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.
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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: ->->->->->->->->->->->-> :
If the tension increases steadily from 0 between 8cm to 10cm, and ends with 3N at 10cm, then the minimum tension would seem to be 0N and the maximum tension would seem to be 3N. The average tension should logically be 1.5N.
• How much work is required to stretch the rubber band from 8 cm to 10 cm?
answer/question/discussion: ->->->->->->->->->->->-> :
I would assume that the work required to stretch the rubber band would be from 8cm to 10 cm would be 3N.
Work is obtained by multiplying force by displacement in the direction of the force. 3 N is a force.
You would multiply the average force, which you found to be 1.5 N, by the 2 cm displacement to get work 1.5 N * 2 cm = 3 N * cm.
• During the stretching process is the tension force in the direction of motion or opposite to the direction of motion?
answer/question/discussion: ->->->->->->->->->->->-> :
I think that the tension would be the force opposite to the direction of motion.
• Does the tension force therefore do positive or negative work?
answer/question/discussion: ->->->->->->->->->->->-> :
The tension would seem to do negative work.
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?
I am starting to get confused here, even though I thought I understood what was happening. I think that the work done by the tension force is -.06J.
answer/question/discussion: ->->->->->->->->->->->-> :
• 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: ->->->->->->->->->->->-> :
If Fnet*’ds=’dKE, then -3N*.02m= -.06J
As the rubber band 'snaps back', the tension force exerted on the block is in the direction of motion, so the work would be positive.
• At this point how fast will the domino be moving?
answer/question/discussion: ->->->->->->->->->->->-> :
KE=.5 m v^2, or -.06J = .5 * .02kg * v^2
-.06J=.01kg/v^2
v^2=.01kg/-.06J
v^2= -.167m^/s^2
I am clearly incorrect here, because in order to solve for v I would need to take the square root of a negative number. Help!
You're not doing badly. See my notes. See also the link provided below.