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PHY 231
Your 'cq_1_15.1' 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?
answer/question/discussion: ->->->->->->->->->->->-> :
The minimum tension occurs when the rubber band is at/around the length of 8 cm. Thus, the minimum tension is 0 Newtons at 8 cm. The maximum tension, 3 Newtons, occurs when the rubber band is 10 cm in 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: ->->->->->->->->->->->-> :
Because it is conservative, it can regain all of its tension; it is reversible. The elastic potential energy is equal to the work required to stretch the rubber band. Because its average force is 1.5 N (( 0 N + 3 N)/2 ), it would take .03 Joules to stretch the rubber band. Thus, its elastic potential energy is equal to .03 Joules at the 10 cm length, because it has an average force of 1.5 N applied over a .2 m distance.
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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: ->->->->->->->->->->->-> :
We can use the formula for Kinetic Energy, which is equal to the Force multiplied ( applied ) over a displacement, the work. KE = 1/2 mv^2. If the KE is equal to the PE, and it is 100% conservative, the KE is equal to .03 Joules. Thus, the final velocity will be about 1.73 m/s.
Calculations:
W = F * 'ds = KE = 1/2* m * ((v)^2)
1.5 N * .02 m = 1/2 * .02 kg * v^2
.03 J = 1/2 * .02 kg * v^2
.06 J = .02 kg * v^2
3 m^2/s^2 = v^2
+-sqrt. ( 3 m^2/s^2 ) = v
+ sqrt. ( 3 m^2/s^2 ) = v
v = 1.73 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: ->->->->->->->->->->->-> :
At 0 m/s, the domino will be at its highest height.
0 = 1.73 - 9.80t
-1.73 = -9.8 t
( -1.73/-9.8 ) = t
.1765 s= t
t = .177 s
Assuming it starts from a position of 0 m; the origin is its position. Thus, it rises .153 m above where it started.
position ( p ) = 1.73t - 4.90t^2
p = 1.73 ( .177 ) - 4.90 ( .177^2 )
p = .30621 - .1535121
p = .153 m ( above where it started )
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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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The PE, the work in this 100% conservative system, is equal to the slope of the force vs. position graph. Thus, the appropriate interval is from the position 0 m ( where it started ) to .153 m.
The force would be the slope of the PE vs. position graph. The units of the slope of a force vs. position graph are N / m, not N * m.
The PE change is the area beneath the force vs. position graph, which corresponds to average force * displacement.
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35 Minutes
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&#Good responses. Let me know if you have questions. &#