PHY 201
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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?
answer/question/discussion: ->->->->->->->->->->->-> :
If the rubber band starts the tension at the 8 cm mark and went to a length of 10 cm, then the maximum tension would be 3 N. The minimum tension would be 0 N, which is at the 8 cm length.
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• Assuming that the tension in the rubber band is 100% conservative (which is not actually the case) what is its elastic potential energy at the 10 cm length?
answer/question/discussion: ->->->->->->->->->->->-> :
The displacement is 2 cm, and the tension is 3 Newtons. We know the force and the displacement, so we can get the elastic potential energy of the tension by: 3 N / .02 m = 150 Joules.
N / m doesn't give you Joules.
You don't get work by dividing force by displacement.
Note also that 3 N is the max, not the average. force.
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• If all this potential energy is transferred to the kinetic energy of an initially stationary 20 gram domino, what will be the velocity of the domino?
answer/question/discussion: ->->->->->->->->->->->-> :
The kinetic energy is the equal and opposite of the potential energy so: 150 J PE = -150 J KE. So, I can use the work energy theorem equation to solve for the final velocity: v = sqrt(2 * 150 / .02 kg) = 122.47 m/s. So, the average velocity would be: 61.24 m/s.
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• If instead the rubber band is used to 'shoot' the domino straight upward, then how high will it rise?
answer/question/discussion: ->->->->->->->->->->->-> :
I need to find the displacement. I have the change in velocity, so I need the find the acceleration of the domino, so the acceleration is the Fnet / m = 150 J / .02 = 7500 m/s^2. So, the displacement is the change in velocity divided by the acceleration: 122.47 m/s / 7500 m/s^2 = .016 meters.
if you divide `dv by accel, you get the time interval, not the displacement. Note that the units are (m/s)/(m/s^2) = m/s * s^2/m = s.
These answers does not seem right to me.
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For University Physics students:
Why does it make sense to say that the PE change is equal to the integral of the force vs. position function over an appropriate interval, and what is the appropriate interval?
answer/question/discussion: ->->->->->->->->->->->-> :
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30 minutes
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Solution
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