what impact does friction have on the movement of an object?
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It slows it down and more effort is required to complete the work.
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Ah, time to dig out the physics from last year.  If I remember correctly:\n\nStatic friction is when the force of friction is greater than the force applied; thus basically what amount of force it takes to move the object from rest.\n\nKinetic friction is when you apply a force for an object to keep sliding.  In this, it "breaks free" of your push and continues sliding.\n\nAnd the force needed to initially break the static friction (to get it moving) is much more than what is needed to keep it going.\n\nHope this helped a bit.  Just remember that static friction is associated with objects at rest and kinetic is objects in motion, and don't get the two mixed up. :*P\n\nHave fun with physics!
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I assume, since this is in "physics", that you are referring to the frictional force exerted on an object by a fluid through which it travels.\n\nIf, by "definition", you were wondering what it is, then the above paragraph should provide a simple answer. For more information, check out:\nhttp://en.wikipedia.org/wiki/Drag_%28physics%29\n\nIf, on the other hand, you were wondering about the mathematical definition of drag, the equation is\n\nFd = -1/2 * (rho) * v^2 * A * Cd\n\nwhere Fd is the force of the drag, rho is the density of the fluid through which the object in question travels, v is the velocity of the object in question in relation to the fluid, A is roughly the area of the object as viewed from the line of motion (e.g. a bullet's area would be a circle), and Cd is the drag coefficient, which is dependent on the surface material, texture, and shape of the object.\n\nFor more information about this formula, see:\nhttp://en.wikipedia.org/wiki/Drag_equation
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In the absence of friction, the acceleration of an object down a 12-degree slide would be\n\n9.81 * sin 12, or around 2.04\n\nInstead, your obejct is accelerating at 1.22.\n\nThe loss of acceleration due to friction is\n\n(9.81 * sin 12) - 1.22\n\nand the coefficient of friction is thus\n\n[(9.81 * sin 12) - 1.22] / (9.81 * sin 12)
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Depends also if it is muscle or fat.  The muscle can overcome the inertia & air friction.
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to the road through wear caused by friction
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The force of friction is dependent on two things, the Normal force between the "object" and the "surface" and the coefficient of friction between the two.\n\nThe normal force on a level surface is equal the weight of the object.  You are given the mass, from this, you can easily calculate the force of weight (and this F_n).\n\nYou are given the coefficient of friction, greek letter mu.\n\nThe force of friction can be found as,\nF_f = F_n * mu\n\nThe force of friction is always in the direction opposite that of the motion.  So if your traveling "forward", the frictional force is pointed "backward".\nWhat are the two components in the formula that affect the magnitude of the frictional force?  Just the normal force (which is dependent on mass and gravity) and the coefficient of friction.  For all intensive purposes, mu is constant, as is the mass and gravity of the object.  Therefore, the force of friction is constant.\n\nYou are given in initial velocity.  You are trying to find how far (change in distance) the car will travel while this constant force is being exerted on it before the car comes to a stop.\nThere are several ways this could be found.  I think the energy method would be the easiest, but I will cover the kinematic method as well....\nThe car has some initial kinetic energy when the problem starts out.  KE = 1/2 mv^2.  After the car comes to a stop, the car has zero KE...therefore some work had to be done on the system to change the energy (we assume a level road).  The change in KE equals the work done.  Work is also equal to the product of the force applied over the distancd applied.\nW = F * d\nSince the frictional force is what is causing this change in energy (which we know, it was calculated earlier), and we also know the change in KE, we can simply solve for the distance, d.\n\nThere is also another possible solution.\nWe know that the frictional force will cause an acceleration.\nF = m*a\nWe know that an acceleration will cause a change in velocity.  We know ,\nv_f^2 = v_i^2 + 2ad\nwhere v is velocity (i is initial, f is final), a is acceration, and d is distance.  Simple solve for d.
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It has to do more with the material of the ground rather than with the surface. It all depends on it's elastic coeficient. The least elastic, the more it will bounce because of the transmission of the kinetic energy.
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Nice trick, combining multiple questions into one. I'll answer one q, no. 3.\nThe 300N force along the slope doesn't contribute to the normal force so has no effect on friction. The normal force is m*g*sin(slope) = 100 * 9.81 * 5/20. Your picture\nhttp://i34.photobucket.com/albums/d130/MATTRAINES/q4.jpg?t=1164668264\ndoesn't give a value for coefficient of friction, so if we assume it's 1, then Ff = the normal force.
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FRICTION .... causes a Polarized( + or - ) energy buildup .... which is released in the form of lighting....\n\nAs when you bush your hair and the hair is attracted to the Brush....  it's the same thing.. ONLY millions of times stronger...
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You cannot get away from friction.  Material properties are also not going to be exact.
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Friction and erosion.
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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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Rolling friction : http://webphysics.davidson.edu/faculty/dmb/PY430/Friction/rolling.html\n\nSliding friction: http://www.mtsu.edu/~phys2010/Lectures/Part_2__L6_-_L11/Lecture_10/Friction/friction.html
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D. The road.\n\nIf you draw a free-body diagram you'll see a frictional force vector pointing in the direction of motion.
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Displacement.  If you mean its change in position *divided* by the time it took, that would be its average velocity.
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If you examined almost any surface under very high magnification, what looked smooth to the naked eye would look like endless mountain ranges in all directions.  If you then place two such surfaces together using weight or another force to join them and tried to move one surface relative to the other, the opposing "mountains" would have to slide up and over each other or even break (shear) the mountain tops.  The resulting resistance to motion is called friction.  When mountain tops are sheared, the energy forcing the motion is converted to extremely high local heat at the molecular level as bonds between millions of atoms and molecules are broken.  Friction causes heat and can quickly destroy an engine bearing that loses its lubrication.\n\nThe high heat of friction can actually cause a lubricant (oil, etc.) to break down and bond (as olifins) to the metal parts making smoother surfaces, especially when the parts are held apart by a film of circulation oil that carries away excess heat to a heat exchanger.  Grease (which is often a mixture of soap and oil) is thicker than oil and can fill the microscopic mountain valleys so that the parts slide on opposing layers of grease reducing friction and heat.\n\nWhen you push on a heavy crate, you must apply enough force to overcome the static friction (sometimes called stiction) and start the crate moving.  Continued motion creates dynamic friction generating a constant supply of heat.\n\nIf you place any object on a board and lift one end, all materials (combinations of board (incline) and object (load) will slide before the board reaches an angle of 45 degrees.  The higher the board angle, the higher the coeficient of friction.  This method can be used to calculate the coefficients of friction for various combinations of materials (board and load).  Ice on teflon has perhaps the least coefficient of friction and is very slippery.  The objects and board should not serve as cogs or act as adhesives to get valid results.\n\nFamily friction (strife) is the only kind of frictioon that can be reduced by flowers or a hug and a kiss.  That may cover most of the other types of friction, right?