query_assign9

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11:21:14 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 --> I believe that we can find the work done by multiplying the distance by the force exerted, in otherh words: `dW = F `ds. So rearranging to solve for F, we get 'dW/`ds = F.

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11:21:49 ** 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 --> We haven't gone over work yet in class, but I looked up the problem set to help me answer this question. Is this what I am supposed to be doing?

The problem sets are part of the assignments, so you did exactly what you need to do.

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11:26:29 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 --> We would multiply the force exerted on the object by the distance, as long as there was no energy loss.

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11:28:17 **`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, so work done by and on the system should be opposite, and therefore the work plus the kinetic energy change should equal zero.

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11:31:16 Why is KE change equal to the product of net force and distance?

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RESPONSE --> Kinetic energy is the energy of motion, so if an object has a force exerted upon it and this causes a change in position, then its energy of motion would be the product of those two variables.

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11:34:05 ** 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. I see how this is derived, but we have not covered any of this material yet. I found it under Idea 8 of the synopsis and memorize this sections.

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11:38:02 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 --> The change is not equal because there is friction and air resistance when one pushes an object. So there is a difference.

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11:39:59 ** 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 --> Ok.

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Good work, considering that the questions were mostly related to introductory problem set questions that don't seem to appear on the Assignments page, and to a set of Class Notes that aren't assigned until next assignment.

I apparently made a series of editing errors when on the Assignments page. I'm sorry for the confusion.

Let me know if you have questions.