#$&* course 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, directed downward. Gravity also acts on m1 which is balanced by the upward force of table on this mass, so the forces on m1 make no contribution to Fnet. Acceleration=net force/total mass = 9.8 m/s^2 * m2 / (m1+m2), again in the downward direction. The change in gravitational PE is equal and opposite to the work done by gravity. This is the definition of change in gravitational PE. If the mass m2 descends distance `dy then the gravitational force m * g and the displacement `dy are both downward, so that gravity does positive work m g `dy on the mass. The change in gravitational PE is therefore - m 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. STUDENT COMMENT I forgot that PE = m * g * 'dy. And I did not think that the table exerting force on the mass took it out of the system. I understand the idea though. INSTRUCTOR RESPONSE the table doesn't take the mass out of the system, but it does counter the force exerted by gravity on that mass so the total mass of the system is still the total of the accelerating masses, but the net force is just the force of gravity on the suspended mass, (since the system is said to be frictionless, there is no frictional force to consider) SYNOPSIS The change in gravitational PE is equal and opposite to the work done by gravity. This is the definition of change in gravitational PE. If the mass m2 descends distance `dy then the gravitational force m * g and the displacement `dy are both downward, so that gravity does positive work m g `dy on the mass. The change in gravitational PE is therefore - m g `dy. As you say, `dw_noncons + `dPE + `dKE = 0 If `dW_noncons is zero, as is the case here (since there are no frictional or other nonconservative forces present), then `dPE + `dKE = 0 and `dKE = - `dPE. In this case `dPE = - m g `dy so `dKE = - ( - m g `dy) = m g `dy. The signs are confusing at first, but if you just remember that signs are important these ideas will soon sort themselves out. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qHow would friction change your answers to the preceding question? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - RESPONSE: Cf. Atwood machine comments above: ... - - Here, ([m1]g * [coefficient of friction between m1 and table]) would work against the direction of motion... - - ... So this force would be in the direction toward the opposite side of the table (for the portion of the system being pulled horizontally) and then in the upward direction (for the portion of the system being pulled downward) -- ... - - - ... again, note that the tension on the string represents the *distribution* of force *throughout* the system rather than the *application* of force *to* the system -- ... - - ... such that this would be subtracted from [(m2)g + (weight of string below pulley so far)] when calculating the net force on the system described above... - - ... and both acceleration and work done would be reduced accordingly because the net force accelerating the mass and doing the work is less. 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): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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: - RESPONSE: The slope of a graph of the relationship between force and stretch (force vs. stretch OR stretch vs. force; see the following) can be viewed from two different perspectives, as expressing the rubber band's spring constant (""To stretch the rubber band by X[:X = distance = (L^1)], one must apply Y[:Y = force = (M^1)(L^1)(T^-2)]."" or ""Stretching the rubber band by X will involve the application of force Y and the exertion by the rubber band of counter-force -Y."") or as expressing its elastic modulus (""To get the rubber band to exert Y[:Y = (counter)-force = (M^1)(L^1)(T^-2)], you must stretch the rubber band by X[:X = distance = (L^1)]."" or ""Applying Y will result in stretching the rubber band by X.""), depending on whether a) one's subjective goal is to stretch the rubber band or to exert force and b) whether one is more worried about not getting the desired amount of counter-force or about breaking the rubber band, ... - - And the ""force vs. stretch"" graph, with stretch as the independent variable and applied (same as counter-exerted but with opposite sign) force as the dependent variable, will take the former form (sorry!), in which the slope represents the spring constant... - - - ... although a ""force vs. stretch"" graph might seem a bit counterintuitive because the easiest way to generate a force-vs.-strech graph is probably to a) start by graphing stretch vs. force by i) exerting force F (e.g., via masses on a hanger of known mass) and ii) measuring `dL[ength] and then b) plot the inverse of that function...
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Given Solution: `a** If we ignore thermal effects &&&&& Grrrr, forgot about this... &&&&&, which you should note are in fact significant with rubber bands and cannot in practice be ignored if we want very accurate results &&&&& Idea for extra credit if a question like this appears on an exam: ""Q: Name a real-life catastrophe caused by the failure to account for the effects of temperature on the elasticity of a polymer. [Optional hint(s): 1) The polymer in question took the form of a flat, relatively thin ring. 2) The catastrophe occurred in the 1980s 2b) and resulted in seven deaths. A: The Challenger disaster.""] &&&&&, 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. ** STUDENT QUESTION I am still a little confused about if the work is done by the rubber bands, or if the work is done one the rubber bands. Would you explain the difference? INSTRUCTOR RESPONSE This example might be helpful: If you pull the end of an anchored rubber band to the right, it exerts a force to the left, in the direction opposite motion, so it does negative work during the process. You, on the other hand, pull in the direction of motion and do positive work on the rubber band. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - RESPONSE: - - The slope of the force-versus-stretch graph represents the spring constant for the reasons discussed in the preceding Response; if we were to plot stretch versus force, the slope would represent the elastic modulus. - - As for whether the area under the curve represents the work done, that depends on what the meaning of ""force [as used to label the Y-axis]"" is (apologies to Bill Clinton if necessary). - - - If ""force"" refers to the amount of force that one exerts on a given rubber band in order to stretch it to that length, then the integral does represent work done, and the work in question is the work that the person stretching the rubber band has done *on* it. - - - If ""force"" refers to the amount of ""counter-force"" that a rubber band is exerting against one's hand and muscles when one is maintaining the stretch on a rubber band at a level corresponding to the length in question, then no work is being done *yet*, but the integral represents the work that the rubber band *can* do *on* something else (e.g., one's hand and arm, if one relaxes them and thereby removes the force in opposition to that of the rubber band), i.e., elastic potential energy, the work that can be done *by* the rubber band.
