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course Phy 201
`q001. A stretched rubber band is used to set a system in motion. The system comes to rest. For each system described below identify the conservative forces and nonconservative forces acting on the system, the directions of these forces, the direction of the displacement through which each force acts, the work done by each of these forces and whether that work is positive or negative, and state how the kinetic and potential energies change as a result.System 1: A rotating strap loaded at various points with weights, resting on a domino.
Cons - grav (-)
Noncons - air resistance (-)
PE and KE (+)
Ds is (-)
dW (+)
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System 2: A wood block on the tabletop, across which it slides.
Cons - grav (-)
Noncons - air resistance and friction (-)
W is (+)
PE and KE (+)
Ds (+)
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System 3: An arrow which is shot straight up.
Cons - grav (-)
Noncons - friction and air resistance (-)
W is (+)
Ds (+)
KE and PE (+)
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System 4: A rotating strap loaded at various points with weights and a two sheets of tissue paper each weighted down at the bottome and attached near the ends of the strap, resting on a domino.
Cons - gravity (-)
Noncons - air resistance and friction (-)
W - (+)
KE and PE (-)
Ds (-)
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System 5: A toy car released on a slightly inclined tabletop.
Cons - gravity (-)
Noncons - air resistance and friction (-)
W - (-)
KE (-) and PE (+)
Ds (+)
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System 6: A pulley with unbalanced weights supported by a string over both sides.
Cons - gravity (-)
Noncons - air resistance and friction (-)
W - (-)
KE (-) and PE (+)
Ds (+)
See the link at the end of this document for commentary on these questions.
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`q002. A rubber band chain experiences tensions of 0 N, 1 N and 1.6 N when its 'stretch' is respectively 0, 5 cm and 10 cm.
What is your best estimate of the work done to stretch in the chain from 0 'stretch' to the 10 cm 'stretch'?
W = force * ds
W = 1.6 N * 10 cm = 16J
1.6 N is the maximum force exerted. It isn't applied for the whole interval. So you have significantly overestimated the work.
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Upon release the rubber band accelerates a rotating system from rest. The rotating system consists of a thin strap whose mass we will regard as negligible and whose length is 20 cm, with two 50-gram magnets on each end.
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When the rubber band has separated from the system its angular velocity is 10 radians / second (about 570 degrees / second). What is its kinetic energy?
KE = ½ mv^2
KE = ½ (.1g)(10 rad/s) = 5 J
This is a rotating system; you would use 1/2 I omega^2.
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What percent of the work required to stretch the rubber band is transferred to the kinetic energy of the system?
31 %
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How far does would a magnet travel in one 360-degree revolution, if it wasn't slowing down?
360 d/s / 570 d/s = 0.63 s
The answer to 'how far' would be in units of distance, e.g., cm, m, miles, etc.. It wouldn't be in seconds.
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How far would it travel in a second?
About 350 degrees
You're given the angular velocity, which is about 570 degrees / second. That's more than 350 degrees in a second.
However 'how far' would be measured in units of length, not angle.
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What therefore is its speed?
350 d/s^2
Speed would be in units of distance / time.
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What is its kinetic energy at that speed?
KE = ½ mv^2
KE = 2/2 (10g)(570 d/s)^2 = 1,624,500 g*d/s
570 deg / sec is an angular velocity, not a velocity. It would be used with moment of inertia, and you would want to use the angular velocity in radians / second to avoid an inconvenient conversion of units.
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If instead of transferring its energy to the rotating system the rubber band was used to 'shoot' the two magnets (now separated from the strap) straight upward, how much gravitational PE would they gain before coming to rest, and how high would they be at that point?
PE = -(Wcons)
Wcons = 9.8 m/s^2(.1kg) = .98 J
F = m a; multiplying mass by acceleration gives you force, not work.
A kg m/s^2 is a Newton, not a Joule.
PE = -0.98 J
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If the rubber band was used to 'shoot' the magnets across the table, where they slide against a frictional force equal to 30% of their weight, how far would they travel?
Ffriction = (9.8 m/s^2)(.1 kg)(.3) = 0.294 J
That would be 0.294 kg m/s^2, which is 0.294 Newtons.
You would still need to figure out how far the magnets would travel.
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`q003. An 'ideal' rubber band has a force vs. stretch graph which passes through the origin and has slope .8 Newtons / centimeter.
What is its tension when its stretch is 10 cm?
.8n/cm * 10 cm = 8N
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How much work does it take to stretch it from position 0 cm to position 10 cm?
W = fnet * ds
W = .8N * 10 cm
W = 8 J
A N * cm is not a Joule. A Joule is a N * m.
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In terms of the symbol x, what is its tension when its stretch is x?
W = .8N * x cm
You wouldn't use W to stand for tension. W would ordinarily indicate work.
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In terms of x, how much work does it take to stretch it from the 0 cm position to position x?
W = .8N * x cm (x)
Good reasoning but the units are N/cm, not N * cm. The unit N / cm would be multiplied by x^2 so you would end up with energy units, but they might not be Joules.
Also .8 N/cm * x is the maximum force exerted during displacement x, not the average force. The average would be .4 N/cm * x.
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What is its elastic potential energy at the 10 cm position, and at position x?
PE = -(.8N * x cm (x))
The work required to stretch would be positive.
During the stretch the tension does negative work.
The work to stretch is .4 N/cm * x, the work done by the presumably conservative tension is -.4 N/cm * x.
So change in PE, being the negative of this, is -.4 N/cm * x.
You haven't calculated this for the 10 cm stretch.
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`q004. A force is directed along a line in the xy plane. Between two points of the line the 'rise' is 15 cm and the 'run' is 20 cm.
The line segment between these points is the hypotenuse of a right triangle whose legs are the 'rise' and the 'run'. What is the length of that segment?
15^2 + 20^2 = c^2
C = 25 cm
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What percent of that length is the 'rise'?
25* sin (.75) = 33 %
The rise is 15 cm, which is 60% of the 25 cm hypotenuse.
No sine or cosine is required to figure this out.
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What percent of that length is the 'run'?
25* cos(0.75) = 25%
The 20 cm run is 80% of the 25 cm hypotenuse.
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Suppose the force is 50 Newtons.
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If the y component of that force is in the same proportion to the force as the 'rise' to the hypotenuse, what is the y component?
50*15 = 750/20 = 37.5
Rise = 37.5cm
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If the x component of that force is in the same proportion to the force as the 'rise' to the hypotenuse, what is the x component?
20x = 750
X = 66.6
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Estimate the angle between the 'run' leg of the triangle and the hypotenuse.
47 degrees
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Self-critique (if necessary):
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Good work on many of these questions.
See the document at http://vhcc2.vhcc.edu/ph1fall9/2012-13/121008_solnpaoienrnkgfasp.htm for complete solutions to and commentary on these questions.