Phy 231
Your 'cq_1_16.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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A rubber band has no tension until it reaches a length of 7.5 cm. Beyond that length its tension increases by .7 Newtons for every additional centimeter of length.
What will be its tension if its endpoints are at the points (5 cm, 9 cm) and (10 cm, 17 cm) as measured on an x-y coordinate system?
answer/question/discussion:
Endpoint at (5 cm, 9 cm) and (10 cm, 17 cm), would give the rubber band a magnitude of:
10 cm - 5 cm and 17 cm - 9 cm = (5 cm, 8 cm)
using the pythagorean theorm:
magnitude = sqrt( (5 cm)^2 + (8 cm)^2 ) = sqrt( 25 cm^2 + 64 cm^2 ) = sqrt( 89 cm^2 ) = 9.43 cm
9.43 cm - 7.5 cm = 1.93 cm
1.93 cm * 0.7 N/cm = 1.351 N
What is the vector from the first point to the second?
answer/question/discussion:
as calculated above: < 5 cm, 8 cm >
What is the magnitude of this vector?
answer/question/discussion:
as calculated above: 1.93 cm
The vector < 5 cm, 8 cm > has magnitude sqrt(89 cm^2) = 9.4 cm, approx..; you found this above.
What vector do you get when you divide this vector by its magnitude? (Specify the x and y components of the resulting vector).
answer/question/discussion:
5 cm / 1.93 cm = 2.59, and 8 cm / 1.93 cm = 4.15
< 2.59, 4.15 >
What vector do you get when you multiply this new vector by the tension?
answer/question/discussion:
2.59 * 1.351 N = 3.5 N
4.15 * 1.351 N = 5.6 N
< 3.5 N, 5.6 N >
What are the x and y coordinates of the new vector?
answer/question/discussion:
as calculated above: < 3.5 N, 5.6 N >
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10 mins
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You had one basic error, which is the only thing that caused most of your subsequent answers to be wrong; you were doing the right things throughout except for this one step. Please submit a revision if there's anything at all you don't understand in the discussion at the link below.
&#At least part of your solution does not agree with the solution and comments given at the link below. You should view the solution at that link and self-critique as indicated there.
Solution
This link also expands on these topics and alerts you to many of the common errors made by students in the first part of this course. &#