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Anderson Video - Impulse Problem

Professor Anderson
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>> Hello class. Professor Anderson here. Let's try a simple impulse problem. And we are going to apply a force to a mass. And we want to see how fast that mass will be moving after some amount of time. And let's say that this is a frictionless surface that it's on. And we're going to apply a force in this direction. And we're going to do it for some amount of time: Delta T. If we applied this force on an object that starts at rest -- and we can say the Vi equals 0. We are looking for the final speed of the block. And we will say that we are given some mass. We are giving the amount of time that the force is applied. And we are given the force F. All right. And just for kicks, we can try some real numbers. Let's say that M is 2 kg, Delta T is 10 seconds, and F is 5 Newtons. And let's see if we can figure out what Vf is. So, it's -- should be obvious to you that if I'm going to push on this thing to the right, it's going to end up moving to the right. So, at some later time, it's got some Vf to the right. It started at rest. Vi is equal to zero. And now, how do we calculate this? Well this is an impulse problem. So, we go back to our definition of impulse. Impulse is MVf minus MVi. And that is equal to the integral of Fdt. But if f is a constant that whole time, then this becomes F Delta t. All right. Let's solve this thing for Vf. MVf minus Vi, equals F Delta t. I'm going to stop writing the vector arrows just for simplicity. Vi we said was 0 so that goes away. So Vf is F Delta t, divided by M. And now we know all those numbers. All right. F is 5 Newtons. Delta t is 10 seconds. M is 2 kg. All of those are SI units so we don't have to write the units down. Okay. And we get 25 meters per second. Good. Let's say that we make it slightly more complicated now. And let's add some friction to the problem. All right. Let's add some friction to the problem here. We'll make the surface kind of rough. And we will say that uK is some small value maybe it's 0.1. And let's ask the same question now. What is Vf? Well, when we go to our impulse equation: J. What we said was this is M, Vf minus Vi, which is the integral of F dt. But what F do we really put in there? Okay. There's obviously more than one force at play here. There is not only the pushing force F, but there is friction that's acting against us. Right. Friction is trying to prevent our movement. And so this really needs to be the net force. And you need to include the net force in that equation. And in this case, we have two opposing forces. So, J is no longer just F. It is in fact F minus the friction force. F is to the right. We know that the friction force is to left. It's trying to oppose the motion. Both of those are constant. And so, they can come out of the integral. And so, we can write this as F minus Fk Delta t. And we know from before that that's equal to M V final if it started from rest. VI is equal to zero. And so, we can write Vf is equal to F minus f sub k all over M. And I'm going to multiply that by Delta t. But we also know what Fk is. Fk is uk times a normal force which is just the weight of the thing; mg. And so, this simplifies a bit. One of the Ms cancels out right there. And we get F over M minus ukg. All of that, times Delta t. Whenever you get an answer like this, you should always double check that it gives you the right units. Okay. And that it gives you the right limits. So, if uk goes to 0, this is the manner for frictionless. We should be back to where we were originally. Which is good. This term goes away. We get F over M times Delta t which is what we had before. And now you can plug in some numbers, and let's see how it works out. All right. Let's plug in some of our numbers and see what we get. So, we said that F was 5 Newtons, the mass was 2 kg, uk we said 0.1, g is of course 9.8, and we do this for 10 seconds. So, we are all in SI units there. Five over 2 is 2.5. 9.8 times 0.1 is 0.98. We're going to multiply this whole thing by 10. And 0.98 is pretty close to 1. So, that would be 2.4 here. And we're going to multiply it by 10. So, it's 24 meters per second. Now that is certainly less than we had in the no friction case. Which is good. Right. With no friction, we had a Vf of 25 meters per second. And now with a little bit of friction, not a lot, we get Vf is equal to 24 meters per second. And obviously as you increase the friction, you're going to increase this number. You going to slow that object down even more and this number will keep on dropping. And in fact, if you make the friction high enough, you can get to a number where this goes to 0. And in that case, we would be pushing along at a constant velocity. The force that you're applying is exactly opposite to the friction that's trying to slow it down. Questions about this approach? Everybody okay with this? Okay.