09Does this make sense? Well, let's see. If somebody was sitting on this block, we know, intuitively, it's much harder now to get it to slide. Why is that? Well, that's because the normal force would have increased. The block is now pressed more into the floor, and so you have higher value for the normal force, and as a result, the frictional force also increases, so the frictional force is proportional to the normal force, and that proportionality constant is what we call the coefficient of kinetic friction, and why do we call it kinetic?

1:44Well, that's because the block is already sliding. Whenever objects are sliding, we use the coefficient of kinetic friction. It's a constant for a given pair of surfaces, and you can see the property of this force now. As long as the block is sliding, look, the magnitude of this force is going to be a constant, but once the block comes to a stop, now, it's no longer sliding, we no longer use this.

2:08In fact, once the block comes to a stop, the force of friction disappears, so as long as it's sliding, the force of friction is a constant, and what about the direction of the force? Well, it opposes sliding motion, so in this particular case, the block is sliding to the right, and so the direction of the frictional force would be to the left.

2:27Okay, so now that we understand the frictional force, its magnitude and direction, how do we tackle this problem? Well, we can draw a free body diagram to figure out what the net force is in the horizontal, and then by using Newton's second law, we can connect it to the acceleration in the horizontal, so from there, we can get the expression for the acceleration, and then once we have that, we can find the expression for the velocity, and from there, we can find the expression for the position, so that's going to be our line of attack.

2:58So let's do this, and it would be a great here to see if you can pause and try to draw a free body diagram for this block yourself first. All right, let's do this. Before we do that, we have to choose one direction to be positive, we have to put our coordinates and choose one direction to be positive.

3:14Let's choose the right side to be positive, and we'll also assume the block is at zero initially. At time equals zero, the block is at zero. Okay, so now let's draw the free body diagram. What are the different forces acting on it? Well, in the vertical, I know that there's a force of gravity, so that's MG, and the floor is pushing up on it.

3:32That's the normal force. Okay, what about in the horizontal? In the horizontal, we know there's a frictional force acting. What direction is it? It always opposes sliding motion, or attempted sliding, for that matter. This object is already sliding to the right, so the frictional force will be to the left, and we know its magnitude, it's going to be mu K times FN.

3:50Okay, since we're interested in the acceleration in the horizontal, we can now find the net force in the horizontal, which is just this force. There's only one force acting in the horizontal. We can use a Newton's second law to connect it to the acceleration. Newton's second law says net force equals mass times acceleration, and from this, I can calculate what the acceleration in the horizontal is.

4:10But before I do that, I need to know what the normal force is going to be, because the force of friction depends on that. So to do that, I need to first apply Newton's second law in the vertical, so can you pause the video and apply Newton's second law in the vertical and figure out what the normal force is going to be?

4:29All right, let's do this, so in the vertical, let's try to do this in our head, in the vertical, the acceleration is zero, so the net force in the vertical must be zero. In other words, the two forces must cancel out, so the normal force must be exactly equal to the force of gravity, and therefore, the normal force must also be MG.

4:48Okay, now that I know what the normal force is, it's MG, I can plug this in over here, and I have now figured out what the force of friction is going to be. All right, so let's apply Newton's second law in the horizontal now. We would get, in the horizontal net force is the force of friction, but it's negative because we've chosen right side to be positive, and the frictional force is to the left, so the negative force of friction should equal mass times the acceleration in the horizontal, which is just AX of T, so that's our equation in the horizontal, and notice we have gotten rid of the vector sign, because we're using signs to represent the direction.

5:23So from this, acceleration is just F by M, negative F by M, and if we plug in for the force of friction, it's going to be mu K MG, so you'll get negative mu K MG, divide by M, the mass cancels out, and as a result, we get the acceleration to be negative mu K G.

5:40We have found the expression for the acceleration. Notice that it's a constant! It does not depend on time. That makes sense, because the net force is a constant, it does not depend on time, so we'd expect the acceleration to be a constant as well. We see the acceleration is a negative direction.

5:57That makes sense, we expect this block to slow down, so we expect the acceleration to be in the opposite direction of the velocity. Okay, can we plot an acceleration time graph over here? Yeah, what would it look like? Well, it's a constant, so it's going to be a straight line, and it's negative, so it's gonna be below the time axis, so it's gonna be minus mu K G, and it's gonna be a constant, so there we have it.

6:21We have the expression for the acceleration. Okay, now, how do we calculate velocity from the acceleration? Well, we know from kinematics that change in velocity is the integral, the definite integral of acceleration, so let's use this to find the expression for velocity.

6:41Okay, what is change in velocity? Change in velocity is the final velocity minus the initial velocity. Final velocity, we'll just call it as the velocity at some time, T, and initial velocity is the velocity at time T equals zero, so we have T minus, we have zero in general, and what would that equal to?

