course Phy 231 6/21 10 012. `query 12
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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. 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. 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): OK Self-critique rating #$&* OK ********************************************* Question: `qHow would friction change your answers to the preceding question? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The acceleration of the system will be less when friction is taken into account. Also, the net force will be less because it is going against the force of gravity. The new formula would be (m2*9.8m/s^2) – friction. Confidence rating #$&*: 2
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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): OK Self-critique rating #$&* OK ********************************************* Question: `qExplain how you use a graph of force vs. stretch for a rubber band to determine the elastic potential energy stored at a given stretch. Your solution: When you have the graph, you take the integral of the equation from the lowest to the highest stretch. The number you will get is the total energy stored in the stretch (PE). This is actually the work needed to stretch the band, but that energy is stored (conservative force). Confidence rating #$&*: 2
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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 Self-critique rating #$&* OK ********************************************* Question: `q Does the slope of the F vs stretch graph represent something? Does the area under the curve represent the work done? If so, is it work done BY or work done ON the rubber bands? Your solution: The slope of the graph is the force constant. You multiply it by the distance stretched and you get the amount of force at that distance. The area under the curve is the work done to the system to stretch the rubber band, so it would be the work on the bands. Confidence rating #$&*: 2
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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 pretty well understand the really simple force problems like with the weights on either side of a pulley and things, but when the questions get more complicated I have trouble setting them up. For example, I really did not understand the Introductory Problem Set #14. I was not sure how to set it up or what it was asking. ?????Do you have any tips for setting up and understanding problems like the Introductory Problem Set #13 and are we going to have more practice problems?????????? &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* OK "