course phy 201 cƩٽassignment #012
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21:44:07 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 --> Fnet = 6 * 9.8 Fnet = 58.8J f = ma f/m = a Fnet = F1 - F2 Fnet = 88.2 - 58.8J you must include the force of the first mass on the second since it is one system Fnet = 30J PE = mgy w= - PE w = F * d w = 58.8 * dy confidence assessment: 1
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21:48:27 ** 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 --> Im guessing since friction is neglected then m1 willl move without any resistance and because the force downward and normal force is balanced then that is why this mass does not contribute to fnet. F= ma since I got the Fnet wrong I answered this problem wrong as well but i did have it set up right. PE = mgy self critique assessment: 2
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21:49:25 How would friction change your answers to the preceding question?
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RESPONSE --> this would cause an additional force to be incorp. into the equation confidence assessment: 3
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21:50:11 **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 --> self critique assessment:
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21:53:22 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 --> we could find work by measuring the total area under the curve of this graph, hence we would be finding PE at the same time confidence assessment: 3
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21:58:46 ** 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 --> I left out an important exception and realized thermal effects had such an effect on this set up. I am confused because it looks like we used the same amount of energy for W, PE and KE. I am going to guess that since we are dealing with a rubber band energy is conserved and since we are stretching it will a mass and the mass has motion that is why PE moves into KE self critique assessment: 2
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21:59:41 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 --> confidence assessment:
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22:04:08 ** 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 --> area = work rubber band does work on object = + work on object if force is in the direction of motion if the object stretches the rubber band then the obj is doind work on the rubber band it will be + if the force is the same direction of displacement then work will be + self critique assessment: 2
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22:06:36 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> i need to review problems dealing with KE and PE I am still not comfortable with these variables and how they are associated with + and - work etc. self critique assessment: 2
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