course PHY 121 1/4 012. `query 12*********************************************
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Given Solution: ** 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. ** STUDENT COMMENT: I don't understand why m1 doesn't affect the net force. Surely it has to, if mass1 was 90kg, or 90g, then are they saying that the force would be the same regardless? INSTRUCTOR RESPONSE: m1 has no effect on the net force in the given situation. Whatever the mass on the tabletop, it experiences a gravitational force pulling it down, and the tabletop exerts an equal and opposite force pushing it up. So the mass of that object contributes nothing to the net force on the system. The mass does, however, get accelerated, so has a lot to do with how quickly the system accelerates. Also if friction is present, that mass has a large effect on the magnitude of the frictional force. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understood all of it except the part about the change in PE.. in the give answer in the statement PE decreases by - m2 g * `dy, does g = gravity? I assume it doesn’t stand for grams and gravity would make sense, I just wanted to make sure.
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Given Solution: **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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn’t write the equation I should have. ------------------------------------------------ Self-critique Rating: ---- ********************************************* Question: 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There is an direct relationship between PE and amt. of stretch in a rubber band. As the amount of stretch increases, there is a greater PE; as the amount of stretch decreases, the PE required is smaller. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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): I got this completely wrong.. I need to go back and look at the lesson on this and see if I can figure out the correct graph. ------------------------------------------------ Self-critique Rating: ---- ********************************************* Question: 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The slope of the force versus stretch graph would be rise over run, equaling force/stretch. This tells use the force per stretch ratio (if that makes sense?). The area under the curve represents work done on the rubber bands since they aren’t doing work by themselves (I think). confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: "