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Question '(6.25 points) Frank is driving his car on a flat country roed ind aboutto 055 21 Dieoe section of the road; Amazingly You knaw that the rafs of the o...

Question

Question '(6.25 points) Frank is driving his car on a flat country roed ind aboutto 055 21 Dieoe section of the road; Amazingly You knaw that the rafs of the oune s {azdte coefficient of static friction between the car tTES and tne {031t07i G5t3 information what is the maximum speed t0 the neatest 0nl na Frt0jal the curved section of the road without skicd ng? (Do not motde Deuninte answer)Your Answer;AnswerAcrnFaa 3 8 (6.251 polnts) Sunb 16o0r3- Ree Fhn teft Question belt around E SCRC the

Question '(6.25 points) Frank is driving his car on a flat country roed ind aboutto 055 21 Dieoe section of the road; Amazingly You knaw that the rafs of the oune s {azdte coefficient of static friction between the car tTES and tne {031t07i G5t3 information what is the maximum speed t0 the neatest 0nl na Frt0jal the curved section of the road without skicd ng? (Do not motde Deuninte answer) Your Answer; Answer AcrnFaa 3 8 (6.251 polnts) Sunb 16o0r3- Ree Fhn teft Question belt around E SCRC the asterold! Calculate % the orbatal asteroid In [ OQuvot from the Suns foljns 6.4 distance that the asteroid [ (AUI 7#1988 Lornr 'assuming E m/s by {0l[ha. 5 * 1011 , The miss AU=15 Data: 131 Nm? /kg? 10 Your Answer Answer pownts} (6.25 0l ulb Wnjt 'shows $ sct € Question = 'RS _ ot Jm below = figure = Dalks h = The black ( bals? Karge ser 0f 0f this



Answers

What is the maximum speed at which a car can round a curve of $25-\mathrm{m}$ radius on a level road if the coefficient of static friction between the tires and road is $0.80$ ?
The radial force required to keep the car in the curved path (the centripetal force) is supplied by friction between the tires and the road. If the mass of the car is $m$, the maximum friction force (which is the centripetal force) equals $\mu, F_{N}$ or $0.80 \mathrm{mg}$, this arises when the car is on the verge of skidding sideways. Therefore, the maximum speed is given by $$ \frac{m v^{2}}{r}=0.80 \mathrm{mg} \quad \text { or } \quad v=\sqrt{0.80 \mathrm{gr}}=\sqrt{(0.80)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(25 \mathrm{~m})}=14 \mathrm{~m} / \mathrm{s} $$

So we're told that when a vehicle goes around a corner around a curve um the force that needs to be exerted to keep the vehicle on this curve path is just from friction. And so the path it's a flat road. So we just need we have the friction force and the friction forces. The coefficient of friction times the weight of the vehicle. And so that needs to be um well for it to not skid. This thing needs to be greater than this. So at the onset of skidding these two things are equal to one another. And so they want us to know they want to they want us to find the velocity at which skidding occurs. So that's what we call that Visa bar. And so we can see here that just setting these two things equal to one another, we have that M's cancel out. And then solving for V. P. R. We get that that is the square root of mu g. Times the radius radius of the circle that were going around. So we can see that obviously as the radius gets smaller, the speed at which we were slipped gets smaller. It's likewise as the friction gets lower and lower so that's why we want you know going around a tight corner we are more likely to slip and you know for the tire system on the road. And so they give us an example here of music. Was your four and G. In 33 ft per second squared. R. 32 3 per second squared. And our wasn't had recovered had a radius of 400 ft. So we're plugging out all in. We wind up with 71.6 ft/s is how fast we could be going before we start to slip slide out. And that turns out to be about 48 8 MPH. So depending on what kind of road this is obviously if it was on an expressway, this would not be good because these seem like reasonable values just seems like a reasonable value here. Um You know, probably so low, but you want to be careful and so we really would not want to have a radius that was like on a flat curve That was 400 ft. We definitely want to have to make a make a much larger curve um so that this would be much higher value.

