course Phy 121
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13:35:33
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RESPONSE -->
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13:36:59 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 --> by the following equation: F='dW/'ds
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13:37:11 ** 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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13:38:17 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 would take the net force and divide it by the distance to obtain the KE change
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13:42:32 **`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 had to go back and review that one a little bit...After doing the problem sets I was still a little unsure I fully understood....but I try and study this a little more.
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13:44:40 Why is KE change equal to the product of net force and distance?
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RESPONSE --> The product of the net force and distance gives the multiples of the KE change. -- Also relates to some of the basic equations that we have been working with
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13:45:11 ** 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
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13:48:45 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 --> Not to sure
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13:52:13 ** 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 --> Got is....For some reason I was thinking too much on this one....when in fact the KE is in relationship to the net force ... I guess I was trying to figure this out in the perferct world view. But I do understand that other forces..friction, wind resistance, etc. would effect the change in KE
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13:53:03 Query pendulum force proportionalities (video exp #9) Describe how we conclude that for small displacements the force tending to restore a pendulum to equilibrium is proportional to the displacement
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RESPONSE --> This was listed as not currently assigned on my assignment sheet...
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13:53:08 ** Since displacement over length equals the ratio of the effective suspended mass over mass of pend., and this ratio is equal to Force over weight, then displacement is also proportionate to the force. **
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RESPONSE --> ok
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13:53:14 Explain how your common sense leads you to believe in the plausibility of the hypothesis that the force tending to restore a pendulum to its equilibrium position is in fact in the same proportion to the weight of the pendulum as the displacement to the length of the pendulum.
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RESPONSE --> ok
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13:53:18 ** Since weight is the measurement of gravity acting on mass, then the weight implies some force, and the length of pendulum implies some distance over which gravity works, then there should be a set proportion. **
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RESPONSE --> ok
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13:53:22 Describe how we conclude that for small displacements the force tending to restore a pendulum to equilibrium is in the same proportion to the weight of the pendulum as the displacement to the length of the pendulum.
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RESPONSE --> ok
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13:53:28 ** Intuitively, the further you pull back the greater the restoring force. As you pull back the ratio x / L increases. As you pull back the ratio F / (M g) = restoring force / pendulum weight also increases. The proportionality between the two says, basically, that if you pull the pendulum back a distance equal to 5% of its length then the restoring force is 5% of the pendulum's weight. This proportionality applies very well to small displacements. As displacement gets larger the linearity of the relationship deteriorates. More rigorously: The force m * g exerted on the suspended mass m is the force exerted by gravity, and with a small correction for friction is equal to the displacing force. This force is equal and opposite to the restoring force. The weight of the pendulum is the force M * g exerted by gravity on its mass M. If x / m = L / M then x / L = m / M = m * g / M * g = displacing force / weight. **
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RESPONSE --> ok
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