course phy 121 Please address my questions ãÐI·ç̯ãOèò³¯…}í¨éZúÀÝúŸäÝöÇT„assignment #009
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11:18:54 ** Knowing the distance `ds and the work `dW we note that `dW = F * `ds; we solve this equation and find that force is F=`dw/`ds **
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. Answer popped up by accident, but I knew it anyway. self critique assessment: 3
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11:20:35 If we know the net force exerted on an object and the distance through which the force acts how do we find the KE change of the object?
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RESPONSE --> If the object is moving in the same direction as the force and it isn't losing any energy to anything then dW = KE so, KE = F * ds confidence assessment: 1
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11:25:00 **`dW + `dKE = 0 applies to the work `dW done BY the system and the change `dKE in the KE OF the system. The given force acts ON the system so F `ds is work done ON the system. The work done BY the system against that force is `dW = -F * `ds. When you use the energy equation, this is the work you need--the work done BY the system. **
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RESPONSE --> I knew there was some ""if"" there, I just wasn't sure what it was. I knew that the work done ON the system (dW) wasn't the same as the work done BY the system (KE). So, now I think that dW = F*ds is work done ON the system KE = - F *ds I hope That is right self critique assessment: 1
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11:26:45 Why is KE change equal to the product of net force and distance?
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RESPONSE --> Because the net force X distance is the amount of energy put into the object by some other force, but once it is transfered to the object it becomes the kinetic energy of that object. confidence assessment: 0
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11:30:39 ** It comes from the equation vf^2 = v0^2 + 2 a `ds. Newton's 2d Law says that a = Fnet / m. So vf^2 = v0^2 + 2 Fnet / m `ds. Rearranging we get F `ds = 1/2 m vf^2 - 1/2 m v0^2. Defining KE as 1/2 m v^2 this is F `ds = KEf - KE0, which is change in KE. **
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RESPONSE --> Oh, I wasn't even close, I didn't use an equation, just reasoning. I have forgotten why KE is defined as 1/2 m v^2 ?
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11:37:42 When I push an object with a constant force, why is KE change not equal to the product of the force I exert and the distance?
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RESPONSE --> Well, five mins ago I thought it was :) An object exerts force equal and opposite to force exerted on it. So when the force is exerted on an object the F x the ds = the work done ON the object, the work done by the object is the opposite. But I am still having trouble with this, If the object is moving in the same direction as the object was pushed wouldn't the kinetic energy be the same as the energy behind the push (disregarding any energy it loses due to friction, etc.)? confidence assessment: 1
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11:40:36 ** Change in KE is equal to the work done by the net force, not by the force I exert. When I push an object in the real world, with no other force 'helping' me, there is always at least a little force resisting my push. So the net force in this case is less than the force I exert, in which case the change in KE would be less than the product of the force I exert and the distance. If another force is 'helping' me then it's possible that the net force could be greater than the force I exert, in which case the change in KE would be greater than the product of the force I exert and the distance. It is actually possible for the 'helping' force to exactly balance the resisting force, but an exact balance would be nearly impossible to achieve. ANOTHER WAY OF LOOKING AT IT: If I push in the direction of motion then I do positive work on the system and the system does negative work on me. That should increase the KE of the system. However if I'm pushing an object in the real world and there is friction and perhaps other dissipative forces which tend to resist the motion. So not all the work I do ends up going into the KE of the object. **
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RESPONSE --> I was tricked by the wording. But then what does the object exerting an equalt and opposite force relate to? I have written down that dW + dKE = 0, so that would mean that dKE has to be the opposite of dW, but if they are in the same directions, it doesn't seem possibe.
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