Assignment12finalquerry

course phy201

Having a little trouble following text and questions

???????????assignment #012

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012. `query 12

Physics I

11-23-2007

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12:41:20

How would friction change your answers to the preceding question?

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RESPONSE -->

confidence assessment: 0

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12:41:38

**Friction would act to oppose the motion of the mass m1 as it slides across the table, so the net force would be m2 * g - frictional resistance. **

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RESPONSE -->

self critique assessment: 3

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12:42:33

Explain how you use a graph of force vs. stretch for a rubber band to determine the elastic potential energy stored at a given stretch.

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RESPONSE -->

Take the length of stretch divide by the force exerted on the rubber band.

confidence assessment: 2

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12:43:10

** If we ignore thermal effects, which you should note are in fact significant with rubber bands and cannot in practice be ignored if we want very accurate results, PE is the work required to stretch the rubber band. This work is the sum of all F * `ds contributions from small increments `ds from the initial to the final position. These contributions are represented by the areas of narrow trapezoids on a graph of F vs. stretch. As the trapezoids get thinner and thinner, the total area of these trapezoids approaches, the area under the curve between the two stretches.

So the PE stored is the area under the graph of force vs. stretch. **

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RESPONSE -->

It would be the area of the graph.

self critique assessment: 3

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12:43:49

STUDENT QUESTIONS: Does the slope of the F vs stretch graph represent something? Does the area under the curve represent the work done? If so, is it work done BY or work done ON the rbber bands?

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RESPONSE -->

The area is a representation of the work done bt the rubber band.

confidence assessment: 2

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12:44:14

** Slope isn't directly related to any physical quantity. The area is indeed with work done (work is integral of force with respect to displacement).

If the rubber band pulls against an object as is returns to equilibrium then the force it exerts is in the direction of motion and it therefore does positive work on the object as the object does negative work on it.

If an object stretches the rubber band then it exerts a force on the rubber band in the direction of the rubber band's displacement, and the object does positive work on the rubber band, while the rubber band does negative work on it. **

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RESPONSE -->

self critique assessment: 1

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More detail in solutions and self-critiques is recommended. See my notes on other assignments regarding this.

end of document

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