Question

The Coefficient of kinetic friction between the tires of your car and the roadway is "μ". (a) If your initial speed is "v" and you lock your tires during breaking, how far do you skid? (b) What is the stopping distance for a truck with twice the mass of you car, assuming same initial speed and coefficient of kinetic friction.

I'm stuck on (b), my mass cancels out every time, but that doesn't seem right

Answer 1

\(\sf\Large\color{orange}{(a)}\)
First let's find the acceleration. That's what changes the velocity.

I guess we should have a more specific goal, though. Our reason to find acceleration comes from \(v^2=v_0^2+2a\ \Delta d\). We can solve for \(\Delta d\). It's even easier when you realize that \(v\), the final velocity, is just \(0\). So let's go through the work quick.
\(0^2=v_0^2+2a\ \Delta d\\\implies -v_0^2=2a\Delta d \\\implies -\dfrac{v_0^2}{2a}=\Delta d\)


Onto acceleration.
\(F=m\ a\implies a=\dfrac{F}{m}\)
So, we'll just say the only force is that of friction. Force of friction depends on the normal force, \(F_n\), and the \(\mu\). Specifically, \(F=F_n\mu\). If you forbet the order ever, just know that the Friction force increases with greater \(F_n\) and \(\mu\). Now we'll substitute in \(F\) to find the acceleration. So, we have \(a=\dfrac{F_n\ \mu}{m}\).

Now let's put our \(a\) in to find \(\Delta d\) in a new way.
\(-\dfrac{v_0^2}{2\frac{F_n\ \mu}{m}}=-\dfrac{v_0^2\ m}{2F_n\ \mu}=\Delta d\)
One step further, we'll use \(F_n=m\ g\)
\(=-\dfrac{v_0^2\ m}{2m\ g\ \mu}\\=-\dfrac{v_0^2\ \cancel m}{2\cancel m\ g\ \mu}\\=-\dfrac{v_0^2}{2g\ \mu}=\Delta d\)

\(\sf\Large\color{green}{(b)}\)
You're right. Mass cancels out. So, the stopping distance will be the same, no matter what the mass is.

How to make sense of this? Well, the mass in the numerator was from the acceleration. The mass in the denominator was from the normal force which contributed to the friction force.

This can be tied to the conceptual physics, but I won't tie them myself, yet.
\(\bullet\) A greater mass will be have greater inertia - if there was the same friction force, the truck would be harder to stop.
\(\bullet\) A greater mass, though, means greater friction force.
\(\bullet\) The inertia increases proportional to the change in mass, just like the force of friction does. So it gets harder to stop just as much as there is a stronger force to stop it. By "it," I guess I mean the increasing mass.

\(F=m\ a\) is the part concerning inertia, where as
\(F=F_n\ \mu=m\ g\ \mu\) is the part concerning friction force.


\(\sf\large \color{blue}{Extra:}\)
\(F=F\)
\(m\ a=m\ g\ \mu\)
\(\implies a = g\ \mu\)
Acceleration depends on gravitational acceleration and the coefficient of kinetic friction, I guess.


I encourage anyone to double check me! I am tired, at the moment! I hope this helps - take care!