course PHY 121 1/4 014. `query 14*********************************************
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Given Solution: The acceleration of the mass is a = F_net / m, so the velocity of the object changes by amount `dv = a * `dt = F_net / m * `dt. Since the initial velocity is zero, this will also be the final velocity: vf = F_net / m * `dt. From this and the fact that acceleration is constant (const. net force on const. mass implies const. acceleration), we conclude that vAve = (v0 + vf) / 2 = (0 + (F_net / m) * `dt) / 2 = F_net * `dt / (2 m). Multiplying this by the time interval `dt we have `ds = vAve `dt = (F_net * `dt) / (2 m) * `dt = F_net `dt^2 / (2 m). If we multiply this by F_net we obtain F_net * `ds = F_net * F_net * `dt^2 / (2 m) = F_net^2 * `dt^2 / (2 m). From our earlier result vf = F_net / m * `dt we see that KE_f = 1/2 m vf^2 = 1/2 m ( F_net / m * `dt)^2 = F_net^2 * `dt^2 / (2 m). Our final KE, when starting from rest, is therefore equal to the product F_net * `ds. Since we started from rest, the final KE of the mass on this interval is equal to the change in KE on the interval. We call F_net * `ds the work done by the net force. Our result therefore confirms the work-kinetic energy theorem: `dW_net = `dKE. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I messed up in the second step of the problem; I should have used vAve = `ds/`dt, NOT `dv. That would involve using an extra step to find vAve, which would just equal (vf+vi)/2. I got lost somewhere towards the end, and cant tell if I was on the right track or if I was close to the given answer. I think my answer was right?
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Given Solution: ** The system does positive work at the expense of its kinetic and/or potential energy. The work done by the system against all forces is `dW_net_BY. `dW_net_BY is equal and opposite to `dW_net_ON, which is in turn equal to `dKE, the change in the kinetic energy of the system. We conclude that `dW_net_BY = - `dKE. The change in KE is equal and opposite to the work done by the system against the net force acting on it. To consider the role of PE, we first review our formulation in terms of the work done ON the system: `dW_net_ON = `dKE. The work `dW_net_ON is the sum of the work done on the system by conservative and nonconservative forces: `dW_net_ON = `dW_cons_ON + `dW_NC_ON and `dW_cons_ON is equal and opposite to `dPE, the change in the system's PE. Thus `dW_net_ON = `dW_NC_ON - `dPE so that `dW_net_ON = `dW_cons_ON + `dW_NC_ON becomes `dW_NC ON - `dPE = `dKE so that `dW_NC_ON = `dPE + `dKE. Since `dW_NC_BY = - `dW_NC_ON, we see that -`dW_NC_BY = `dPE + `dKE so that `dW_NC_BY + `dPE + `dKE = 0. Intuitively, if the system does positive work against nonconservative forces, `dPE + `dKE must be negative, so the total mechanical energy PE + KE of the system decreases. (Similarly, if the system does negative work against nonconservative forces that means nonconservative forces are doing positive work on it, and its total mechanical will increase). As usual, you should think back to the most basic examples in order to understand all these confusing symbols and subscripts (e.g., if I lift a mass, which you know intuitively increases its gravitational potential energy, I do positive work ON the system consisting of the mass, the conservative force of gravity acts in the direction opposite motion thereby doing negative work ON the system, and the work done BY the system against gravity (being equal and opposite to the work done ON the system by gravity) is therefore positive). The equation -`dW_NC_BY = `dPE + `dKE isolates the work done by the system against nonconservative forces from the work it does against conservative forces, the latter giving rise to the term `dPE. If the system does positive work against conservative forces (e.g., gravity), then its PE increases. If the system does positive work against nonconservative forces (e.g., friction) then `dPE + `dKE is negative: PE might increase or decrease, KE might increase or decrease, but in any even the total PE + KE must decrease. The work done against a nonconservative force is done at the expense of at least one, and maybe both, the PE and KE of the system. (In terms of gravitational forces, the system gets lower or slows down, and maybe both, in order to do the work). If nonconservative forces do positive work on the system, then the system does negative work against those forces, and `dW_NC_ON is negative. Thus -`dW_NC_ON is positive, and `dPE + `dKE is positive. Positive work done on the system increases one or both the PE and the KE, with a net increase in the total of the two. (In terms of gravitational forces, the work done on system causes it to get higher or speed up, and maybe both.) STUDENT RESPONSE WITH INSTRUCTOR COMMENTARY: The work done by a system against nonconservative forces is the work done to overcome friction in a system- which means energy is dissipated in the form of thermal energy into the 'atmosphere.' Good. Friction is a nonconservative force. However there are other nonconservative forces--e.g., you could be exerting a force on the system using your muscles, and that force could be helping or hindering the system. A rocket engine would also be exerting a nonconservative force, as would just about any engine. These forces would be nonconservative since once the work is done it can't be recovered. STUDENT RESPONSE WITH INSTRUCTOR COMMENTS: The work done by a system against conservative forces is like the work to overcome the mass being pulled by gravity. INSTRUCTOR COMMENT: that is one example; another might be work to compress a spring &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I need to review this concept. I get the general idea but am not confident in the details. ------------------------------------------------ Self-critique rating: ---- ********************************************* Question: `qclass notes: rubber band and rail How does the work done to stretch the rubber band compare to the work done by the rubber band on the rail, and how does the latter compare to the work done by the rail against friction from release of the rubber band to the rail coming to rest? