Ditujukan untuk meningkatkan kualitas proses dan hasil perkuliahan Fisika di tingkat Universitas
18: Review of Lectures 6 through 15
Here you see the topics that will be covered by this exam--
twice as much material as last time.
And, of course, the material is also more difficult.
I will touch upon most of these topics today--
and I can't go into great depth, for obvious reasons.
We don't have the time.
And I want to emphasize that what is not covered today does not mean at all that it will not be on the exam.
I want to ask your attention for these two key concepts: the work-energy theorem, which tells you that if an object moves from A to B that the work done on that object is the kinetic energy at point B minus the kinetic energy at point A.
This always applies both for conservative forces, like gravitational and spring forces, but it also holds for nonconservative forces such as friction.
Friction can remove kinetic energy.
It turns kinetic energy into heat, and that is perfectly fine in the work-energy theorem.
However, the conservation of mechanical energy only holds for conservative forces.
There you have the conservation of the sum of potential energy and kinetic energy, and now, of course, you cannot afford to lose kinetic energy through heat, because then the conservation of mechanical energy would not hold, and so that can only be used exclusively in the case of conservative forces.
Let's start with a simple example of an incline at an angle theta.
I have here an object, mass m, friction coefficient kinetic is mu k, and the static friction coefficient equals mu s.
This point here, let that be point A, and let the bottom of the incline be B, and let the distance between them along the slope be l.
And the first thing that you want to do with a problem like this, you want to make what we call a free-body diagram.
That means you want to draw all the forces on that object.
Clearly there is gravity, which is mg.
And then there is the normal force perpendicular to the surface.
The surface pushes up, and we call this N, the normal force.
The object wants to slide down.
Friction holds it up.
So there is also a frictional force.
And these are the only three forces that are on it.
So with a free-body diagram, you won't know more.
However, I want you to appreciate the fact that the force from the surface onto this object is, of course, the vectorial sum between these two, and that is what's called normally the contact force.
And that contact force better be exactly the same as mg, but in opposite direction, otherwise there could never be equilibrium.
Of course, we always split it into perpendicular directions because of the way that we analyze this.
There is no acceleration in the y direction, only in x direction when it starts moving, and that's why we split it.
But, of course, the vectorial sum is really the force with which the surface pushes onto that object.
All right, let's suppose now that we increase the angle theta until the object starts to slide.
So now I'm going to decompose these forces.
In the x direction I have here mg sine theta, and the component of gravity in the y direction equals mg cosine theta.
Since there is no acceleration in the y direction, the normal force must be also mg cosine theta.
I lifted up the incline to the point that it is just about to start sliding.
And so that means that the frictional force is a maximum value possible, which is the static friction coefficient times N, and therefore that is mu of s times mg times the cosine of theta.
And when will that happen? When the frictional force and mg sine theta are exactly equal.
If I go then one hair further, it will start to slide, and so that's the case when this equals mg sine theta, and so you lose your mg, and you find that mu s equals tangent theta.
That's when it will happen.
This is a way that you can measure the static friction coefficient.
Alternatively, if you know the static friction coefficient, you can predict at what angle that will occur.
Now, at that situation that it's just hanging by its thumbs, so to speak, by its fingernails, we touch it very lightly, we blow on it--
and it starts to slide.
And now I'm interested in knowing what the acceleration is downhill.
The reason why it will be accelerated, that this friction coefficient goes down now to mu k, and so now there is a t force along the slope in the plus x direction, and so I can write down now Newton's Second Law: ma, which is the force...
the net force in the positive x direction, would be equal mg times the sine of theta minus the frictional force, but now this becomes mu k mg cosine theta.
I lose my m, and so the acceleration in the positive x direction--
that's the way I defined positive x direction--
equals g times the sine of theta minus mu of k times the cosine of theta.
That's acceleration, and now a natural question would be, what is the speed at which it reaches point B? Well, what we would have done early on in the course, we would have said, "Well, if I know the acceleration, "and I know the distance, "and I know the initial speed is zero, "then clearly that distance l must be one-half a t squared, t being the time that it takes to go from A to B." And so t would be the square root of 2l divided by a.
What speed would it have at B? Well, the speed at B would be a t--
early on 801.
And so that would become, if we multiply this by a, the square root of 2al.
And if you want to see that in all glorious detail, then of course you'll have to take that a and you have to multiply that by 2l and take the square root.
So we get 2gl times the sine of theta minus mu of k times the cosine of theta and the whole thing to the power one-half.
That is the speed at point B.
I, however, would not do it that way.
I don't find it very elegant.
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