6:58Well, that equals to the integral from where to where? Well, we are going from T equals zero to T, those are the bounds, from zero to T, and that would be the acceleration, and notice I have chosen T prime over here, and the only reason to do that is because we're already using, you're using T as an upper bound, so just to not confuse it with the variable T, I've chosen a different variable, T prime.

7:21You can choose any variable you want, whether you call it T or T prime or tau, whatever, doesn't matter, so I've just changed the variable to T prime, just to make sure it's not confused with the upper bound T. All right, so now we can plug in. What is the final velocity?

7:36Since we're only considering in the X direction, our final velocity is going to be VX of T. Our initial velocity is V zero, and that equals the integral from zero to T. What is the acceleration? What is the function of acceleration? What is A of T prime? Well, we know acceleration is just a concept, it's negative UKG, so it's gonna be negative UKG times DT prime.

7:59So we can now integrate this, and we can find an expression for the velocity. Again, great idea to pause the video and see if you can try this yourself. Okay, let's do this. We can just pull this out of the integral, because this is a constant, so if you do that, we get minus mu K G integral of zero T DT prime.

8:18What's the integral of one with respect to T prime? Well, that's just T prime, so that's going to be T prime, but of course, there'll be bounds, so zero to T, and then how do we solve this? Well, we first plug in the upper bound minus the lower bound. If you plug in the upper bound, you'll get T, lower bound with zero, so it's gonna be T minus zero, it's just gonna be T, so the right hand side just becomes minus mu K G times T, and that would lead us to this equation, so we have found the expression for velocity, velocity as a function of time.

8:48Again, what would the velocity time graph look like? Well, at T equal to zero, the initial velocity is V zero. That's what we had, so we'd start with some V zero, and then as time progresses, that velocity reduces, and you can see there is a minus sign over here, and eventually, it goes to a stop, and it's a linear equation, so it's gonna be a straight line, so eventually, the velocity goes to zero, and the slope of that straight line is this number, minus mu K G.

9:18And if you look at this, you might say, "Hmm. Hey, this is a familiar equation." We know V of T equals V zero plus AT, where this is A, so this is the kinematic equation for constant acceleration, of course, because the acceleration is a constant. You may be like, "Well, we could have directly used that."

9:35Yes, we could have, but this is the most general way to do this. This method would work, even when the acceleration is not a constant. That's why we did it in the most general way over here. All right, now let's see if we can find the position function from the velocity function.

9:52How do we do that? Well, we know that the change in position is the definite integral of velocity, so let's plug in and see if we can find the expression for position. So what is change in position? Well, it's the final position, which we'll call X of T, minus the initial position, which we can call X of zero in general, and that equals the integral.

10:11What are the bounds from zero to T? And again, because the upper bound is T, we're gonna use a different dummy variable over here. Instead of V of T, we'll just call it V of T prime, so it's gonna be V of T prime, VX of T prime, because we're dealing with X direction, times DT prime.

10:26Okay, so let's plug that in. What is VX of T prime? Well, we know what VX of T is. That's just V naught minus mu K GT, so VX of T prime is going to be V naught minus mu K GT prime. So if you plug that in, this is what you'd get, and now, let's try to simplify.

10:44So what happens to the left hand side? Well, X of T is X of T, that's what we're gonna find out, but what is X of zero? Well, it's the initial position, and remember, we chose our coordinates in such a way that our initial position, X of zero, is just zero. At time T equals zero, the position was zero, so we can just make this zero, so that's the left hand side, so left hand side is just X of T, but what is the right hand side?

11:07We now have to evaluate this integral. Let's do this step by step. What's the integral of V naught with respect to T prime? Well, V naught is a constant, so we can pull it out, and the integral of one with respect to T prime is, again, T prime, so we'd get V naught T prime to begin with minus mu K G, which is a constant, which can be pulled out, and then we have to integrate T prime DT prime.

11:30What's the integral of T prime DT prime? Well, it's a half T prime square, so we get minus half mu K GT prime squared, and of course, we have to consider the bounds, so we substitute the upper bound, that's T, and when we substitute the lower bound, we get zero, so this expression just reduces to V naught T minus half mu K GT squared, and there we have it.

11:54That is the expression for the position, and again, if you look at this carefully, you'll see that is the kinematic equation. The position function, the position at any time T would be equal to initial position, which is zero in this case, plus V naught T plus half AT squared.

12:13A over here is negative mu K G, so what would the position function look like? Well, there's a T square over here, so this is a quadratic equation, so when you plot a quadratic equation, you get a parabola, so we get a parabola. So there you have it. We start with the free body diagram, calculate the net force, equate it to the acceleration using Newton's second law.

12:38From acceleration, we calculate velocity, and from velocity, we calculate position.