course Phy 231 wȁassignment #009
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16:28:26 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 know that work equals the force * displacement which can be represented by the equation: `dW = F * `ds To find the force we just manipulate the equation to get F = `dW / `ds confidence assessment: 3
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16:28:29 ** 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 --> self critique assessment: 2
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16:31:48 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 --> The given force is acting on the system so the work done can be found by the equation: `dW = F * `ds The work going against that force is: `dW = -F * `ds And we know that `dW + KE = 0, so the KE will equal the `dW = -F * `ds confidence assessment: 3
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16:31:52 **`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 --> k self critique assessment: 2
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16:38:18 Why is KE change equal to the product of net force and distance?
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RESPONSE --> We know from newton's 2nd law that Acceleration = Force / Distance and this also comes from one of the motion equations: Vf^2 = Vo^2 + 2*a*`ds Substituting in the acceleration: Vf^2 = Vo^2 + 2 * (Force/Distance) * `ds Apply some algebra to get: F `ds = 0.5m vf^2 - 0.5m Vo^2 We know that KE = 1/2mv^2 we get F `ds = KEf - KEo, which is the change in kinetic energy confidence assessment: 3
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16:38:22 ** 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: 2
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16:40:37 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 KE equals the work done by net force, the force you exert is just one of the forces on the system, not the net force. Not all of the work you do on the system is necessarily kinetic energy therefore it won't be equal. confidence assessment: 3
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16:40:46 ** 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: 2
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