Usually, the vector component of the contact force between two surfaces which lies in the plane of the surfaces, while the surfaces are moving relative to each other. The force's magnitude is proportional to the magnitude of the normal force.

A common myth perpetrated by high school physics teachers is that the proportionality constant, the coefficient of kinetic friction is in fact a constant for a given pair of materials. This approximation does make the math much easier to work with, and is reasonably accurate for low speeds. However, if the relative velocity of the two surfaces gets above some speed (which varies depending on what the surfaces are), a better approximation is that the Kinetic Friction Force is proportional to the Normal Force AND to the Velocity.

Newton's first law states that "an object set in motion will tend to remain in motion, unless it is acted on by some other force." If you push your chair hard, it will move. However, it will eventually stop moving. This is because the force of its motion is counteracted by the force of friction.

Kinetic friction is the frictional force that affects objects that are already in motion. In order to get the object moving in the first place, a force must be applied that exceeds the force of static friction. This is discussed elsewhere. To determine how far a sliding object will go, the force of kinetic friction has to be determined. A thrown or launched object is also subject to kinetic friction in the form of air drag. All surfaces of an airborne object are rubbing against the air, and this produces friction. The force of kinetic friction tends to be somewhat less than that of static friction, but it fluctuates slightly as the object continues to move.

The force of kinetic friction varies with the weight of the object and the degree of roughnessbetween the bottom of the object and the surface it's resting on. In more specific terms, it's equal to the normal force times the coefficient of kinetic friction for the two surfaces. The normal force is equal to the weight of the object. In other words, heavier objects will stop moving faster, and so will objects with rougher surfaces. Your chair slides much further when it's on an ice rink versus when it's on your carpeted floor. When you are sitting in the chair, it slides even less before stopping.

For an example of kinetic friction at work, consider a large box being pushed across the floorfrom one of its sides at a perfect horizontal. In this case, the box is being actively pushed, so it's gaining speed. Imagine that the box contains 50 kg of concrete and the box itself weighs 1 kg. Suppose that it's being pushed with a force of 250 N to the right.2. Today we are interested in how fast the box will be moving after it's been pushed like this for 10 seconds.

Remember that the force on an object is equal to the object's mass times its acceleration, and the sum of the forces on an object determine what happens to it. So we need to figure out the net force on the object, after the frictional force (the only opposing force, besides negligible air drag) has been calculated and subtracted.

The weight of the box pushing down also needs to be considered. Weight is equal to mass times the acceleration imposed by gravity. We calculate W = mg = (51 kg)(9.8 m/s2) = 499.8 N, in the downward direction (positive* y-axis).

The normal force is equal to the weight, but acts in the opposite direction. So we calculate N = -499.8 N, in the upward direction (negative y-axis).

To determine how fast the object is actually going to be moving, we need to figure out the force of kinetic friction on the object. This is equal to the normal force times the coefficient of kinetic friction for the two surfaces. In this example, we'll assume a plastic box being pushed across a smooth wooden floor. The easiest way to figure out the coefficient of kinetic friction is to consult a table of these values for different materials or conduct an experiment. From a table, a value for wood-on-plastic is around .4**. So the force of friction here can be calculated as f = μN = (.4)(-499.8 N) = -199.92 N.

Summing all the forces on the object on a particular axis gives us the net force. In this case, all we really need to do is add the horizontal pushing force and the frictional force together. However, the frictional force is negative (it goes in the opposite direction), so we wind up subtracting it (adding a negative to a positive is the same as subtracting the negative as if it were positive). ΣF = F - f = 250 N - 199.92 N = 50.08 N.

Force is equal to mass times acceleration, so we divide the net force by the mass of the box to determine its acceleration: a = F/m = 50.08 N / 51 kg = 0.98 m/s2. The equation for the velocity of an object related to its acceleration and time spent in motion is as follows: v = at + v0, where v is the velocity, a is the acceleration (assuming it is constant), t is the time in seconds that it has been accelerating, and v0 is the velocity it was at before it began accelerating. In this case, there was no initial velocity because the box was not moving before we started pushing on it. So the velocity can be calculated as v = (.98 m/s2)(10 s) + 0 = 9.8 m/s.

The box will be travelling at 9.8 m/s after 10 seconds. If we then let it go, the force of friction would slow the box down by a = F/m = 199.92 N / 51kg = 3.92 m/s2, and we could determine approximately how long it would take to slide to a stop by applying the same equation above, but by solving for t where v equals 0 (no motion) and a equals 3.92 m/s2. Note that in this case v0 is not zero, but is equal to 9.8 m/s, the velocity of the box at the moment we stopped pushing.

We calculate 0 = (3.92 m/s2)t + 9.8 m/s and solve for t, which turns out to be: t = -v0 / a = -9.8 m/s / -3.92 m/s2 = 2.5 s. The box will stop moving after 2.5 seconds due to the force of kinetic friction. Other equations would allow us to determine how far the box would have moved, but it's pretty simple. I would "leave it as an exercise for the reader", but I really doubt anyone would bother.

* This makes the axes upside-down, but it's convenient to think about it this way. ** This value is nowhere near accurate, strictly speaking.

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