cq_1_141

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Phy 121

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 **

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: ->->->->->->->->->->->-> :

 The tension increases from ) N at 8 cm to 3 N at 10 cm. Average tension= (0 N+3 N)/2= 1.5 N

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• How much work is required to stretch the rubber band from 8 cm to 10 cm? 

answer/question/discussion: ->->->->->->->->->->->-> :

'ds=10 cm-8 cm= 2 cm

Average Force (from above)= 1.5 N

Work exerted=force*'ds

=1.5 N* 2 cm= 0.03 J

 

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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: ->->->->->->->->->->->-> :

 The stretching force is in the direction of motion

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The tension in the rubber band tends to pull its ends together. The end that is being stretched will exert a force toward the opposite end, in the direction opposite the motion.

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• Does the tension force therefore do positive or negative work?

answer/question/discussion: ->->->->->->->->->->->-> :

 The tension force does negative work since the rubber band is exerted by the force.

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The question is whether the tension force is in the direction of the displacement or not.

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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: ->->->->->->->->->->->-> :

 +0.03 J (from above)

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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: ->->->->->->->->->->->-> :

 'dW_net='dKE

+0.03 J='dW_net

So the Ke increases by 0.03 Joules (originally at rest then goes to 0.03 J)

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• At this point how fast will the domino be moving?

answer/question/discussion: ->->->->->->->->->->->-> :

m=0.02 kg KE=0.03 J

 KE=1/2 mv^2

v=sqrt(2 * KE/m)

v=sqrt (2*0.3J/0.02kg)

v=1.7 m/s

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[ extended discussion of T vs. L and T vs. x including graphs at linked document to be provided ]

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20 min

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Good, but see my notes on the nature of the tension force and the work it does.

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