course PHY 201 p˷耲âÑ‹úõåÿÖ¸WЗ³Ü«÷ÚMäåz¿assignment #009
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14:50:55 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 --> We can find the force exerted by the object by multiplying the distance the object is pushed with the work done by the pushing force. confidence assessment: 3
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14:51:27 ** 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 --> Oops.. solved for the wrong variable. I understand now.. self critique assessment: 2
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14:52:17 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 can find the KE change of the object by multiplying the net force exerted on an object by the distance through which the force acts upon it. confidence assessment: 3
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14:52:35 **`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 --> okay self critique assessment: 3
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14:53:28 Why is KE change equal to the product of net force and distance?
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RESPONSE --> Becuase of no KE is given off in the process, either by friction or some other factor, then all of the energy is still with the object the whole time confidence assessment: 3
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14:53:35 ** 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: 3
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14:54:11 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 --> because it is pushed, which is a factor that causes friction by the surface it's pushed on. confidence assessment: 3
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14:54:19 ** 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 self critique assessment: 3
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