course Phy 121 δٞowVv~՟ӥassignment #009
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00:13:53 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 --> Energy = force * distance To find the force you would divide energy by distance. confidence assessment: 3
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00:14:40 ** 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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00:15:44 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 --> To find the kinetic energy, you would multiply the force and the distance. confidence assessment: 3
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00:18:32 **`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 --> I didn't know at all about the work done by the system or work done on the system. This information is kind of complex, but I am starting to get it. self critique assessment: 2
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00:20:27 Why is KE change equal to the product of net force and distance?
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RESPONSE --> KE change is going to take into account the net force which is what is being done on the system and by the system. confidence assessment: 2
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00:22:51 ** 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 --> I never thought about applying this to the equations for uniform acceleration. But now I see how by just substituting what we already know how we can solve for a bunch of different things. self critique assessment: 2
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00:23:57 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 has to take into account the force being exerted on the object as well, or the net force. confidence assessment: 2
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00:24:50 ** 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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