#$&*
Phy 201
Your 'cq_1_15.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_15.1_labelMessages **
Copy the problem below into a text editor or word processor.
• This form accepts only text so a text editor such as Notepad is fine.
• You might prefer for your own reasons to use a word processor (for example the formatting features might help you organize your answer and explanations), but note that formatting will be lost when you submit your work through the form.
• If you use a word processor avoid using special characters or symbols, which would require more of your time to create and will not be represented correctly by the form.
• As you will see within the first few assignments, there is an easily-learned keyboard-based shorthand that doesn't look quite as pretty as word-processor symbols, but which gets the job done much more efficiently.
You should enter your answers using the text editor or word processor. You will then copy-and-paste it into the box below, and submit.
________________________________________
A rubber band begins exerting a tension force when its length is 8 cm. As it is stretched to a length of 10 cm its tension increases with length, more or less steadily, until at the 10 cm length the tension is 3 Newtons.
• Between the 8 cm and 10 cm length, what are the minimum and maximum tensions?
answer/question/discussion: ->->->->->->->->->->->-> :
just guessing that the minimum tension would be at 1N at the 8cm length.
Making the maximum tension 3N at 10cm length, since it increases by 2cm.
#$&*
• Assuming that the tension in the rubber band is 100% conservative (which is not actually the case) what is its elastic potential energy at the 10 cm length?
answer/question/discussion: ->->->->->->->->->->->-> :
3N * .02cm = .06 Joules PE
#$&*
@&
3 N is the maximum force exerted on that interval, and is exerted only when the length of the rubber band is 10 cm.
As the length of the rubber band returns to 8 cm, the force decreases to 0.
It's no more appropriate to use the 3 N force, the maximum force experienced, than the 0 N minimum force as a basis for calculating the work.
How would you revise your estimate to take this into consideration?
*@
• If all this potential energy is transferred to the kinetic energy of an initially stationary 20 gram domino, what will be the velocity of the domino?
answer/question/discussion: ->->->->->->->->->->->-> :
.06 Joules
Stationary 20gram domino at 0m/s = 0 Joules?
20g = .02kg
.06 Joules / .02kg = 3 m^2/s^2
Maybe the square root of 3?
#$&*
@&
The kinetic energy is 1/2 m v^2.
If 1/2 m v^2 is .06 Joules then, since m = .02 kg, what is v?
In general, 1/2 m v^2 = KE, which can be solved for v in terms of KE and m.
*@
• If instead the rubber band is used to 'shoot' the domino straight upward, then how high will it rise?
answer/question/discussion: ->->->->->->->->->->->-> :
.02kg * -9.8m/s^2 = -.196N acting in the opposite direction of motion of the domino
It would be the upward force minus .196N…. not sure where to go to get ‘ds
#$&*
@&
If the domino rises to height h, how much work is done by this force?
What height h would be necessary for the work done to be .06 Joules?
*@
For University Physics students:
Why does it make sense to say that the PE change is equal to the integral of the force vs. position function over an appropriate interval, and what is the appropriate interval?
answer/question/discussion: ->->->->->->->->->->->-> :
#$&*
*#&!
@&
You've got most of it, but will need some revisions to complete everything correctly.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
*@