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Kai C.

(I) A constant friction force of 25 N acts on a 65-kg skier for 15 s on level snow.What is the skier's change in velocity?

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Aditya P.

(I) A 7150-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s. A 3350-kg load, initially at rest, is dropped onto the car. What will be the car's new speed?

Muhammed S.

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welcome to our second example video looking at special sources of centripetal force. So centripetal force in the vertical direction is a little different from centripetal force that we consider before that was all horizontal. For example, if you consider a ball being slung around at the end of a string, or if we consider our loop the loop again when we draw, these are free body diagram zehr going to look different. For example, straight up in the case of the string have f t down, but I also have f g down when it's exactly down. I would have FT up and F g here. If it's exactly to the side, I would have FTN and F G straight down similarly over here with the loop the loop, we would have normal four straight up and M g down or at the top would have normal force down plus M g on the side. We have normal force, then mg. You can see that these are pretty much the same situation is just that on the one we're using Normal force and on the other were using force of tension. But mathematically they'll turn out to be exactly the same. So if we were to write out some equations of motion here, let's say we dio here, here and here, we would say Okay, well, for for a I've got f t minus F g has to be equal to m B squared over r for B. I have negative f t minus F g is equal to negative MV squared over R because in this case, centripetal acceleration is pointed straight down. And then for C notice that I have centripetal acceleration in ft in the same direction. But F g is perpendicular, which means it wouldn't show up in the same equation I'd have f t is equal to m B squared over our Okay, So what is going on here, then? Well, it's easy to see on the left hand side that we actually have a tangential force, which means that the rate of rotation theme Alfa there's an Alfa here. So Omega is changing as we go around the circle. It's very interesting, so we have a tangential component to our force. But then we also have a centripetal component and that centripetal components changing here so two things could be happening either one it could be changing how fast it's moving or to the tension in the string might be changing if you've ever done this with a ball and swing it around, you can also do this with a bucket if you take a bucket with some water in the bottom and you swing it around really fast, Um, then you can keep the water there in the bottom of the bucket, and you can feel this is it goes around and around and around. It's almost easier because notice that we have. If I were to solve for F T. In both these cases, I'd have F T schools MV squared over our plus F G and then down here when I saw for I'll have F t is equal to m V squared over r minus F g. So the tension force is going to be different if we're maintaining a constant velocity here now. In truth, um, probably several of these variables are changing. Maybe these changing a little bit tensions changing a little bit could even be changing a little bit. But these air, at least the relationships that we have set up here, so we know how they relate to each other. And it's our job to read the problem carefully and see what's happening So we can interpret how we need to use these relationships, which again are just Newton's second law using a centripetal acceleration here now we could if we wanted to, for example, in and see we could then go on to say, Oh, well, there's a tangential force which is equal to mass times tangential acceleration and in this case, that be equal to M. G. So our tangential acceleration would be equal to G eso. So we would need to consider issues like this that we do definitely have a change in velocity because we have a tangential acceleration in order to do all this requires some heavy duty calculus, so we're not going to do it here. And it's generally not covered in a physics 101 course. Um, but if you go on to classical mechanics or later style physics like that, you'll see it

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