Ҡ̘Ww Physics I Class Notes 02-11-2006
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13:23:06 Do the data seem to indicate, within the limits of the errors inherent in the experiment, that acceleration is the same on a constant incline regardless of where the cart is or how fast it is going?
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RESPONSE --> The acceleration would not be the same on a constant incline because the higher the incline is the quicker it will accelerate.
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13:25:33 ** The usual results do show that the acceleration is the same, within about 10%, for any section of the ramp. Since different sections can have different lengths, which are associated with different average and final velocities, this indicates that within these limits the acceleration does appear to be constant. **
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RESPONSE --> I dont think i understand what the question is asking. I understand that the acceleration is the same for any section of the ramp at the same level. That makes since but if you raise the ramp increase the slope then the acceleration will change. It will be the same from any position but just adding the washer to the botton will increase the acceration.
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13:26:34 Where does the force that accelerates the cart come from?
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RESPONSE --> F is the force that is exerted on the cart by gravity.
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13:27:04 ** The force is the result of the gravitational force exerted on the cart by the Earth. On an incline a component of this force acts in the direction down the incline. The greater the incline the greater the proportion of the total gravitational force that acts down the incline. **
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RESPONSE --> right... Yes, the gravity is pulling the cart in a downward position.
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13:29:30 List all the forces that act on the cart, and discuss how they affect its motion.
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RESPONSE --> Gravity from the earth is pulling on the cart in a downward motion. The weight of the car is also pulling the cart in a downward position.
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13:30:45 ** gravity pulls straight down, but the ramp isn't able to push straight up. That leaves a component of the gravitational force parallel to the ramp; the greater the slope the greater this component. Friction opposes motion. If there is a force holding the car back, then it needs to be included in the list, as it is in your list. **
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RESPONSE --> Friction trys to stop the car from moving down the ramp.
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13:32:57 Explain why a `ds is equal to the change in 1/2 v^2.
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RESPONSE --> the equation of motion tells us that the result of the acceleartion through tht known distance is to increase v^2 by 2 a'ds or increase 0.5v^2 by a 'ds.
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13:34:09 ** By the fourth equation of motion vf^2 = v0^2 + 2 a `ds. Thus a `ds = (vf^2 - v0^2) / 2. Since vf^2 - v0^2 is the change in v^2, it follows that a `ds is half the change in v^2. So a `ds is proportional to v^2. **
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RESPONSE --> right since vf^2-v0^2/2 = v^2/2 = 'ds
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13:35:27 Explain in detail why we expect Fnet `ds to be proportional to the change in v^2.
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RESPONSE --> Since the net force on the cart is directly proportional to its acceleration a, a graph of the change in v^2 vs. Fnet `ds should also give a straight line.
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13:37:52 ** a `ds is proportional to the change in v^2 (see the preceding question), and Fnet * m = a. So the change in Fnet * `ds is the change in m a * `ds, which for a given object is proportional to the change in v^2. **
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RESPONSE --> F is force and m*a= F If you multiply the force time the change in displacement you would do that on the other side of the problem which would be m*a*'ds which equals v^2
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