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phy 201
Your 'cq_1_14.1' 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_14.1_labelMessages **
Seed 15
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, and what do you think is the average tension?
answer/question/discussion: ->->->->->->->->->->->->
1 newton and 3 newtons witht e avg tension being 1.5 newtons
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• How much work is required to stretch the rubber band from 8 cm to 10 cm?
answer/question/discussion: ->->->->->->->->->->->-> :
1 newtons* .08 m=.08 joules
2 newtons *.09m= .18 joules
3 newtons*.1=.3 joules
.3+.18+.08 =.56 joules
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The rubber band exerted force over a distance of 2 cm, from the 8 cm to the 10 cm length.
There was no interval, for example, of .08 m, or 8 cm, over which it exerted a nonzero force.
What average force was exerted over the 2 cm interval, and how much work was therefore done on this interval?
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• During the stretching process is the tension force in the direction of motion or opposite to the direction of motion?
answer/question/discussion: ->->->->->->->->->->->-> :
opposite to the direction of motion
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• Does the tension force therefore do positive or negative work?
answer/question/discussion: ->->->->->->->->->->->-> :
negative work
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The rubber band is released and as it contracts back to its 8 cm length it exerts its tension force on a domino of mass .02 kg, which is initially at rest.
• Again assuming that the tension force is conservative, how much work does the tension force do on the domino?
answer/question/discussion: ->->->->->->->->->->->-> :
-.56 joules *.02= -.01
Not really sure what Im doing
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You have the right idea, but you should be multiplying average force by displacement, not work by displacement.
You should also be including units in your calculations (.02 would be .02 meters, so your result would be -.01 Joule * meters; that's not the right unit, which could be useful in figuring out what you should have done).
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• Assuming this is the only force acting on the domino, what will then be its kinetic energy when the rubber band reaches its 8 cm length?
answer/question/discussion: ->->->->->->->->->->->-> :
back to .08 joules
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• At this point how fast will the domino be moving?
answer/question/discussion: ->->->->->->->->->->->-> :
.5mv^2=KE
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This is the right relationship to use.
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.08/.5(.02)= v^2
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You don't have units on these quantities; if you did you would see that they aren't all correct.
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Sqrt(8)=2.8 m/s
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*#&!
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You're on the right track but will need to modify most of your work in order to get the right quantities into the right places.
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