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The work done to stretch the rubber band should be available in the same amount for the rail. The work done by the rail is the same as the friction it must overcome. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The work done to stretch the rubber band would in an ideal situation be available when the rubber band is released. Assuming that the only forces acting on the rail are friction and the force exerted by the rubber band, the work done by the rail against friction, up through the instant the rail stops, will equal the work done by the rubber band on the rail. Note that in reality there is some heating and cooling of the rubber band, so some of the energy gets lost and the rubber band ends up doing less work on the rail than the work required to stretch it. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: ---- ********************************************* Question: `qWhy should the distance traveled by the rail be proportional to the F * `ds total for the rubber band? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `dWnet = Fnet *`ds. So these two quantities relate to the work done. Assuming that the conditions of the system are constant, the amount of work done influences the distance traveled. The same amount of work put into the rubber band should be the work done on the rail when it moves in the opposite direction (if I phrased that correctly?). For example, it could be that the more work done on the rubber band, the farther the rail travels; or, the less work done on the rail, the shorter distance the rail travels. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Bottom line: The system accelerates from zero to max KE then back to zero, defining a phase for which `dKE is positive and a phase for which `dKE is negative. The system starts and ends at rest so the total `dKE is zero. Thus F * `ds between the initial state of rest and max KE must be equal and opposite to F * `ds between max KE and the final state of rest. During the second phase the net force is the frictional force, which is assumed constant, i.e., the same no matter how far the rubber band was pulled back. The only thing that can vary with pullback is therefore the coasting displacement, which is therefore proportional to the F * `ds total for work done by the rubber band on the system. More details: The F_`ds total for the rubber band is the work done to accelerate the rail to its maximum velocity v_max. Let's denote this by F_ave * `ds, where F_ave is understood to be the average force exerted by the rubber band (the rubber band force is at its maximum when the rubber band is pulled back, and decreases to 0 as it 'snaps back', accelerating the rail; so it makes sense to talk about the average rubber band force) and `ds is the displacement through which this force acts (i.e., the displacement from release until the rubber band loses contact with the rail). While in contact with the rail the rubber band exerts its force in the direction of the system's motion and therefore does positive work. So F_ave * `ds is positive. The block then coasts to rest subject to the force of friction, which acts in the direction opposite motion and therefore does negative work. Assuming the frictional force f_frict to be constant, and using `ds_coast for the coasting displacement, the work done against friction is f_frict * `ds_coast. For simplicity of notation we will neglect the presence of the frictional force during the first phase, while the rubber band is in contact with the 'rail'. It isn't completely accurate to do so, but if the displacement during this phase is small compared to the coasting distance the error is small. A comment at the end will indicate how to easily modify these results. We will also neglect any other forces that might be acting on the system, so that the net force for the first phase is just the rubber band force, and for the second phase the net force is just the frictional force. Now, during the first phase the rail's KE changes from 0 to 1/2 m v_max^2, where m is its mass, so by the work-KE theorem F_ave * `ds = `dKE = 1/2 m v_max^2. During the second phase the rail's KE changes from 1/2 m v_max^2 to 0, so that f_frict * `ds_coast = -1/2 m v_max^2. Thus F_ave * `ds = - f_frict * `ds_coast so that the coasting displacement is `ds_coast = - (F_ave * `ds) / f_frict = (- 1 / f_frict) * F_ave * `ds. F_ave and f_friction are in opposite directions, so if F_ave is positive f_frict is negative, making -1 / f_frict negative and `ds_coast = (-1 / f_frict) * (F_ave * `ds) indicates a direct proportionality between `ds_coast and F_ave * `ds. The above relationship tells us that the coasting displacement is proportional to the F * `ds total for the force exerted by the rubber band. To correct the oversimplification of the given solution, if that oversimplification bothers you, you may proceed as follows (however if you find you don't completely understand the preceding you shouldn't confuse yourself with this until you do): To account for the frictional force while the rubber band is in contact with the rail, assuming that the frictional force is also present during the first phase, we can simply replace `ds_coast with `ds_coast + `ds. The f_frict * (`ds_coast + `ds) will be the actual quantity that is proportional to F_ave * `ds for the rubber band. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK.. I think ------------------------------------------------ Self-critique rating: "