course phy 121 ã“p³´·žÏ‹öü„±{¸{ªûÁŸassignment #009
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13:01:04 Introductory prob set 3 #'s 1-6 If we know the distance an object is pushed and the work done by the pushing force how do we find the force exerted by the object?
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RESPONSE --> Since work is the product of force and displacement, we would simply divide the work by the distance, and the quotient would be the force. confidence assessment: 3
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13:01:20 ** 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 --> ok self critique assessment: 3
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13:05:38 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 --> KE = 'dW, and W='ds * f, then KE = 'ds * f We will multiply force by distance travelled to get the KE change of the object. confidence assessment: 2
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13:06:10 **`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 --> ok self critique assessment: 3
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13:07:00 Why is KE change equal to the product of net force and distance?
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RESPONSE --> KE change is equal to the product of net force and distance, because KE = work done, and work done is equal to net force times distance. confidence assessment: 2
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13:07:15 ** 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 --> ok self critique assessment: 2
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13:07:55 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 --> KE change is not equal to the product of the force I exert and the distance because some energy dissipates. confidence assessment: 2
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13:09:04 ** 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 understand that mine is not the only force being exerted. The system is subject to forces that push both with me and against me, though it is possible (but unlikely) that they could balance. self critique assessment: 2
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