Asst 12 Query

course pHY 201

I accidently hit 'enter' before I pasted my work. Here it is!

No problem. Very easy for me to delete.

Physics I

03-02-2007

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15:38:39

Query set 3 #'s 13-14 If an object of mass m1 rests on a frictionless tabletop and a mass m2 hangs over a good pulley by a string attached to the first object, then what forces act on the two-mass system and what is the net force on the system? What would be the acceleration of the system? How much would gravitational PE change if the hanging mass descended a distance `dy?

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RESPONSE -->

Forces: Fgrav on mass 2. Fgrav on mass 1 which balances the force exerted by the table on mass 1.

Fnet = Fgrav on mass 2

a = Fnet / (mass 1 + mass 2)

PE = -dy

confidence assessment: 2

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15:57:42

** The net force on the system is the force of gravity on the suspended weight: Fnet = m2*9.8m/s/s

Gravity also acts on m1 which is balanced by force of table on m1, so this force makes no contribution to Fnet.

Acceleration=net force/total mass = 9.8 m/s^2 * m2 / (m1+m2).

If the mass m2 descends distance `dy then gravitational PE decreases by - m2 g * `dy.

COMMON MISCONCEPTIONS AND INSTRUCTOR COMMENTS:

The forces acting on the system are the forces which keep the mass on the table, the tension in the string joining the two masses, and the weight of the suspended mass. The net force should be the suspended mass * accel due to gravity + Tension.

INSTRUCTOR COMMENT:

String tension shouldn't be counted among the forces contributing to the net force on the system.

The string tension is internal to the two-mass system. It doesn't act on the system but within the system.

Net force is therefore suspended mass * accel due to gravity only

'The forces which keep the mass on the table' is too vague and probably not appropriate in any case. Gravity pulls down, slightly bending the table, which response with an elastic force that exactly balances the gravitational force. **

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RESPONSE -->

I dont understand why Fnet = 9.8 m/s^2 * m2 / (m1+m2) and not 9.8 m/s^2 * (m1+m2). Wouldn't m2 / (m1+m2) be the average mass rather than the total mass?

As you say , a = Fnet / (mass 1 + mass 2), where Fnet = Fgrav on mass 2.

The force exerted by gravity on mass 2 is m2 * g.

So a = m2 * g / (m1 + m2).

m2 / (m1 + m2) is the proportion of m2 to the total mass of the system; the greater this proportion the closer the acceleration will be to g. If m1 is very small the acceleration is close to g. If m2 is very small the acceleration is close to 0.

I should have included the entire equation of PE = m2 * g * `dy rather than PE = `dy which doesn't make sense by itself.

self critique assessment: 2

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15:59:45

How would friction change your answers to the preceding question?

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RESPONSE -->

Forces acting on the system would be -Ffrict, Fgrav on mass 2, Fgrav on mass 1 which balances the force exerted by the table on mass 1.

Fnet = Fgrav on mass 2 - Ffrict

confidence assessment: 2

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16:00:11

**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 -->

ok

self critique assessment: 3

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16:06:54

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 -->

The graph shows us how much force is necessary to stretch a rubber band. As the band stretches, it is gaining PE which can be seen on the graph as the area beneath the curve at a given stretch.

confidence assessment: 2

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16:12:55

** 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 -->

ok

self critique assessment: 3

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16:14:19

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 beneath the curve represents the PE/ work done to stretch the rubber band. This is work done on the rubberbands.

confidence assessment: 2

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16:15:47

** 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 -->

ok I didn't mention that the rubber band does negative work on the object.

self critique assessment: 3

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16:16:23

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

It helped me understand PE a little bit better. But I am still very confused and need clarification on KE and PE.

self critique assessment: 2

In a nutshell, KE is energy of motion, equal to 1/2 m v^2.

PE is stored energy that can be converted to energy of motion, as in an object held at some height in a gravitational field, or a stretched rubber band.

You will see much more about KE and PE in upcoming assignments.

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Good. See my notes and let me know if you have questions.