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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. ** STUDENT QUESTION Okay, so are you saying that the rubber band could either be doing work or getting work done on it? I believe I understand this, but just wanted to double check. INSTRUCTOR RESPONSE Yes, and that depends on whether the rubber band is being stretched, or contracting. When it is being stretched positive work is being done on the rubber band. After being released the rubber band does positive work on the object to which its force is applied. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK PREVIOUS STUDENT RESPONSE TO REQUEST FOR COMMENTS There is a whole lot of stuff concerning Newton’s laws of motion and there applications to force and acceleration. They will take some serious application to master. I understand what potential energy is, I understand that it is decreasing as kinetic energy increase, but I don’t understand how to measure it. Its like an invisible force, and the only relation to which I can apply it is in the context of gravity. If we have a 1kg object and we hold it 5meters off the ground, then according to the equation above PE = m*g*`dy this would be PE = 1kg * 9.8m/s^2 * 5m = 49 kg * m^2/s^2 my algebra is so bad but I still cant see this contributing to a useful measurement or unit. I don’t know how to swing it so it’ll give me a newton, PE has to be measured in newtons right because it is indeed a force? Ohhh I get it now!! I remember, a kg times a m/s^2 is a newton, and a newton times a meter is a Joule!!! So this is a valid measurement, which would make that equation valid, the potential energy for the above circumstance would be 49 Joules then. INSTRUCTOR RESPONSE Very good. Remember that F_net = m a If you multiply mass m in kg by acceleration a in m/s^2, you get the force in Joules. Of course when you multiply kg by m/s^2 you get kg m/s^2. This is why a Newton is equal to a kg m / s^2. Work being the product of a force and a displacement will therefore have units of Newtons * meters, or kg * m/s^2 * m, which gives us kg m^2 / s^2. Query Add comments on any surprises or insights you experienced as a result of this assignmen &&&&&- RESPONSE: The last question was really interesting, in part because of its terminological ambiguity: If I hadn't thought of the person stretching the rubber band, then I wouldn't have thought about the elastic modulus, and if I hadn't thought of the rubber band pulling back, then I wouldn't have thought of the spring constant. In turn, the process of sorting out which graph's slope corresponds to which property -- whose most effective strategy involved visualizing cause-and-effect relationships in terms of visualized experiments ""tweaked"" by small but significant variations in accompanying instructions (e.g., ""put a given amount of mass [on + in] the hanger and then see how far the band has stretched"" as opposed to ""add masses until the band reaches a given length and then see what their total is"") -- helped me figure out the relationship between the elastic modulus and the spring constant: For any given spring, their values seem to be reciprocals: F = k(`dL), such that k = F/(`dL); and Modulus:Strain is proportional to (`dL)/F [varies with cross-sectional area, but then again, so would the spring constant; remember that we're holding constant the identity of the spring in question].) &&&&& ###" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: PREVIOUS STUDENT RESPONSE TO REQUEST FOR COMMENTS There is a whole lot of stuff concerning Newton’s laws of motion and there applications to force and acceleration. They will take some serious application to master. I understand what potential energy is, I understand that it is decreasing as kinetic energy increase, but I don’t understand how to measure it. Its like an invisible force, and the only relation to which I can apply it is in the context of gravity. If we have a 1kg object and we hold it 5meters off the ground, then according to the equation above PE = m*g*`dy this would be PE = 1kg * 9.8m/s^2 * 5m = 49 kg * m^2/s^2 my algebra is so bad but I still cant see this contributing to a useful measurement or unit. I don’t know how to swing it so it’ll give me a newton, PE has to be measured in newtons right because it is indeed a force? Ohhh I get it now!! I remember, a kg times a m/s^2 is a newton, and a newton times a meter is a Joule!!! So this is a valid measurement, which would make that equation valid, the potential energy for the above circumstance would be 49 Joules then. INSTRUCTOR RESPONSE Very good. Remember that F_net = m a If you multiply mass m in kg by acceleration a in m/s^2, you get the force in Joules. Of course when you multiply kg by m/s^2 you get kg m/s^2. This is why a Newton is equal to a kg m / s^2. Work being the product of a force and a displacement will therefore have units of Newtons * meters, or kg * m/s^2 * m, which gives us kg m^2 / s^2. Query Add comments on any surprises or insights you experienced as a result of this assignmen &&&&&- RESPONSE: The last question was really interesting, in part because of its terminological ambiguity: If I hadn't thought of the person stretching the rubber band, then I wouldn't have thought about the elastic modulus, and if I hadn't thought of the rubber band pulling back, then I wouldn't have thought of the spring constant. In turn, the process of sorting out which graph's slope corresponds to which property -- whose most effective strategy involved visualizing cause-and-effect relationships in terms of visualized experiments ""tweaked"" by small but significant variations in accompanying instructions (e.g., ""put a given amount of mass [on + in] the hanger and then see how far the band has stretched"" as opposed to ""add masses until the band reaches a given length and then see what their total is"") -- helped me figure out the relationship between the elastic modulus and the spring constant: For any given spring, their values seem to be reciprocals: F = k(`dL), such that k = F/(`dL); and Modulus:Strain is proportional to (`dL)/F [varies with cross-sectional area, but then again, so would the spring constant; remember that we're holding constant the identity of the spring in question].) &&&&& ###" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!