course Phy 241 Ez]M{aަ㳵[assignment #009
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13:26:09 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 --> Work equals the force times the displacement.
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13:26:44 ** 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 --> Yes we can solve for F by dividing `dw by `ds.
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13:27:38 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 --> We can multiply the force and the displacment and get work. Work will be the same as the KE.
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13:29:55 **`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 --> the force was on the system so to calculate the work done by the system it is equal and opposite so the negative sign must be there
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13:30:25 Why is KE change equal to the product of net force and distance?
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RESPONSE --> the change in the KE is the work of the system
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13:33:18 ** 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 yes if we use one of the equations from the unifrom accelerated motion and substitue F=ma and KE=.5mv^2 we can come up with F`ds=Kef-KE0
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13:34:16 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 you are doing work on the system, so `dw=-F `ds
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13:37:14 ** 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 the force you exert is only one part of the net force, friction and other factors can have affect on the total net force.
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