When Does Mass Matter? I have been trying to wrap my head around this idea for a month or so, and I am not really sure how mass plays into the velocity of a moving body.  I have not taken physics in a while, but one thing I remember clearly is my professor harping on mass does not matter.  But that does not make sense to me.  For example, if I apply a few Newtons of force to a pencil, I can easily slide it across a driveway, but if I apply the same force to a car it would not budge.  Another thing my professor said was, "more mass hauls a**," i.e. a body in motion stays in motion, which I get because it takes more force to stop something with more mass.
So now here is where my question begins.  I have been trying to figure out how to calculate the force needed to get a specific final velocity accounting for both gravity and friction.  As per my example above with the pencil and the car, the results I am getting do not match what I was expecting.  One of the reasons I think this is happening is because I don't have the right equations.  I thought that this was an inertia problem, but I may be wrong, or maybe my conception of the physical world is wrong and I just can't get past my preconceived ideas about mass and motion.
Here is the scenario I am working with:  


*

*I am trying to push a 1000 kg car

*The car is initially at rest

*The coefficient of friction is .80


If I want to car to move at 10 m/s, how hard to I need to push it?
After it is going 10 m/s, I want to push it again and make it go 20 m/s, how hard do I need to push it now to achieve that?
Now that I am not pushing it, how long will it take to slow down to 0 m/s?
 A: The mass of an object in a physics problem doesn't matter to that objects behavior when all the forces that act on it are fractions of the objects weight so that the acceleration has the from $$\vec{a} = \frac{\vec{F}}{m} = \frac{\vec{k}m}{m} = \vec{k} \,,$$ for some vector $k$.
This is true of idealized projectiles and idealized objects siding down ramps without friction (or with friction in the usual intro class form of $\mu F_N$ when the normal force is some fraction of the weight of the same object). For Atwood's machines the ratio of the masses matters, but the scale of the masses does not (a Atwoods machine with 1 and 2 kilograms weights has the same acceleration as one with 1 and 2 ton weights).
It's not true of projectiles with air resistance factored in, or if you're trying to hold a book up by pushing it into a wall with a horizontal force. In the first case the drag (air resistance) is proportional to cross-sectional area and to the square of speeds but not to mass. In the second case the applied force is a property of an external agency and not of the mass of the book. (Note that in the last case the minimum force that will successful hold the book up is proportional to the mass, but that means that the mass does matter.)

As a matter of practice, the mass of the object drops out in a large fraction of exercises you find early in introductory treatments, so your instructor has noticed a common theme, but I think (s)he may have overgeneralized in hopes of getting a point across. A easy thing to do in teaching, but one that should be avoided.
A: Let's assume that the wheels aren't able to turn. That means the frictional coefficient is kinetic in nature because the car is skidding across the surface...assuming you already overcame static friction.
The frictional force is as follows:
$$f=u_{k}*m*g$$
You apply a force that opposes and overcomes the kinetic frictional force that is resisting your effort such that the car accelerates according to the following setup:
$$F_{applied}-u_{k}*m*g = m*a$$
From that, the acceleration of the car is defined by:
$$a = \frac{F_{applied}-u_{k}*m*g}{m}$$
Then use this equation for final velocity where $V_{original} = 0$:
$$V_{final}-V_{original} = a*t$$
Now substitute the $a$ equation:
$$V_{final} = \frac{F_{applied}-u_{k}*m*g}{m}*t$$
If you want the final velocity to be $10$ $m/s$ you get this:
$$10 m/s = \frac{F_{applied}-u_{k}*m*g}{m}*t$$
From the equation you can see that if you apply a force greater than the kinetic frictional force, the car will accelerate and eventually hit $10$ $m/s$ at time $t$ according to the equation. You decide how much force you apply based upon how long you want to wait to reach $10$ $m/s$.