For this problem, on the topic of force and motion were shown in the figure, an overhead view of a car's part as it travels toward the wall, The driver begins to break when the wall is 107 m from the car. And were given that the car's masses 1400 kg and its initial speed 35 m/s. The coefficient of static friction is 0.5. And we went to assume that the car's weight is distributed evenly on the four wheels. We want to find in magnitude of static friction required to stop the car just as it reaches the wall, the maximum possible static friction, The coefficient of kinetic friction between the tires and the road is 0.4, and you want to find the speed of the cow will hit the wall. Would and we want to find the magnitude of the frictional force that would be required to keep the kind is certainly part of radius D at the given speed. V. Not so that the car moves in a quarter circle and then parallel to the wall. And lastly, we want to know if the required force is less than the static frictional, the maximum static frictional force so that a circular part is possible. So firstly we'll use kinda Matics to find the deceleration of the car and we'll use the equation V squared. He's able to renounce squared plus to a. D. And so the final speed of the car zero. This is equal to its initial speed 35 m/s squared Plus two times acceleration a into 107 meters. This gives the deceleration of the car a to b minus five 0.72 m/km2. And so the force of friction required to stop the car. Little F. Is equal to um times the magnitude of the acceleration by Newton's Second law. This is the mass the car 1400 kg And the magnitude of the acceleration 5.72 m. Let's go wait a second. And so the frictional force has magnitude eight Times 10 to the power three Newton's for part B. We want to find the maximum possible static frictional force F. S. Max. The maximum possible static friction is equal to the coefficient of static friction. New s times the weight of the car. Mg. And we are given this to be 0.5, The mass 1400 Kg. And we know the acceleration due to gravity is 9.8 m/km2. And so this gives us the maximum possible value for static friction To be 6.9 Times 10 to the three Newton's for part C. We want to Find the speed that the cow will hit the wall if the coefficient of kinetic friction between the sliding tires and the road is 0.4. Now we know that the kinetic frictional force F. K. Is equal to the coefficient of kinetic friction, UK tens mg. And the deceleration is able to minus mu k times G. And so therefore the speed in which the car hits the wall V. Is equal to the square root of the north squared plus two A. D. Squared, using the same kind of magic equation as above. And this is the square root of 35 m/s squared minus two Times 0.4 times 9.8 m/km2 Times a distance of 107 meters. And so the speed at which the car will hit the wall is 20 m/s for part D. You want to find the magnitude of the frictional force that is required to keep the kind of circular part of radius D. At its given speed. Now the force required to keep the motion circular if our is equal to um Vinod squared over the radius of the orbit are This is the mass of the car, 1400 kg times its speed, 35 m per second squared, Divided by 107 meters. And so to keep the kind of circular orbit until it travels parallel to the wall Would require a force of 1.6 Times 10 to the four newton's. And lastly for part E, we want to know if the required force is less than the maximum static frictional force so that a circular part will be part possible. And we can see that this force is much greater than the maximum static frictional force. So this implies that no circular part is possible.

According to the given problem. I can write a question asked, happen is equal to m omega. Also I can like kmt e greater than equal to M B squared by our So the is smaller than equal to underwrote K R and D. So be max is equal to under route. He uh minimum G on solving it for the we know that radius of curvature for a car at any point tax why is given by R is equal to Model of one plus divide by dx squired All to the Power three x 2. Bye. The square by by dx square. So for the given carve, I can write the value of divide by the X is equal to a by alpha cause act by alpha and the square by by dx square is equal to minus a by alfa square signed X Y el palm on solving it further, I can also write the value of R is equal to one plus a square by Alfa Square bosses square act by Alpha. All to the power three x 2 by everybody. Alpha is square. Sign. Acts by Alpha. For minimum value of our odd minimum value of art. That's viol five is equal to five x 2. So corresponding radius of curvature can be written edge. Our minimum is equal to I'll play square by a mx is equal to who tended kg by a multiplication. Al part

