Lab 4 Answers

Lab 4 Answers - values are higher then the calculated...

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1. The relationship shown on the graph of a exp vs. F a is a direct relationship. That is as F a increases a exp increases, which is what one would expect. As the cart becomes lighter and lighter and the force pulling it grows it should accelerate at a faster rate. a=f/m if f increases and m decreases a will increase. 2. Graph 2 of A exp vs Total Mass shows an inverse relationship. As Total Mass grows A exp decreases. This is expected as the force pulling the cart remains constant, so the heavier the cart becomes the slower it’s acceleration. a=f/m, so as m increases a will decrease. 3. See Worksheet Attached. 4. My ∆a exp and ∆a calc do not explain the variations in my values. The percentage difference for part I was 4.55% and for part II it was 8.09%. 5. Looking at the variations in my data I think the assumption of negligible friction needs to be revised. I find it odd that my acceleration
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Unformatted text preview: values are higher then the calculated values, as I would think friction would slow acceleration down. As I think about it, if friction slowed the cart more for V1 then for V2 you would end up with higher acceleration values. That is, if overcoming static friction slowed the carts start and effected our V1 times, we would end up with artificially high values for our acceleration. Another idea is that the effects of friction would show up more in V1 because T1 is a larger value, so it really wouldn’t be an instantaneous velocity. By the time the carts velocity was measured it was much higher. Perhaps friction would effect T2 less as the time interval was so much shorter. These are some of my ideas, but bottom line I find a large difference in a exp and a calc . I would have to blame either the effects of friction or experimental error....
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This note was uploaded on 06/23/2008 for the course PHY 101 taught by Professor Petrou during the Spring '08 term at Erie CC.

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