Does mass affect speed of a sliding object with friction? Sorry if this has been already asked, but I've been looking around google for a while and couldn't find an answer suitable. I'm a beginning physics student so pardon the dumb question.

Let's say we let a block slide down a ramp of angle $\theta$. I know the component down the ramp is equal to $mg \sin \theta$ and the component normal to the ramp is mg cos theta. Since $F = ma$, $mg \sin \theta = ma$, and the masses cancel right? But this is without friction. So my question is
Does mass affect the speed of an object (down a ramp for example, or even in free fall) when there is friction/air resistance?
I guess it would be written as $F = ma = mg \sin \theta - uF_N = u mg \cos \theta$. So mass still doesn't matter right?
One more question. Let's say we have an object moving at a constant velocity on a rough surface with friction, so some force is applied. Will adding mass to the object slow it down? Common sense says yes, but why?
 A: *

*In the example of the incline plane that you have provided, the mass
does not affect the speed, because the only friction force present
is proportional to the object's weight. However, oftentimes
significant dissipative forces are proportional to the velocity of
the object --- for example, if an object if freely falling through a
viscous fluid. You could model such situation by the equation below:
$$ F = ma = mg - kv $$
Here, the coefficient $k$ of the drag force is some parameter that
could depend, for example, on the object's shape. We cannot simply divide
through by $m$ to solve for $a$; moreover, we have a differential equation, since
$a$ is the derivative of $v$. If you solve it, you would indeed get
that $v$ is a function that depends on mass. We first notice that
when the object reaches the velocity large enough for the drag force
to cancel $mg$ completely, the object will maintain that velocity,
since there will be no acceleration. If we substitute $a = 0$ in the
equation, we get that this terminal velocity should equal to
$v_{terminal} = \frac{mg}{k}$. We could actually solve this
differential equation by using a substitution function $u = g -
    \frac{k}{m}v$ and the fact that $a = v'$. We get then:
$$ u' = -\frac{k}{m}u  $$
Guessing a solution $u = C_{1}e^{-\frac{k}{m}t} + C_{2}$, where
$C_1$ and $C_2$ are some constants, we get that 
$$ v(t) = \frac{mg}{k} - C'e^{-\frac{k}{m}t} - C'' $$
Where some constants $C'$ and $C''$ depend on our initial
conditions; if the initial velocity of the object is zero, the
function that would work is:
$$ v(t) = \frac{mg}{k}(1 - e^{-\frac{k}{m}t}) $$
Where we set $C' = \frac{mg}{k}$ and $C''= 0$, so that the boundary
conditions $v_{initial} = 0$ and $v_{terminal} = \frac{mg}{k}$ are
met. We ended up with a velocity function that depends on time and the mass
of the object.

*As for the second question: let's track back to when we set the object in motion. We know that a coefficient for static friction is greater that the coefficient of kinetic friction: $\mu_{static} > \mu_{kinetic}$. This ensures that objects resting on rough services don't start moving from a slightest touch, but rather require an initial "jerk". Hence the initial force with which we will be pushing an object to start it moving, $F_{initial}$, will be greater than the eventual force $F_{final}$, to provide that "jerk": $F_{initial}$ is slightly greater than $\mu_{static}mg$, and $F_{final}$ is just equal to $\mu_{kinetic}mg$ to ensure constant velocity. We only need to exert $F_{initial}$ for a short amount of time to get the object going; but in that short time, it accelerates, and the magnitude of the acceleration depends on the force: the larger the mass of the object is, the larger the static friction is, and the larger $F_{initial}$ has to be. That implies a greater acceleration in the same amount of time --- and hence a greater constant velocity.

A: Sliding down of an object on a friction less inclined plane is independent of its mass but if it has friction than the value of mass matter