Thank you. Consider a scenario call this scenario Part One. So in Part one, we have this nebulous mass, which is my attempt at drawing an asteroid Absolutely one of the better, more accurate drawings I've made. So we have force that acts on this asteroid. All this asteroid is traveling some displacement distance and rest. Variety of questions about it. First we want to consider um, which direction is this force in? So the asteroid's close down. Okay, so it stands to reason. Yeah, that's a little bit read skeptical of the youth drawing. It stands to reason that the work must be done in the opposite direction. Okay. Mhm. Now if I pushed in say the same direction that the asteroid is already going, it's conceivable that would be pushed forward. No. Is the factor of course is a positive or negative. Now, you know, the properties of factors vector that points outward that is considered a positive vector, whereas a vector that points inward that's considered negative. This is just by convention through a vector addition or multiplication vector that is negative. It's being subtracted if it's pop pointing outward it's positive. In this case our voice for force it's directed on the asteroid. Therefore points inward. It is negative party. Alright, okay. How does what type of energy changes as the object slows so familiar with the con on survey. Thank you. Energy is conserved throughout a system in which kinetic energy and potential energy are considered not not a whole lot else. So there's potential energy doesn't mean anything because there's no gravitational force out there. On the other hand, kinetic energy still means something. So traditionally kinetic energy on earth may as well just be a um half the mass times velocity squared. However in space you have a change of velocity so it's V. Two you want squared So we know that this component is changing as a result of the asteroid getting hit. So Connecticut is the energy. That is the fact that let's consider part day. Okay, what is the relation? Okay. How is how does the work done by the forest change the objects energy? Well, we refer to our answer from part from previous part. It changes kinetic energy and the effect was that this asteroid slowed. So the effect is that it decreases kinetic energy. Yeah. Now let's worry about cartoons cartel were given a variety of bits and pieces of information, all of which are still related to this asteroid. Mhm. All right. Mass of the asteroid is 45 times and fourth kilograms. The velocity of the initial velocity 7100 m per second. Uh This is the speed velocity Excuse me before. Well, the asteroid is hit by this other object or the sports tax on it. So after after for the velocity goes down to 5500 meters per second. So, let's find the work done by this force. Mhm. Yeah. Okay. So, conservation of energy equation. Kinetic energy plus opportunity and potential energy. You're well, you're not consider potential energy here because there is none gravity. Nothing change in kinetic energy 20 Yeah. All right, mm hmm. Now, let's consider what happens to the change in kinetic energy. Actually, it's not really equal to here. No, thanks. Yeah. Oh, no, I don't understand the kinetic energy. Part of the kinetic energy would be transferred to whatever is doing the force. So, that's why the kinetic energy would be concerned with this case here. But we are just considering the change in kinetic at energy of the asteroid. Okay. Call that has changed kinetic energy subscript asteroid. That's a better way to put it. Mhm. The change in kinetic energy of the asteroid. All right, let's go. So, the train kinetic energy of this at Detroit, it's called changing K E. Subscript A. Mhm should equal one half mask. I'm change velocity while she squared. So we know that the work done by this force is this change in kinetic energy. So let's just say changing the change in kinetic energy of the asteroid people is going to have massive times change in the last week. Let's go ahead and enter our quantities for this. Yeah, you should find its uh 4.5 cents the floors kilograms 7000 grip. It is 5500 for a second. Zero over there. A little bit difficult to read. Here we go. Anyways. Thanks for that. Should find that your solution for changing needs equal 2 36 mega drools. Remember this mega here. That's just in our six and it's being multiplied by about 36 to give us that money baggage rules. Now for the last part of this question, we're asking what is the force that actually this asteroid? Given that the distance traveled, it's about 1.8. Find 10 to 6 which is a long dream. That's a bit long. Okay, so we know that or force times distance. So, if you want your answer for force you take the work down or the change in energy in this case to buy a bike. All right, Acting that you subsequent day in my original location. Indeed, I did force acting on the object. Let's say change in. Yeah, Yeah. Let's go ahead and write this fire equal. So that just leaves us with our answer from the previous part, which was 36 meghan over 1.8 times 10 to the six. Yeah. Meters. Which if you calculate it should come out to be Newton's. Yeah. I like your answer for part B.


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