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course Phy 241
December 11 around 5pm. I'm resubmitting this lab, it never showed up in my portfolio.
101108`q001. Answer the following:
Sorry for not posting this earlier. I needed to be able to access the class notes to get all the necessary information.
That information included the following:
Between cart position 63 cm and 73 cm there is no significant tension in the rubber band (some tension is required to support the weight of the rubber band; for the moment this will be ignored).
The rubber band chain begins exerting tension at the 73 cm length, with tension increasing by about .3 Newtons for every centimeter beyond this position.
At the 80 cm position the system is in equilibrium.
If the cart is released from the 63 cm position it comes to rest for an instant at the 86 cm position, before being accelerated back toward the rubber band.
The coefficient of rolling friction is between .01 and .02, and will for the moment be considered negligible.
What is the meaning of the area of a 'graph trapezoid' on a graph of F_net vs. x?
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Area=work done by force during displacement (cm)
Int (F ds, s, a, b)=’dKE
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For this ideal situation, why should the two areas therefore be equal?
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The work-energy theorem says because the work done BY the net force on the system equals the change in KE.
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The gravitational PE of the system decreases as the suspended mass descends. It should be clear that since the gravitational force on this mass remains constant, every centimeter of descent corresponds to the same decrease in gravitational PE, and the graph is therefore linear.
The original value of the gravitational PE depends on the position with respect to which you have chosen to measure that PE. If you measure with respect to the floor, the original and final PE will both be positive, and the graph below might well represent this choice of reference position. Had you chosen to measure from the top of the table, the both values of the PE would be negative, and graph would have been below the x axis. However for any choice of reference position, the 'rise' between the two points would have been the same, since the change in PE does not depend on the reference point.
`q002. What does the slope of this graph represent?
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Fnet vs x graph: (change in net force)/(change in displacement) average ROC net force WRT displacement
PE vs. x graph: (change in PE)/(change in displacement) average ROC PE WRT displacement
The average rate of change of PE with respect to position is (change in PE) / (change in position). Since `dPE = -F_cons _ave * change in position, i.e., work done by the conservative force,
slope = (change in PE) / (change in position) = - F_cons_ave * change in position / change in position = - F_cons_ave.
The slope of the PE vs. position graph on an interval is equal and opposite to the average conservative force on that interval.
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The elastic PE of the rubber band chain is originally zero. The chain is not stretched, and contains no PE.
The chain remains unstretched until the position of the cart reaches 73 cm, so the PE remains zero up to this point.
The chain then begins to stretch. With every centimeter its tension increases, to that each centimeter corresponds to a greater increase in elastic PE. The graph of PE vs. x therefore increases at an increasing rate, as indicated in the figure below.
`q003. Using the information given at the beginning, determine the elastic PE of the rubber band chain at the 80 cm and the 86 cm positions. Assume an ideal rubber band, whose force (unlike that of a real rubber band) is wholly conservative in nature.
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80cm: elastic PE=-80J
86cm: elastic PE=-86J
The elastic force exerted by the rubber band is in the ideal case conservative because it’s experienced again in the opposite direction upon release. The work done against the elastic force can be recovered.
The rubber band begins exerting tension at length 73 cm, and tension increases by .3 N for each additional centimeter.
So at 80 cm the magnitude of the tension is .3 N / cm * 7 cm = 2.1 N, and at 86 cm the tension has magnitude .3 N/cm * 13 cm = 3.9 N.
So from 73 cm to 80 cm the magnitude of the tension increases linearly from 0 to 2.1 N, averaging 1.05 N. The tension acts in the direction opposite the displacement so it does work (-1.05 N * 7 cm) = -7.35 N * cm on the system. The PE change is equal and opposite to this work, equal to +7.35 N * cm = .074 Joules, approximately.
A similar calculation between 73 cm and 86 cm yields a PE change of about 19 N * cm, or .19 Joules.
The zero-force position is regarded as the equilibrium position, and .3 N / cm is the approximate force constant. So it would also have been possible to calculate PE using 1/2 k x^2, with x = 7 cm then with x = 13 cm.
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Between the 63 and 73 cm positions, the KE of the system will increase as its gravitational PE decreases. So the KE graph will for this interval be a straight line whose slope is equal but opposite to that of the graph of gravitational PE vs. x.
Beyond the 73 cm position the elastic PE of the system begins to increase, and continues to increase more and more quickly. The gravitational PE continues to decrease at a constant rate, at first decreasing more quickly than elastic PE increases. So the total PE will for a time continue to decrease, but as the elastic PE builds it does so more and more slowly, until at a certain point the elastic PE is increasing just as fast as the gravitational PE is decreasing. The elastic PE continues to increase at an increasing rate, so beyond this point the total PE begins to increase.
