10-3

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course Phy 201

`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 dominos make the rb chain stretch 40 cm and a domino weighs 17.3 g

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17.3 grams is not a weight, it is a mass.

Below you calculate the mass of a single domino, correctly obtaining .166 N.
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Show how you analyzed your data.

We calculated Fnet and found the area under a trapezoid.

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Give your conclusions.

Fnet = 0.0173 kg * 9.8 m/s^2 = 0.166 N

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This would be the force exerted by gravity on a single domino.

The tension in the rubber band chain supported five dominoes against gravity.

Once you have the correct tension at the 40 cm stretch, you can calculate the work required to stretch it, which would correspond to the area beneath a force vs. stretch or a force vs. length graph.
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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

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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.

In order corresponding with 10-1 lab. Measured in degrees/s.

From 30 cm 66.3, 82.57, 57.7, 73.5, 62.79. From 15 cm 113.5, 70.97, 80, 76.27, 73.21. From 7.5 cm 76.51, 97.83, 74.02, 70.59, 102.25.

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These are reasonable results, but you didn't include your data in this document so your results can't be immediately verified.
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By referring to your 10-1 document I can verify that these appear to be the correct average angular velocities.

I would have been appropriate to copy your data from the preceding document.
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The question, however, asked you to calculate the angular velocity the system would have had, assuming the energy stored in the rubber band was converted to angular kinetic energy.
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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 final = ½ (1500 kg)(40)^2 = 1,200,000 J

KE initial = ½ (1500kg)(50 mph)^2 = 1,875,000 J

@&
The units of this calculation would be kg * miles^2 / hours^2, not Joules.

kg * miles^2 / hours^2 is a unit of energy, but not a standard unit. You would need to do the appropriate conversions to get this energy in Joules. A Joules is a kg * m^2 / s^2.
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dKE = 675,000

dWcons = (9.8 m/s)(1500kg)(-25m) = 367,500

@&
The acceleration of gravity is 9.8 m/s^2, not 9.8 m/s.

Note that with this correction the units of your calculation will be

m/s^2 * kg * m = kg * m^2 / s^2.

So this energy is in Joules.
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How much work do nonconservative forces do on the truck?

dWnet = dWcons + dWnoncons

dWnoncons = 307, 500 J

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If all your energies had been in Joules, you would have gotten the right answer here. As it is you are adding and subtracting energies is kg miles^2 / hour^2 with energies in Joules.
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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?

1500 kg (0.02) = 30

1500 kg - 30 = 1470 kg

W = Fnet *ds

W = (1470kg)(0.309)(1000m) = 454230 J

@&
1470 kg is not a force, it's a mass.

It's not clear what .309 means. It isn't given with units and I don't see any other mention of it in this document.

In any case the frictional force is 2% of the weight of the truck, and this result would be multiplied by 1000 m.
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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?

W = Fnet *ds

W = 1500kg)(0.309)(1000m) = 463,500 J

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1500 kg is a mass, not a force; again unclear on the .309.

Above you got

dWnoncons = 307, 500 J

Because of unit inconsistencies this isn't correct, but the procedure by which you got it was otherwise correct.

Once you correct your units and that preceding result, and correctly calculate the work done by the frictional force, which is part of `dW_noncons, you will be able to calculate the work done by the remaining nonconservative forces.
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What average force did those other nonconservative forces have to exert during the truck's 1000 meter displacement?

463,500 J - 454,230J = 9270 J/1000m

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For reasons listed above 9270 J isn't correct, but when you do get the correct result you'll be correct in dividing it by the 1000 m displacement.
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9.27 J/m

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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?

Fgrav = 9.8m/s^2 (2000kg) = 19600m/skg

19600m/s kg (0.02) = 392 = Ffriction

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This is a correct calculation of frictional force, except that it's m/s^2 * kg.

This unit is the same as kg * m/s^2, or Newtons.
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dWcons = 19600 m/s kg (25) = 490,000J

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You don't have all your units here, but you're multiplying the right numbers and once you get the units in place you will get kg m^2 / s^2, which is Joules.
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all other dWnoncons = 470,400 N

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You haven't found `dKE so you're probably not yet in a position to put all this together, but you're on the right track.
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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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@&
Lots of good work, but also a number of errors. Please make the corrections you can make 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.

