misogynisticfeminist said:
ahh alright, maybe the moon was a bad example because there's stuff about the gravitational field. what if i carried the sofa around the world, at constant velocity?
Let's take this one step at a time. First, we'll simplify a bit and say you
pushed it the same distance as the circumference of the world (i.e. a straight line instead of a circle) at a constant velocity. If we neglected friction, you would only need:
W=\frac{1}{2}mv^2
in order to get the sofa up to that velocity. Notice there's no depedence on distance. Why? Well, remember Newton's First Law. An object that's in motion stays in motion unless acted upon by an unbalanced force. All you'd have to do is get it up to that velocity and it would keep going all by itself. The above energy is just the kinetic energy of an object with mass m and velocity v. If it started out with no energy (v=0), then the above must equivalent to the work you did on it (where else would it get the energy).
Next, let's add some friction. For friction on a horizontal surface, the first thing you have to worry about is
static friction. There is a nice discussion of that
here (jump to mine or Doc Al's post). The basic idea is that it resists your force up until a breaking point, after which the friction gives and you can freely move the object. This means, to get the sofa started, you need:
F > \mu_smg
where \mu_s is the coefficient of static friction. This alone wouldn't change the work required to move it the distance you requested, it would only mean that you'd have to use a larger force to do it. However, there's also
kinetic friction. What this does is produce a constant force in opposition to your force once you already have the object moving. The magnitude of this opposing force is given by
F_f = \mu_kmg
where \mu_k is the coefficient of kinetic friction. This means that the total force on the object once you've passed the static friction barrier is
F_{tot}=F_0 - F_f
where F
0 is the force that
you are a applying. If we use Newton's second law and say that the object's acceleration is the force divided by its mass, this equation becomes
a=\frac{F_0}{m}-\mu_kg
Remember that a
constant velocity corresponds to zero acceleration, so let's solve for the force we would need to apply to keep it at a constant velocity for the whole trip:
F_0=\mu_kmg
Notice that this is just the force of kinetic friction. In order to have zero net force on the object, your applied force must equal the resisting force.
So let's calculate the work that comes out of all this, remembering that
W=Fd
Since static friction is usually stronger than kinetic friction, we're going to have to apply two forces, one to get it up to speed (F
1) and another to keep it going against friction (F
0). The total energy at the end of the application of F
1 is going to be the same, but you'll have to do more work because friction will be working against you all along the way. Thus, the work that
you do to get it up to speed is
W=F_1(\frac{E_f}{F_{tot}})
where the term in parentheses is just the total distance traveled by the sofa in accelerating from 0 to v. This can be expanded to
W_1=F_1\frac{\frac{1}{2}mv^2}{F_1-F_f}=\frac{1}{1-\frac{F_f}{F_1}}\frac{1}{2}mv^2
Notice that this is the same as before, except with an extra term out front that takes into account the work you do against friction. This is not it, however. We also must do work to keep the sofa going at a constant speed. This is a bit easier and is simply given
W_2=F_0d=F_fd
where d is the distance you push the sofa. We'll assume that the total distance is much larger than the distance you push it in accelerating to v. So now, finally, we have our expression for total work:
W=W_1+W_2=\epsilon E_k+F_fd
where
F_f=\mu_kmg
and
E_k=\frac{1}{2}mv^2
and
\epsilon=\frac{1}{1-\frac{F_f}{F_1}}
Notice I've still neglected air drag, circular motion, and complications brought in by
carrying the sofa. I had hoped to address those, but this took longer than I thought, so maybe I'll come back to it later.