The total energy of the system remains constant (remember we're talking about the ideal case here). So as the PE of the system decreases, its KE increases. Then as the PE begins to increase, its KE decreases. So as the cart passes the 73 cm position, the KE of the system will continue to increase, but more and more slowly until it peaks at the point where the PE begins to increase. The
At first the elastic PE increases slowly and the gravitational PE decreases more quickly than the elastic PE increases. At some point the two rates are equal in magnitude, and beyond this point the elastic PE builds more quickly than the gravitational PE decreases. The graph of KE vs. x will therefore increase at a constant rate from the 63 to the 73 cm position (corresponding to the linear decrease in PE), then will continue to increase but at a decreasing rate (corresponding to the continuing but slowing decrease in PE) before peaking and beginning to decrease at an increasing rate (corresponding to the increasingly rapid increase in PE).
At the 86 cm point the system is at rest, so the KE will again be zero.
`q004. In your own words:
Clearly, beyond the 73 cm point the elastic PE increases. Why does it do so at an increasing rate?
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The total PE will, for a short time, decrease until elastic PE builds. It does so, more slowly, until a specific point where the elastic PE is INCREASING, is as fast as the DECREASING PE grav. The elastic PE, therefore, continues to increase at an increasing rate. (Greater force with every cm displaced)
Beyond the 73 cm point the rubber band begins exerting an elastic force and doing work against the rest of the system. Its force is regarded as conservative (not completely so but we analyze this as an ideal case). When a conservative elastic force does negative work, the definition of PE implies directly that the elastic PE increases.
Since the elastic force is increasing, elastic PE is increasing at an increasing rate, with respect to position.
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At what point does total PE stop decreasing and start increasing? Hint: think of how the F_net vs. x graph is related to the graph of the total PE.
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As soon as the elastic PE builds
Gravitational PE continues to decrease as elastic PE builds. Since elastic PE builds slowly at first, the net PE will continue to decrease until the elastic force becomes equal to the gravitational force, at which point the increasing elastic force will begin to exceed the gravitational force. The elastic force acts in the direction opposite the gravitational force, so beyond this point the elastic PE increases faster than the gravitational PE decreases, and the system begins to slow.
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`q005. Using the information given at the beginning, which includes the force constant for the rubber band and the mass of the suspended weight, find the following (the F_net vs. x graph could also be helpful):
The change in gravitational PE between for the interval from 60 cm to 65 cm.
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Decreases
W_grav=mgy_1-mgy_2
=-‘dU_grav
I know how to calculate this, but don’t I have to know the system’s mass? What is it?
At the 80 cm position the system is in equilibrium and the rubber band chain exerts a force of 2.1 N, so the suspended weight must be 2.1 N.
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The change in elastic PE for the interval from 60 cm to 65 cm.
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W_el=-kx or F=-kx, ‘dW=-1/2kx^2
‘dPE=1/2kx
k=(1/2)/(# of rubberbands on chain)
How many rubberbands were on the chain? Or is this even important?
The rubber bands haven't yet begun exerting a force, so elastic PE is zero.
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The change in gravitational PE and in elastic PE for the interval from 75 cm to 80 cm.
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[0.3N for every cm AFTER 73cm]
75cm: (75-73)*(0.3N)=0.6
80cm: (80-73)*(0.3N)=2.1
‘dPE_grav=-F*’dy: (-0.6*75cm=-45) and (-2.1*80cm=-168)
‘dPE_el (increases): I don’t know how to solve any of the elastic PE’s without knowing “k”?!?
You just about have it.
You've used 0.3 N. That quantity should in fact be 0.3 N / cm. This is 'k'.
x in the formula 1/2 k x^2 is the displacement from equilibrium, which occurs at the 73 cm position.
At the 75 cm position, for example, x = 2 cm. You have the force at that position, which is 0.6 N. Average force over the 2 cm displacement from equilibrium to the 75 cm position is therefore 0.3 N and the work is 0.3 N * 2 cm = .006 Joules.
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The change in gravitational PE and in elastic PE for the interval from 80 cm to 85 cm.
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[0.3N for every cm AFTER 73cm]
80cm: (80-73)*(0.3N)=2.1
85cm: (85-73)*(0.3N)=3.6
‘dPE_grav=-F*’dy: (-2.1*80cm=-168) and (-3.6*85cm=-306)
The displacement is only 5 cm. The average force is the average of the two forces you give.
You also need to calculate the change in gravitational PE, which is just -2.1 N * displacement (recall the above note showing that the suspended weight must be 2.1 N)
‘dPE_el:
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The change in KE for each of the three intervals given above:
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‘dKE=1/2mv^2
I think all of these would be equal and opposite to ‘dPE.
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You're doing OK here. Just a couple of adjustments and you'll have it. See my notes.