Physics Help Forum Help with lab

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 Oct 27th 2008, 05:36 PM #1 Junior Member   Join Date: Oct 2008 Posts: 4 Help with lab Hello, I have to do a section for error analysis on my physics lab. For the lab we used a photogate to measure the velocity of an object we slid from under it, and then we also captured the stopping distance using a meter stick. what are examples of sources of errors I can use? Also, there is a part of the lab in the conclusion where I have to tell the acceleration of the object? How am I suppose to do this without us actually having to measure a time in the lab? we measured the velocity, the stopping distance, and then we had velocity squared. thank you for the help in advance
Oct 27th 2008, 06:27 PM   #2
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The "errors" (maybe better to say "uncertainties") can be the appreciation of the meter stick (maybe 1mm in your case) and the appreciation of the photogate (unless you assume it has no uncertainty in it).
And any systematic error (you measured badly for example, or your partner did so). These kinds of errors are harder to evaluate. Generally in labs, one measure many times the same thing and use statistic to chose the most representative value of each value required.
 How am I suppose to do this without us actually having to measure a time in the lab?
You measured a velocity which is a distance divided by a time. As you have the distance and the velocity, you have the time!
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 Oct 27th 2008, 06:42 PM #3 Junior Member   Join Date: Oct 2008 Posts: 4 Oh ty, but how do i get the acceleration? still kind of confused.
Oct 27th 2008, 06:58 PM   #4
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 For the lab we used a photogate to measure the velocity of an object we slid from under it, and then we also captured the stopping distance using a meter stick.
,
 Oh ty, but how do i get the acceleration? still kind of confused.
I don't understand exactly what you've done. (English is not my mother tongue).
Did you let slide an object (with the supposition the slope was frictionless?) on a slope that form an angle with the horizontal?
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 Oct 27th 2008, 07:00 PM #5 Junior Member   Join Date: Oct 2008 Posts: 4 yes, it was frictionless and it was horizontal, basically we had to collect 20 trials in the entry, so the velocity for each was different and along it the stopping distance was different. the point of the lab was to see the relationship in velocity and stopping distance but i dont know how to get the acceleration from the data
 Oct 27th 2008, 07:20 PM #6 Senior Member     Join Date: Apr 2008 Posts: 815 I sure am missing something. If the object was moving on a horizontal frictionless surface, there's no way it would stop (the stopping distance would be infinite, but in reality there is a bit of friction so that it always stops). If that can help you, I guess that the acceleration you are looking for is a constant, say $\displaystyle a$. Then $\displaystyle a(t)=a$. Thus $\displaystyle v(t)=a\cdot t +v_0$ where $\displaystyle v_0$ is the initial velocity. And $\displaystyle x(t)=\frac{a}{2} \cdot t^2 +v_0\cdot t$. I'm sorry if I am of no use, but I'm sure someone else will help you for it. __________________ Isaac If the problem is too hard just let the Universe solve it.
 Oct 27th 2008, 07:27 PM #7 Junior Member   Join Date: Oct 2008 Posts: 4 Ok friction probably affects it :P, but I still don't understand the formula? which time am I suppose to use? I mean the data has a lot of variety.... is the initial velocity just the average of all the velocities? I mean here is a little of my data Velocity Stopping Distance 1.18 .22 1.58 .32 1.27 .295 1.12 .151 1.33 .28 velocity squared 1.392 2.49 1.61 1.25 1.76 sig figs are inaccurate but just an example
Oct 28th 2008, 07:49 AM   #8
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 is the initial velocity just the average of all the velocities? I mean here is a little of my data Velocity Stopping Distance 1.18 .22 1.58 .32 1.27 .295 1.12 .151 1.33 .28 velocity squared 1.392 2.49 1.61 1.25 1.76
In this example, did you push the object 5 times? (This is what I've understood so far).
What does the velocity stand for? Is it the velocity you pushed the object? If yes then it's the initial velocity of the object and its velocity will decrease until the object stops.
I'll do as if it was true. First example :
$\displaystyle a(t)=a$.
$\displaystyle v(t)=a\cdot t+1.18$.
$\displaystyle x(t)=\frac{a\cdot t^2}{2}+1.18\cdot t$. But we know that when $\displaystyle x=0.22$, the velocity is worth $\displaystyle \frac{0m}{s}$.
So let's find at what t it occurs. The best way to do that is to count how much photos took the photogate from the instant you pushed the object till it stops completely. Say it had time to take 20 photos with time intervals of $\displaystyle 0.1 s$. This means the body stopped at $\displaystyle t=2s$.
So now we now that at $\displaystyle t=2s$, the velocity is worth $\displaystyle \frac{0m}{s}$. Thus $\displaystyle v(2)=2a+1.18=0 \Leftrightarrow a=\frac{-1.18}{2}=-\frac{0.59m}{s^2}$ in this case. Do that for all the values you got and you obtain all the accelerations it seems they are asking you.
But still, I don't understand why did you calculate the velocity squared, so I must be missing something. I'm almost sure I don't understand well what you've done.
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