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course PHY 201
11/12 8:25pm I got bogged down on this and had to quit and submit for help. I'm not sure on how to apply the definitions and formulas for the calculations we need to do. Please take a look and let me know what you think. Thanks
The figure below indicates how various quantities `dt, v0, vf, `dv, vAve, `ds, a related to uniform acceleration on an interval, and the additional quantities Fnet, p0, pf, impulse, `dp, `dW_net, KE0, KEf, `dKE obtained as a direct result of Newton's Second and Third Laws, can be paired to reason out other quantities.Every quantity in this diagram can be found based on at least one other pair of quantities, using a few basic definitions and theorems:
· definition of average rate
· definition of average velocity
· definition of average acceleration
· linearity of the v vs. t graph for uniform acceleration
· definition of work and energy
· definition of impulse and momentum
· work-kinetic energy theorem
· impulse-momentum theorem
A cart is attached by one string to a rubber band chain at left, and by another to a suspended mass over a pulley.
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.
Between the 63 cm and 73 cm positions, tension on the left is negligible so the net force on the system is the constant gravitational force on the suspended mass.
Between the 73 and 80 cm positions, tension on the left increases linearly while gravitational force on the suspended mass remains constant, so the net force on the system decreases linearly to 0.
Between the 80 cm and 86 cm positions, tension on the left continues to increase linearly while gravitational force on the suspended mass remains constant, so the net force continues to decrease linearly into negative values.
The class was asked to sketch graphs of F_net, PE_grav, PE_elastic and KE for a system released from rest at the 63 cm position, as it moves to the 86 cm position.
F_net is constant between the 63 and 73 cm position, being just equal to the weight of the suspended mass.
F_net then decreases linearly as the elastic force exerted by the rubber bands increases linearly in magnitude.
For an accurate sketch of the ideal F_net vs. x between these positions, the area between the graph and the x axis for the interval from 63 cm to 80 cm (this area is above the graph) should be equal to the area between the 80 cm and 86 cm positions (this area lies below the graph).
`q001. Answer the following:
What is the meaning of the area of a 'graph trapezoid' on a graph of F_net vs. x?
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The area represents the total change in net force relative to the total change in position.
That would be the interpretation of the slope.
The area represents average force * displacement, or work.
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For this ideal situation, why should the two areas therefore be equal?
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They are equal because Fnet increases and decreases linearly.
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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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It represents the avg rate of change of PEgrav
That would be the avg rate of change of PEgrav with respect to position.
This turns out to be equal to the average force: since `dPE = F_ave * `ds, F_ave = `dPE / `ds, which is the average rate of change of PE with respect to position.
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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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The rubber band began exerting a tension at the 73cm length and increased by .3N per cm beyond this position. So, at the 80cm position, the rubber band is exerting a tension of .21N
PE=1/2kx^2
X=7cm
K=2.1N
=(1/2*2.1N)*49cm^2
=51N*cm^2
Good, but k = .3 N / cm. 2.1 N is the force at the 7 cm position.
So 1/2 k x^2 = 1/2 ( .3 N / cm) * (7 cm)^2 = 7.4 N / cm * cm^2 = 7.4 N * cm.
Again, I’m not sure on how to express the value of PE. I’m not even sure if I am correct on applying the definition of PE. What is the correct way to express PE?
work = force * displacement, so the unit of work is (unit of force * unit of displacement).
N * cm^2 would be (unit of force * (unit of displacement)^2), which is not a unit of work (or of energy, which has the same units as work).
The same comments apply to your next set of calculations:
At the 80cm position, the rubber band is exerting a tension of 3.9N
PE=1/2kx^2
X=13cm
K=3.9N
=(1/2*3.9N)*169cm^2
=330N*cm^2
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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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Is this because it is acting to stop the car from moving so that’s why it is increasing so fast?
The rubber band is exerting greater and greater force, so for every centimeter is does more work.
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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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At the 86cm mark because that’s when the car is stopped and then starts going back to it’s equilibrium point.
When the F_net vs. x graph drops below the x axis, the area beneath the graph starts going negative.
It's the area that is equal to the change in PE, so the PE begins decreasing at that point.
At what x coordinate does the F_net vs. x graph become negative?
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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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PE=m*g*h
m*g=.3N
h=1M
PE=.3N*M
`dPE is equal and opposite to the work done by the gravitational force. That force is indeed .3 N. However the height of the weight above the floor is not equal to its displacement.
What is the displacement, what is the force exerted by gravity, are they in the same direction or not?
What therefore is the work done by gravity?
What therefore is the change in PE_grav?
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The change in elastic PE for the interval from 60 cm to 65 cm.
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PE=1/2kx^2
K=
X=5cm
For this formula, what does k represent and how is it calculated?
See also my previous notes. k is the force constant, which is the slope of the force vs. position graph.
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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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The change in gravitational PE and in elastic PE for the interval from 80 cm to 85 cm.
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The change in KE for each of the three intervals given above:
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KE=1/2mv^2
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You're mostly on the right track. See if my notes can get you rolling again.
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