Work and Energy
10. The first thing to do for this problem is to draw a picture.
Now apply the work equation and solve for q.
11. I prefer to draw a picture for this problem first. This problem does not tell you that the escalator is moving at a constant velocity, but I would make that assumption.
If you set your coordinate system at an angle of 30deg like the example in class, the only force that is doing work is the x component of the weight (mgsinq).
14. This problem is very similar to the previous problem, except they do not tell you how far the crate actually slides. That is the first thing you should do.
Now just use the 225N force and the 2.3m distance to find work.
15. Here is this problem exist the same situation where the only force that is doing work is the x component of the weight (mgsinq).
16. Part "a" is the same problem as before except now there is no need to concern yourself with with angles, because they give you the force parallel to the plane.
17. This is a problem that only involves movement straight up and down, so no components or angles. (yippee!!!)
18. hmmm....... I am not sure what to say. Straight forward comes to mind.
19.
29. If this were an ideal matching, the work in would equal the work out so,
30.
a. Since the worker applies the force of 496N directly to the dolly to move it 2.10m, the work done here is 1040J.
b. Since the dolly raised the refrigerator .85m, we will use that as our x and calculate work.
33. Use the equation for efficiency and plug and chug.
