Open Query 12

course Phy 121

012. `query 12*********************************************

Question: `qQuery 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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Your solution: Fnet=m2*9.8

A=(m2*9.8)/(totalm)

PE= -m2g*y

confidence rating #$&*

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Given Solution:

`a** The net force on the system is the force of gravity on the suspended weight: Fnet = m2 * 9.8 m/s/s. Gravity also acts on m1 which is balanced by force of table on m1, so the forces on m1 make 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.

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Self-critique (if necessary): OK

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Self-critique rating #$&* OK

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Question: `qHow would friction change your answers to the preceding question?

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Your solution: It would result in PE=m2*g-resistance.

confidence rating #$&*

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Given Solution:

`a**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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Self-critique (if necessary): OK

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Self-critique rating #$&* OK

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Question: `qExplain 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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Your solution: The PE is the area under the curve of the graph.

confidence rating #$&*

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Given Solution:

`a** 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. **

Self-critique (if necessary): OK

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Self-critique rating #$&* OK

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Question: `q 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 rubber bands?

Your solution: rise=dF run=dstretch. The area does represent work done.

If the rubber band is pulled by the object then the objects work is positive and the rubber bands is negative, and vice versa for if the object is pulled by the rubber band.

confidence rating #$&*

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Given Solution:

`a** The rise of the graph is change in force, the run is change in stretch. So slope = rise / run = change in force / change in stretch, which the the average rate at which force changes with respect to stretch. This basically tells us how much additional force is exerted per unit change in the length of the rubber band. 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. **

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

I was slightly confused about the area under the curve of the graph. I ultimately understood the concept after some study.

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Self-critique (if necessary): OK

Self-critique rating #$&* OK

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&#Good responses. Let me know if you have questions. &#

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