SAT物理考试－Inclined Planes
we know that when people are placed on slippery inclines, they slide down the slope. We also know that slides can sometimes be sticky, so that when you are at the top of the incline, you need to give yourself a push to overcome the force of static friction. As you descend a sticky slide, the force of kinetic friction opposes your motion. In this section, we will consider problems involving inclined planes both with and without friction. Since they’re simpler, we’ll begin with frictionless planes.
Suppose you place a 10 kg box on a frictionless 30º inclined plane and release your hold, allowing the box to slide to the ground, a horizontal distance of d meters and a
vertical distance of h meters.
Before we continue, let’s follow those three important preliminary steps for solving problems in mechanic
s:
1. Ask yourself how the system will move: Because this is a frictionless plane,
there is nothing to stop the box from sliding down to the bottom. Experience suggests that the steeper the incline, the faster an object will slide, so we can expect the acceleration and velocity of the box to be affected by the angle of the plane.
2. Choose a coordinate system: Because we’re interested in how the box slides
along the inclined plane, we would do better to orient our coordinate system to the slope of the plane. The x-axis runs parallel to the plane, where downhill is the positive x
direction, and the y-axis runs perpendicular to the plane, where up is the positive y
direction.
3. Draw free-body diagrams: The two forces acting on the box are the force of
gravity, acting straight downward, and the normal force, acting perpendicular to the
inclined plane, along the y-axis. Because we’ve oriented our coordinate system to the slope of the plane, we’ll have to resolve the vector for the gravitational force, mg, into its x- and y-components. If you recall what we learned about vector decomposition in
Chapter 1, you’ll know you can break mg down into a vector of magnitude cos 30º in the negative y direction and a vector of magnitude sin 30º in the positive x direction. The result is a free-body diagram that looks something like this:
Decomposing the mg  vector gives a total of three force vectors at work in this diagram: the y -component of the gravitational force and the normal force, which cancel out; and the x -component of the gravitational force, which pulls the box down the slope. Note that the steeper the slope, the greater the force pulling the box down the slope.
Now let’s solve some problems. For the purposes of these problems, take the acceleration due to gravity to be g  = 10 m/s 2. Like SAT II Physics, we will give you the values of the relevant trigonometric functions: cos 30 = sin 60 = 0.866, cos 60 = sin 30 = 0.500.
1. What is the magnitude of the normal force?
2. What is the acceleration of the box?
3. What is the velocity of the box when it reaches the bottom of the slope?
4.
What is the work done on the box by the force of gravity in bringing it to the bottom of the plane?
1. What is the magnitude of the normal force?
The box is not moving in the y  direction, so the normal force must be equal to the y -component of the gravitational force. Calculating the normal force is then just a matter of plugging a few numbers in for variables in order to find the y -component of the gravitational force:
pulleys2. ?
We know that the force pulling the box in the positive x  direction has a magnitude of mg  sin 30. Using Newton’s Second Law, F = ma , we just need to solve for a :
3. What is the velocity of the box when it reaches the bottom of the slope?
Because we’re dealing with a frictionless plane, the system is closed and we can invoke the law of conservation of mechanical energy. At the top of the inclined plane, the
box will not be moving and so it will have an initial kinetic energy of zero ().
At the bottom of the slope, all the box’s potential energy will have been converted into kinetic energy. In other words, the kinetic energy, 1⁄2 mv2, of the box at the bottom of the slope is equal to the potential energy, mgh, of the box at the top of the slope. Solving for v, we get:
4. What is the work done on the box by the force of gravity in bringing it to the bottom of the inclined plane?
The fastest way to solve this problem is to appeal to the work-energy theorem, which tells us that the work done on an object is equal to its change in kinetic energy. At the top of the slope the box has no k
inetic energy, and at the bottom of the slope its kinetic energy is equal to its potential energy at the top of the slope, mgh. So the work done on the box is:
Note that the work done is independent of how steep the inclined plane is, and is only dependent on the object’s change in height when it slides down the plane.
Frictionless Inclined Planes with
Let’s bring together what we’ve learned about frictionless inclined planes and pulleys on tables into one exciting über-problem:
Assume for this problem that —
, w
Now let’s tackle a couple of questions:
1. What is the acceleration of the masses?
2. What is the velocity of mass m after mass M has fallen a distance h?
1. What is the acceleration of the masses?
First, let’s determine the net force acting on each of the masses. Applying Newton’s Second Law we get:
Adding these two equations together, we find that . S
Because , th
. S
. T
. B
Finally, note that the velocity of mass m is in the uphill direction.
As with the complex equations we encountered with pulley systems above, you needn’t trouble yourself with memorizing a formula like this. If you understand the principles at work in this problem and would feel somewhat comfortable deriving this formula, you know more than SAT II Physics will likely ask of you.
Inclined Planes With Friction
There are two significant differences between frictionless inclined plane problems and inclined plane problems where friction is a factor:

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