course Phy 201
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22:49:23 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 --> m2*9.8m/s^2=Force Force of gravity on m1 and m2 Normal force of table and m1 Net force=force of gravity on m2 F=ma a=Force/(m1+m2) The PE of the system will decrease as the altitude of the hanging mass decreases by an amount equal to the work done by gravity on the mass. confidence assessment: 2
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22:50:34 ** 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 --> PE decreases by m2g*`dy. self critique assessment: 2
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22:53:24 How would friction change your answers to the preceding question?
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RESPONSE --> In the presence of friction, there is a frictional force on m1. This force acts in the direction opposite to the motion of the system Net force=Force from earlier-frictional force confidence assessment: 2
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22:53:50 **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 --> Net force=m2*g-friction self critique assessment: 2
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23:02:24 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 --> Work to stretch=area under curve The PE is equal to the area under the curve because you are neglecting heat loss confidence assessment: 2
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23:03:25 ** 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 --> The work is the sum of all the F *`ds for the graph to get trapezoids on the graph. self critique assessment: 2
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23:05:53 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 work required to stretch the bands is the area under the curve. confidence assessment: 1
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23:08:51 ** 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 --> Slope isn't any physical quantity. The area is the work done. If the rubber band pulls against an object, then the force it exerts is in the direction of motion=positive work on the object but the object does negative work on it If the object stretches the rubber band, 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. self critique assessment: 2
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