|
|
| 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:
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 kinetic 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 Pulleys
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
—that is, mass
M will pull mass
m up the slope. Now let’s ask those three all-important preliminary questions:
- Ask yourself how the system will move: Because the two masses are connected by a rope, we know that they will have the same velocity and acceleration. We also know that the tension in the rope is constant throughout its length. Because
, we know that when the system is released from rest, mass M will move downward and mass m will slide up the inclined plane.
- Choose a coordinate system: Do the same thing here that we did with the previous pulley-on-a-table problem. Make the x-axis parallel to the rope, with the positive x direction being up for mass M and downhill for mass m, and the negative x direction being down for mass M and uphill for mass m. Make the y-axis perpendicular to the rope, with the positive y-axis being away from the inclined plane, and the negative y-axis being toward the inclined plane.
- Draw free-body diagrams: We’ve seen how to draw free-body diagrams for masses suspended from pulleys, and we’ve seen how to draw free-body diagrams for masses on inclined planes. All we need to do now is synthesize what we already know:
