course Phy 201 WҠassignment #009
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19:19:02 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 the work done by the force is `dW = F * `ds, we can calculate for force as follows: F = `dW / `ds provided that force is constant parallel to distance.
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19:19:17 ** 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
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19:27:58 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 --> since `dW = `dKE, assuming no dissipation of energy due to friction or air resistance, etc., when work is done on the object then: `dKE = F * `ds
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19:39:32 **`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 I only calculated for the work done on the system (I thought the change in KE of the object would be the work done on it), but it is the work done BY the system that is the KE change for the object, and the work done by the object is the equal and opposite work that was done on it.
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19:52:01 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 work is defined as the product of force and distance, and the energy to do the work here comes from KE, therefore, KE must also equal the product of net force and distance.
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19:54:04 07-27-2006 19:54:04 ** 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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NOTES -------> I don't see this explanation anywhere in the introductory problems so I'm saving this to notes for later review.
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19:55:41 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 --> There are other forces working on the object as well such as friction and air resistance.
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19:56:37 ** 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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