course Phy 121 014. `query 14*********************************************
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Given Solution: `a 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): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `q Define the relationship between the work done by a system against nonconservative forces, the work done against conservative forces and the change in the KE of the system. How does PE come into this relationship? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If the system does positive work on nonconservative forces than KE+PE will be negative and will decrease. And vice versa. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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.) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I wasn’t quite sure how to explain all of this. I understand more fully now, however. Self-critique rating #$&* 3 ********************************************* Question: `qclass notes: rubber band and rail. How does the work done to stretch the rubber band compare to the work done by the rubber band on the rail, and how does the latter compare to the work done by the rail against friction from release of the rubber band to the rail coming to rest? Your solution: Work done by rail=work done by rubberband. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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 #$&* OK ********************************************* Question: `qWhy should the distance traveled by the rail be proportional to the F * `ds total for the rubber band? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The system begins and ends at 0, as does the acceleration. F is constant, so the distance traveled is determined by the tautness of the rubberband. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: • The system accelerates from zero to max KE then back to zero, defining an interval for which `dKE is positive and an interval for which `dKE is negative. • The system starts and ends at rest so the total `dKE, from the beginning of the first interval to the end of the second, is zero. F_net_ave * `ds between the initial state of rest and max KE must therefore be equal and opposite to F_net_ave * `ds between max KE and the final state of rest. • During the second interval 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. During the second interval, therefore, F_net_ave remains constant, so it is the coasting displacement that varies with pullback. The coasting displacement is therefore proportional to the F * `ds total for work done by the rubber band on the system. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK "