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<HW-CHSProblem 5.093 of 10ConstantsPart AWhat is the maximum speed with which a 1200-kg car can round turn of radius 86.0 m on a flat road if the coefficient of ...

Question

<HW-CHSProblem 5.093 of 10ConstantsPart AWhat is the maximum speed with which a 1200-kg car can round turn of radius 86.0 m on a flat road if the coefficient of static friction between tires and road is 65?Express your answer to two significant figures and include the appropriate units:pAValueUnitsSubmitRequest Answer

<HW-CHS Problem 5.09 3 of 10 Constants Part A What is the maximum speed with which a 1200-kg car can round turn of radius 86.0 m on a flat road if the coefficient of static friction between tires and road is 65? Express your answer to two significant figures and include the appropriate units: pA Value Units Submit Request Answer



Answers

Car A uses tires for which the coefficient of static friction is $1.1$ on a particular unbanked curve. The maximum speed at which the car can negotiate this curve is $25 \mathrm{~m} / \mathrm{s}$. Car B uses tires for which the coefficient of static friction is $0.85$ on the same curve. What is the maximum speed at which car $\mathrm{B}$ can negotiate the curve?

Well, we have coefficient of kinetic friction. Milk it that is equal to 0.7. And x relation due to gravity is 9.8 meters per second square and then radius. Well, radius is equal to 65 meter. Ah, we need to find out the speed. Well, the slope off the plane is equal to the friction coefficient. Uh, ladies, Dan Tita is equal to milk a well 10. Today's the slope and slope is equal to coefficient of the kinetic friction. And the velocity is given by, uh, Dean Tita is equal to coefficient or kinetic friction. Well, this equals we square divided by our cross Jeep. Let's call this equation number one now. So far, we then v equals route mu cane Multiply by our multiply by g right now, Vito substitution. Well, by doing substitution, then we is equal to root. Um, you can't. That's 0.7. Well, readies his nine point eat and x relation due to gravity at Sorry. Extradition due to gravity is 9.8 and radius is 65 and therefore speed is a cool too 21 meters per second

Here we have a flat or unbanked curve in the highway has a radius of 170 m. Um a car goes around the curve at 25m/s and we want to figure out what friction force needs to be acting. Two keep the car from sliding. So in the radial direction here we have the friction force acting and then we have the mass, tons of centrifugal acceleration something. No, the velocity, we can say that that's the V squared over. R. Now um if we did a free by diagram in the looking at this way, we'd have the way the normal force and the friction force. So that just tells us that the normal forces, the weight and the friction. The maximum friction force is um us times the weight or the normal force. So that means the maximum friction forces me, us times times MG. So this is the maximum friction force. So that will give us the maximum loss that that will give us the maximum velocity. Okay, so this will give us the minimum us to keep this thing from sliding. So solving from us, we can see that the mass of the car drops out and we get V squared over R. G. Plugging in our values, we get 0.375 which is a very reasonable looking its coefficient of static friction that would be needed to keep this car from sliding. Now they say, okay let's say that the coefficient of friction is one third of this amount. Uh And they ask us how fast can the car be going to not slide? So same problem basically. But now we have a new coefficient new friction force which is caused by a new coefficient of friction and we have a new velocity. Um so we can solve for that velocity here and that's just the square root of the friction force. Times are times g. Again the mass drops out, plugging in our values, we get 2.4 m/s, so if we have a third of the coefficient of friction, then we have to go and be going quite a bit slower uh to keep from slipping. And that would be, you would be that would be expected. We expect this value could be less than this value. If the if the amount of friction force drops and a third of this would be a very low friction force, um that would be a very, very slippery road.

In this problem, a car round sun, a flat circular track of radius 225 m. So let's say this is the circular track on which the car is rounding and radius of this track are is 225 m coefficient of static friction is given, which is 0.65 So we have to draw first force diagram of the car. So this is this is suing, let us a card is going in forward direction or in a poor direction. So this is suing the situation when we are viewing the situation from the, from the top. So what will be the uh what will be the its view when we view the situation from back side? So basically, now I'm going to draw the situation when we are seeing from on backside. So there will be the car and these are the tires which is on the road or the circular track. So this circular track will apply a normal reaction forcing vertically upward direction and there will be weight of the car which will be acting in downward direction. Now due to the centrifugal force, this car will have a tendency to move in outside the track. So to avoid the slipping, a friction force will act in the towards the center direction or in other sense, if we can see that these forces are acting in vertical in vertical direction, so but the circular motion is performing in the horizontal horizontal circle, so there must be a horizontal force which will be providing the centripetal acceleration. So the force should be friction force. Therefore the friction force must act towards the center. So the fourth diagram after the car will be normal reaction force in a poor direction, EMC in downward direction and freaks and force towards the center. No, in part we we have to calculate what is the maximum speed of the car so that it so that it is to stay on the track or it doesn't slip from the track. So for that, what we can see that this friction force is going to provide the centripetal acceleration and normal forces going to balance the MG component. Therefore, in vertical direction, what we can write and must be close to MG. Let us say this is equation when and freaks and force is going to provide centripetal acceleration, therefore freaks. And force will be close to M V squared divided by our. Now we can replace freaks and force with the value of us into and we know that static friction force is always less than equals two mules into. And so here we can replace Fs with the M V. Squared by our term. So what will be the equation M V squared by our will be less than equals two mules into. And so and in the uh equation one, we know that energy calls to MZ Therefore mp square by our should be less than it calls to us into MG. Now, mass can be cancelled from both sides of the expression. So what will be the value of the square? So we square should be less than equals two new S into our into G? Therefore velocity should be less than equals to square root over um us into our end to G. So what is the value of us here? In the problem, it has given that message 0.65 radius of the track is 225 and acceleration due to gravity is 9.8. So velocity has to be less than equals to this quantity, which if you calculate so this terms gives 37.85 m for a second. so velocity should be less than the value of less than the 37.85 m per second. Therefore, maximum permissible velocity. Therefore maximum velocity is 37.85 m per second. So this will be the answer for this problem.

Question No. 22 in this question. It is even though there is a flat wrote and there is a car on it. The friction is present between the tires and the road. So the frictional force is acting like this on the car and this regional forces providing this interpreter force to the uh car moving in a certain part. The reaction force, normal reaction force from the surfaces are, and the way it is acting downwards empty. This is a free body diagram, florida, uh cars moving in a circular part on unbanked road. So we know that if an object moves in a circle part, then centripetal force, alexander object in this condition, the centripetal force is provided by the frictional force so we can write in our general direction. The frictional force is equal to the next interpersonal force on the car and this directional force is equal to meal times, the weight of the car. This is equal to M we square right? Uh Now we can cancel em from both sides and get we is equal to under route mU times are G. R. Artista radius of the simpler path. Now substituting the values we is equal to under root meal is Given a 0.7 into the areas of the path is given us five 12 m and exhalation due to gravity's nine pointed solving this, we get the desired dispute of the car 59 13 meter. What second? This is that required? A speed and free body diagram is already drawn.


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