#$&*
Phy 231
Your 'cq_1_14.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
A rubber band begins exerting a tension force when its length is 8 cm. As it is stretched to a length of 10 cm its tension increases with length, more or less steadily, until at the 10 cm length the tension is 3 Newtons.
• Between the 8 cm and 10 cm length, what are the minimum and maximum tensions, and what do you think is the average tension?
answer/question/discussion: ->->->->->->->->->->->-> :
Minimum tension: 0 N
Maximum tension: 3 N
Average tension = (3 N + 0 N) / 2 = 1.5 N
#$&*
• How much work is required to stretch the rubber band from 8 cm to 10 cm?
answer/question/discussion: ->->->->->->->->->->->-> :
`dW = F*`ds ; F = 3 N ; `ds = 10 cm – 8 cm = 2 cm
`dW = (3 N)*(2 cm) = 6 J
&&&&Corrected:
`dW = Fave*`ds ; Fave = 1.5 N ; `ds = 10 cm – 8 cm = 2 cm ;
(2 cm)*(10^-2 m / 1 cm) = 0.02 cm
`dW = (1.5 N)*(0.02 cm) = 0.03 J
&&&&
#$&*
• During the stretching process is the tension force in the direction of motion or opposite to the direction of motion?
answer/question/discussion: ->->->->->->->->->->->-> :
The tension force is in the opposite direction of motion.
#$&*
• Does the tension force therefore do positive or negative work?
answer/question/discussion: ->->->->->->->->->->->-> :
Therefore the tension force does positive work.
&&&&Corrected:
The tension force is the force applied by the rubber band because the tension force is the pull. When a rubber band is pulled, the tension force is applied in the opposite direction of the pull, or in other words the opposite direction of motion. Since work is being done by the tension force, the work of the tension force is therefore negative.
&&&&
#$&*
The rubber band is released and as it contracts back to its 8 cm length it exerts its tension force on a domino of mass .02 kg, which is initially at rest.
• Again assuming that the tension force is conservative, how much work does the tension force do on the domino?
answer/question/discussion: ->->->->->->->->->->->-> :
F = ma ; m = 0.02 kg ; a = 980 cm/s^2
F = (0.02 kg)*(980 cm/s^2) = 19.6 N
`dW = F*`ds ; F = 19.6 N ; `ds = 8 cm
`dW = (19.6 N)*(8 cm) = 156.8 J
&&&&Corrected:
`dW on = Fave*`ds ; Fave = 1.5 N ; `ds = 10 cm – 8 cm = 2 cm ;
(2 cm)*(10^-2 m / 1 cm) = 0.02 cm
`dW of tension force on= (1.5 N)*(0.02 cm) = 0.03 J
&&&&
#$&*
• 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: ->->->->->->->->->->->-> :
`dKE = `dW
`dKE = 156.8 J
&&&&Corrected:
`dKE = `dW
`dKE = 0.03 J
&&&&
#$&*
• At this point how fast will the domino be moving?
answer/question/discussion: ->->->->->->->->->->->-> :
`dKE = (1/2)*m*(vf^2)
vf = sqrt((2*`dKE) / m)
`dKE = 156.8 J ; m = 0.02 kg
vf =sqrt((2*156.8 J) / 0.02 kg)
vf = 125.22 m/s
&&&&Corrected:
`dKE = (1/2)*m*(vf^2)
vf = sqrt((2*`dKE) / m)
`dKE = 0.03 J ; m = 0.02 kg
vf =sqrt((2*0.03 J) / 0.02 kg)
vf = 1.73 m/s
&&&&
#$&*
** **
** **
This is the revised Seed Question 14.1
Very well done.
I remembered getting this (and other questions, which I see you have also resubmitted) a couple of days ago so I knew this was a revision, so it's no problem here, but for future reference note that it's best to tell me that at the beginning, rather than at the end of the document.