#$&*
Phy 201
Your 'cq_1_14.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_14.1_labelMessages **
A rubber band begins exerting a tension force when its length is 8 cm. As it is stretched to a length of 10 cm its tension increases with length, more or less steadily, until at the 10 cm length the tension is 3 Newtons.
Between the 8 cm and 10 cm length, what are the minimum and maximum tensions, and what do you think is the average tension?
answer/question/discussion: ->->->->->->->->->->->-> :
Minimum tension for 8 cm is 0 N
Maximum tension for 10 cm is 3 N
Average tension=(0 N+3 N)/2=1.5 N
#$&*
How much work is required to stretch the rubber band from 8 cm to 10 cm?
answer/question/discussion: ->->->->->->->->->->->-> :
Displacement=10 cm-8 cm=2 cm
Average force=1.5 N
Work on rubber band=1.5 N*2 cm=3 N*cm or .03 Joules
#$&*
During the stretching process is the tension force in the direction of motion or opposite to the direction of motion?
answer/question/discussion: ->->->->->->->->->->->-> :
During the stretching process the tension force is opposite the direction of motion.
#$&*
Does the tension force therefore do positive or negative work?
answer/question/discussion: ->->->->->->->->->->->-> :
It does negative work
#$&*
The rubber band is released and as it contracts back to its 8 cm length it exerts its tension force on a domino of mass .02 kg, which is initially at rest.
Again assuming that the tension force is conservative, how much work does the tension force do on the domino?
answer/question/discussion: ->->->->->->->->->->->-> :
.03 N
#$&*
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: ->->->->->->->->->->->-> :
dWnet=.03 J=KE
It increases by .03 cm
#$&*
At this point how fast will the domino be moving?
answer/question/discussion: ->->->->->->->->->->->-> :
KE=1/2mv^2
v=sqrt(KE*2/m)=sqrt(.03 J*2/.02 kg)
v=1.7 m/s
#$&*
*#&!*#&!
&#This looks very good. Let me know if you have any questions. &#