course phy 121 2/6 at 9:45 a.m. If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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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. COMMON MISCONCEPTIONS AND INSTRUCTOR COMMENTS: Misconception: The tension force contributes to the net force on the 2-mass system. Student's solution: 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 m1 does, however, get accelerated, so m1 does have a lot to do with how quickly the system accelerates. The greater the mass m1, the less accelerating effect the net force will have on the system. Also if friction is present, the mass m1 is pulled against the tabletop by gravity, resulting in frictional force. The greater the mass m1, the greater would be the frictional force. All these ideas are addressed in upcoming questions and exercises. STUDENT COMMENT I understand the first few parts of this problem, but I am still a little unsure about the gravitational PE. I knew what information that was required to solve the problem, but I just thought the solution would be more that (–m2 * 9.8m/s^2 * ‘dy). INSTRUCTOR RESPONSE Only m2 is changing its altitude, so only m2 experiences a change in gravitational PE. Equivalently, only m2 experiences a gravitational force in its direction of motion, so work is done by gravity on only m2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I wasn't exactly technical producing the equations that were in the answer but I had the general reasoning. ------------------------------------------------ Self-critique rating #$&*:2 ********************************************* Question: `qHow would friction change your answers to the preceding question? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If friction existed it would either add or subtract from the net force depending on in which direction it was being produced. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):net force would be m2 * g - frictional resistance. ------------------------------------------------ Self-critique rating #$&*:2 ********************************************* Question: `qExplain how you use a graph of force vs. stretch for a rubber band to determine the elastic potential energy stored at a given stretch. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: One can measure the amount of the stretch of the rubber band as weights are added to a chain of them. The weights should be measured as well to indicate the force. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Pe is the work required to stretch the band. I never really understood the trapezoids before but apparently the trapezoid on a graph of F versus stretch reflects this. ------------------------------------------------ Self-critique rating #$&*:2 ********************************************* Question: `qSTUDENT QUESTIONS: Does the slope of the F vs stretch graph represent something? Does the area under the curve represent the work done? If so, is it work done BY or work done ON the rubber bands? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I assume this is work done on the rubber bands measured in joules. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): slope = rise / run = change in force / change in stretch ------------------------------------------------ Self-critique rating #$&*:2 Query Add comments on any surprises or insights you experienced as a result of this assignment. Your solution: Contributions of the rubber band being stretched are represented by the areas of narrow trapezoids on a graph of F vs. stretch. The area under the curve between the two stretches represents the PE stored. Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&*:2 "