What are the ramifications of a negative coefficient of kinetic friction?
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It would screw up a lot of calculations and other assigned physical values. Since no friction at all is 0 you would be claiming less friction than none. It is set up the way that it is for much the same reason that there are limits to aerodynamic lift and other forces that are nonsensical when allowed to become either negative or larger than one.
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1. potential\n2. kinetic\n3. it left at 0\n4. friction\n5. the greater the mass, the less  distance\n\nYour source of errors and improvements must come from your lab
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Harmful:\n-means that the rider must exert extra energy to accelerate and thre friction acting on the wheels slow the bicycle down\n-air friction/resistance also acts in a way that requires more energy input to maintain a constant velocity or accelerate.\n\nBenificial:\n-Without friction, the wheels would just spin in place when you pedel, you could not accelerate at all.\n-I dont know about you, but when I ride a bicycle down a big hill, I like the fact that my brakes work...this is due to friction.
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It is pretty hard to model a negative number, since negative objects don't really exist.  \n\nYou could use a balance scale and show that negative numbers are on one side and positive numbers are on the other side.  You achieve the same result by adding four objects to the positive side as you do by removing four objects from the negative side.
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look up stuff like static friction, and your coefficient of friction...
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If you consider a model in which both surfaces which cause friction have crusts and troughs, and their interlocking causes the frictional resistance, then it is explained as follows.\n\nThe lubricant molecules fill up the troughs in both surfaces and make them more or less even, and hence there is less interlocking among the two surfaces, and so less friction.
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Friction and erosion.
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velocity is given by :\nv(t) = -1 + e^(1-t)\n\nacceleration is given by :\na(t) = - e^(1-t)\n\nSo :\na) \nv(3) = e^(-2) -1\na(3) = - e^(-2)\n\nb) at time t = 3, acceleration is negative. The speed of the particle is decreasing.\n\nc) whenever the particle changes direction, it is velocity is momentarily zero.\n\n=> 0 = -1 + e^(1-t)\n=> 1 - t = 0\n=> t = 1\n\nNote that velocity is positive for all times < 1, and is constantly decreasing. For all times > 1, velocity is negative. Thus, the particle changes direction at time t = 1.\n\nd) at time t = 0, \nx(t) = - e\n\nat time t = 1,\nx(t) = -1 - 1 = -2\n\ntotal distance travelled thus far = (e - 2)\n\nat time t = 3,\nx(t) = - 3 - e^(-2)\n\ntotal distance by the particle over the interval 1<=t<=3 is :\n1 + e^(-2)\n \ntotal distance travelled is :\n(e - 2) + (1 + e^(-2))\n= e - 1 + e^(-2)
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Coefficient of friction is steady no matter how fast you go.  The dynamic coefficient will depend on the force of the skier on the incline and the loss of force not attributable to gravity loss.\n\nThe distance lost to friction is 11.2 cos (18) = 10.7 meters.  \n\nThe deceleration took 10.7 m / (12+0) / 2 = 1.8 seconds.\n\nDeceleration rate was 12 / 2 / 1.8 = 3.4 m/s*s\n\nThe dynamic coefficient is the force parallel to the distance divided by force perpendicular (F of skier/F of gravity).\n\nForce is mass times acceleration so mass of skier times 3.4 m/s*s seconds divided by mass of skier times 9.8 m/s*s.\n\nThe answer should be obvious if you understand what I just said.  Else you are doing homework and this would be cheating.
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i got 100N to start the container moving and 80N to keep it moving at constant speed, but i'm not sure if i got it right. What is the answer on the back of your book? if it is the same as mine, i will help you.\n\nstatic friction is how "rough" the surface is between the base of the container and the ground.\n\nnot sure about this one but i think kinetic friction is how rough between a moving object and the ground.\n\nJust think of you are stand on the hill. The static friction keep you still but once you start to slide down, the friction is less.
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Meaning moving continually with no additional force needed to overcome frictional or other energy losses.\n\nNo.  It violates the laws of thermodynamics for some process to occur without an energy loss.  If you don't add energy to a system, eventually it will run out of energy.
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Kinetic friction is equal to the Frictional force divided by the Normal force  (k = Ff / Fn)\n\nStarting at the standard equation F = ma , we need to modify this because there is some friction involved.\n\nThe two forces are our Applied force (Fa) and our Friction force (Ff).  Because Ff is acting in the opposite direction, it will be negative.  Therefore we have Fa - Ff = ma.\n\nThe key to this question is to know that because there is a constant velocity, the acceleration is equal to 0.  \n\nSo Fa - Ff = 0.  However is we re-arrange this, we have Fa = Ff\n\nThis means that the Applied force (Fa) is equal to the Frictional force (Ff).\n\nAlso, the Normal force (Fn) is equal to the Force of gravity (Fg) because we are on a level floor with no vertical displacement.  Remember though that Fg = mg.  g is yor gravitational constant.  Because there is no movement in the vertical direction, it will not be negative.  \n\nSo to solve this question, \n\nk = Ff / Fn\nk = Fa / Fg  (because like I mentioned above Ff = Fa and Fn = Fg)\n\nNow work with the number that you have.  We don't have Fg but we can solve for it using the given mass (12kg) and the constant g = 9.8\n\nk = Fa / mg\nk = 30 / (12)(9.8)\nk = 0.2551020....\n\nNoting your significant digits of 3 decimal places...\n\nThe coefficient of friction is 2.55 x 10 ^ -1\n\nHope that helps you out a little bit.
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The coefficient of friction is the ratio of the force needed to push an object and make it slide divided by the force downward on the object (e.g., weight) resisting motion.  It is used to determine the amount of force that is needed to make a smooth surfaced object of one material slide on a smooth surface of the same or a different material.  For every combination of materials there is a specific coefficient of friction.  Once calculated, the coefficient can be used for various loads (size of object).  Ice on teflon has one of the lowest coefficients of friction.  If you place an object on a plank and raise one end almost every object will slide before you reach an angle of 45 degrees.  Objects that have an adhesive property or are especially rough or have cogs will not have true coefficients of friction.  Once the coefficient of friction is known for a combination of materials (e.g., as determined in a lab) it can be used in many mechanical calculations where friction is present although the motion is more complicated than sliding along a floor.\n\nAn extra effort is sometimes needed to get an object to slide initially that is greater than steady state friction and that is sometimes called stiction.
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It's me again, Dhinesh,your classmate.\nrolling frictin is the friction caused due to rolling. it opposes rolling & it acts on a single point as there is a single pt. which has contact with surface
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time to the minus one.\n\nDo a dimensional analaysis on a rate equation to confirm this.
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both.  the friction, as well as creating its own heat, slows the airflow, which aggrivates the compression along the leading edges.\n\nI'm pretty sure.  don't put this on a physics test and then blame me.
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The net force is 0 Newtons. The force of friction is 100N.\n\nThe key here is that\n1. No acceleration.\n2. Horizontal force only.\n\nThis means that the forces are balanced in the horizontal direction. The object isn't accelerating so net F = ma = 0.