10-3

@& 17.3 grams is not a weight, it is a mass. Below you calculate the mass of a single domino, correctly obtaining .166 N. *@

@& These are reasonable results, but you didn't include your data in this document so your results can't be immediately verified. *@

@& By referring to your 10-1 document I can verify that these appear to be the correct average angular velocities. I would have been appropriate to copy your data from the preceding document. *@

@& The question, however, asked you to calculate the angular velocity the system would have had, assuming the energy stored in the rubber band was converted to angular kinetic energy. *@

@& The units of this calculation would be kg * miles^2 / hours^2, not Joules. kg * miles^2 / hours^2 is a unit of energy, but not a standard unit. You would need to do the appropriate conversions to get this energy in Joules. A Joules is a kg * m^2 / s^2. *@

@& The acceleration of gravity is 9.8 m/s^2, not 9.8 m/s. Note that with this correction the units of your calculation will be m/s^2 * kg * m = kg * m^2 / s^2. So this energy is in Joules. *@

@& If all your energies had been in Joules, you would have gotten the right answer here. As it is you are adding and subtracting energies is kg miles^2 / hour^2 with energies in Joules. *@

@& 1470 kg is not a force, it's a mass. It's not clear what .309 means. It isn't given with units and I don't see any other mention of it in this document. In any case the frictional force is 2% of the weight of the truck, and this result would be multiplied by 1000 m. *@

@& For reasons listed above 9270 J isn't correct, but when you do get the correct result you'll be correct in dividing it by the 1000 m displacement. *@

@& This is a correct calculation of frictional force, except that it's m/s^2 * kg. This unit is the same as kg * m/s^2, or Newtons. *@

@& You don't have all your units here, but you're multiplying the right numbers and once you get the units in place you will get kg m^2 / s^2, which is Joules. *@

@& You haven't found `dKE so you're probably not yet in a position to put all this together, but you're on the right track. *@

@& Lots of good work, but also a number of errors. Please make the corrections you can make 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.

@&
17.3 grams is not a weight, it is a mass.

Below you calculate the mass of a single domino, correctly obtaining .166 N.
*@

@&
These are reasonable results, but you didn't include your data in this document so your results can't be immediately verified.
*@

@&
By referring to your 10-1 document I can verify that these appear to be the correct average angular velocities.

I would have been appropriate to copy your data from the preceding document.
*@

@&
The question, however, asked you to calculate the angular velocity the system would have had, assuming the energy stored in the rubber band was converted to angular kinetic energy.
*@

@&
The units of this calculation would be kg * miles^2 / hours^2, not Joules.

kg * miles^2 / hours^2 is a unit of energy, but not a standard unit. You would need to do the appropriate conversions to get this energy in Joules. A Joules is a kg * m^2 / s^2.
*@

@&
The acceleration of gravity is 9.8 m/s^2, not 9.8 m/s.

Note that with this correction the units of your calculation will be

m/s^2 * kg * m = kg * m^2 / s^2.

So this energy is in Joules.
*@

@&
If all your energies had been in Joules, you would have gotten the right answer here. As it is you are adding and subtracting energies is kg miles^2 / hour^2 with energies in Joules.
*@

@&
1470 kg is not a force, it's a mass.

It's not clear what .309 means. It isn't given with units and I don't see any other mention of it in this document.

In any case the frictional force is 2% of the weight of the truck, and this result would be multiplied by 1000 m.
*@

@&
For reasons listed above 9270 J isn't correct, but when you do get the correct result you'll be correct in dividing it by the 1000 m displacement.
*@

@&
This is a correct calculation of frictional force, except that it's m/s^2 * kg.

This unit is the same as kg * m/s^2, or Newtons.
*@

@&
You don't have all your units here, but you're multiplying the right numbers and once you get the units in place you will get kg m^2 / s^2, which is Joules.
*@

@&
You haven't found `dKE so you're probably not yet in a position to put all this together, but you're on the right track.
*@

@&
Lots of good work, but also a number of errors. Please make the corrections you can make 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.