Before we continue, let’s follow those three important preliminary steps for solving problems in mechanics: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. 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. 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 美国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/s2. 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:
2. What is the acceleration of the box? 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 . Because it is a height h above the bottom of the plane, it will have a gravitational potential energy of U = mgh. Adding kinetic and potential energy, we find that the mechanical energy of the system is:英语作文
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