Why does kinetic friction on an inclined plane have to have a constant velocity?

Answer 1

I'm not absolutely certain what you are getting at here but when I was at school we used to roll or slide objects down a slope where the inclination was just enough to overcome the friction between the ramp and the object. A ball pushed down such a slope will (in theory) have a constant velocity.

Answer 2

If I understand your question correctly, you are asking this: "Why does kinetic friction cause an object to slide down an inclined plane at a constant velocity?"

Actually, it does NOT necessarily do so. Depending on the particular values of the coefficient of friction (μ) and the angle of the incline (θ), a sliding object may do any of the following:

1. It may slow to a stop;
2. It may continue downward at a constant speed;
3. It may increase in speed.

When the object is sliding downslope, there are two forces acting on it in a direction parallel to the slope:

One force is a component of the object's weight, pulling it down the slope. This force has a magnitue of (mg)sinθ, and acts in a downslope direction.

The other force is kinetic friction, which acts in an upslope direction. Its magnitude is (mg)μcosθ.

So the net force acting downslope is:
(mg)(sinθ − μcosθ)

That value may be negative, or zero, or positive, depending on how big θ is, compared to μ

If the value is negative (that is, if tanθ < μ), the sliding object will slow to a stop.

If the value is zero (that is, if tanθ = μ), the object will continue downslope at a constant speed.

If the value is positive (that is, if tanθ > μ), the object will increase its speed as it slides.

Answer 3

the consistent velocity ability that there is not any acceleration and for this reason no internet rigidity appearing on the block, meaning that the rigidity of friction must be equivalent to the element of the rigidity of gravity that's pulling the block down, this element of the rigidity of gravity is talked approximately as "F parallel" because of the fact that's going parallel to the vulnerable airplane, and the formulation to resolve for that's FgsinX the place Fg is the rigidity of gravity and X is the attitude of inclination, that's 25. In different phrases, the friction is comparable to this F parallel. So so as to locate Fg, that's only mg, mass situations gravity. the only situation is which you probably did no longer supply me mass, yet once you have the mass, you only do mgsin25, and that's your answer. in the event that they did no longer furnish you with any mass, see under. If via "kinetic friction" you intended "the coefficient of kinetic friction", that's a distinctive tale. you are able to honestly resolve for that because of the fact in case you paintings in terms of variables, you are able to cancel out Fg and for this reason do no longer desire mass. right here is the way you are attempting this: ok, so I already suggested that F parallel = Ff (the rigidity of friction) because of the fact of this FgsinX = Ff now, Ff is comparable to ?FN the place ? is the coefficient of kinetic friction, and FN is the traditional rigidity. Now, the traditional rigidity(which consequently is F perpendicular) is comparable to FgcosX. that's the rigidity that the ramp exerts up on the block. So when you consider that F parallel = Ff, we combine the two. FgsinX = ?FN FgsinX = ?FgcosX sinX = ?cos X ?=sinX/cosX ?=tanX this implies that the coefficient of kinetic friction is comparable to tan25 that's .466 wish this helps :)

Answer 4

friction doesnt "have" a velocity. What are you asking?

Answer 5

because if it slows down it won't be able to go up.