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course phy201
`q001. In today's lab activity you found the work done by the rubber band used in the preceding class to energize the rotating ramp.Give your data.
5 dominoes bring the rubber band chain to its stretched length of 40 cm. the weight of each domino was 17.3g
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Show how you analyzed your data.
We calculated net force and found the area under the trapezoid
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Give your conclusions.
Fnet= .0173kg (9.8m/s^2)
Fnet=.166N
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This is the force exerted by a single domino.
There are five dominoes, and you still need to figure out the area beneath the force vs. length graph to get the energy.
You've given the stretched length of the rubber band, but not its unstretched length. Assuming it was stretched about 15 cm beyond the length at which it started exerting a force, the work would be around the range of 0.1 to 0.2 Joules.
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If you have collateral observations and/or ideas for extending this investigation, give a synopsis of your observations and/or ideas.
None, that I can put into words that make any sense, maybe a couple of starting points but nothing that adds to the observation or investigation as a whole.
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`q002. Using the moment of inertia of the ramp and dominoes (found in the preceding document) and the energy in your rubber band chain, find the angular velocity of the rotating ramp, assuming that all of the potential energy stored in the rubber band chain is transferred to the ramp and that none of that energy has yet been dissipated by friction.
Ang velocity:
At 30 cm
1-66.3
2-82.6
3-57.7
4-73.5
5-62.8
At 15 cm
1-113.5
2-70.9
3-80.0
4-76.3
5-73.2
At 7.5 cm
1-76.3
2-97.8
3-74.0
4-70.5
5-102.2
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This question could have been stated more clearly, but it was asking you to figure out how fast the system would have been rotating if the energy of the chain all went into the rotation.
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`q003. If my truck has mass 1500 kg and descends through a total rise of magnitude 25 meters while slowing from 50 mph to 40 mph, by how much does its kinetic energy change, and how much work is done on it by the conservative gravitational force?
KE=1/2mv^2
KEo=1/2(1500)(50)^2
Kef=1/2(1500)(40)^2
dKE=675,000 J
Fgrav=367,500
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You don't show the calculation, but this is the correct result for `dW_grav. F_grav = 1500 kg * 9.8 m/s^2 = 14 700 kg m / s^2, or 14 700 Newtons.
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You aren't using units. The units of your calculations would be kg * (miles/hour)^2 = kg mile^2 / hr^2, not Joules.
A Joules is a kg * m^2 / s^2.
kg * mile^2 / hr^2 is, in any case, a unit of energy, so you have done a correct energy calculation. You just put the wrong unit on it.
With proper conversion you would get the energies in Joules.
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How much work do nonconservative forces do on the truck?
dKE=dWnet
dWnet=dWcons + dWnoncons
675,000=367,500+x
dWnon=307,500 J
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Road friction exerts a force equal to about 2% of the truck's weight. If it coasted a distance of 1000 meters along the road, what is the work done on the truck by friction?
1500x .02=30
1500-30=1470
W=FnetxdS
1470(.309)x1000
25/9/9m/s^2=.309
W=454230 J
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You need to use units.
1500 means 1500 kg, so 2% of 1500 would represent 30 kg.
Multiplying 30 kg by 1000 m gives you 30 000 kg * m. This isn't a unit of work.
It's not clear how the .309 ws reasoned out or what its units are.
In the end this calculation ends up with about 300 000 Joules of work.
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Friction is one of the nonconservative forces acting on the truck. What is the work done on the truck by all other nonconservative forces combined?
1500(.309)(1000)
W=463500
454230-463500
9270J for 1000M
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What average force did those other nonconservative forces have to exert during the truck's 1000 meter displacement?
9270J/1000M
Fave=9.27 J/M
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This result would be modified by correcting some of the quantities you calculated earlier, but this would be the correct way to calculate the work.
Note that J / m = (kg m^2 / s^2) / m = kg m/s^2, which is Newtons.
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My 2000 kg station wagon coasts along the same path and its speed decreases from 50 mph to 47 mph. Friction again exerts a force equal to 2% of the vehicle's weight. What average force was exerted by nonconservative forces other than friction?
9.8m/s^2x2000= 19600 x .02= 392
Wcons= 19600
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19600, calculated with correct units, would be 19600 kg m / s^2, or 19600 Newtons.
This is the weight of the car, i.e., the force exerted on it by gravity, not the work done by the conservative force. You do calculate that correctly below as 490 000, which would be 490 000 Joules.
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Ffriction= 392
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This would be 392 Newtons.
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19600(-25)= 490,000 dWcons
490,000-19600= 470,400 N all other non cons forces
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To get the work done by all nonconservative forces you would subtract the change in KE from the work done by conservative forces.
Then calculating the work done by friction, which is part of the work done by conservative forces, you can find the work done by the remaining conservative forces.
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`q004. University Physics.
You did an experiment with a balanced vertical strap, to which two small magnets (mass 3 grams each) were added. The goal was to determine how much energy was transferred from the rubber band chain to the strap.
Give your data.
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Show how you analyzed your data.
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Give your conclusions.
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If you have collateral observations and/or ideas for extending this investigation, give a synopsis of your observations and/or ideas.
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`q005. University Physics.
You were asked to think about how you could determine the energy dissipated per centimeter by rolling friction on the axel.
You were advised to think of everything that could be measured for that system.
Please share your thoughts, and if possible, address the question of how some of the things that could be measured might pertain to the original goal of determining energy dissipated by rolling friction.
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You're on the right track.
You still need to use units throughout your calculation. You've got a lot of units errors, that would at times reveal errors in procedure.
I do recommend that you make at least those revisions you can make fairly easily and quickly, without getting bogged down.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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