Free Fall Lab II
Objective: To verify the following statement: In the absence of all other external forces except gravity, a falling body will accelerate at 9.8m/(s^2).
Data Collection:
Using the strip after a previous lab that used the strip, found the first point and labeled that as our zero or point of origin. We then measured every other mark and calculated its distance in relative to our origin and recorded our measurements on our Excel program. We the found the distance (Δx)between a previous point to the next point on excel using a similar equation as:
t(mid)= t(n) + (1/120)
The apparatus has a weight at a 1.5m high drop with an electric generator and a strip along the column, seen in (fig 1. - 2), is set up in a fashion so when the weight is released, the generator sends a shock on the strip where the weight is every (1/60)s leaving a distinct mark. From there you have a meter stick and a computer opened to excel to chart your results.
Data Collection:
Using the strip after a previous lab that used the strip, found the first point and labeled that as our zero or point of origin. We then measured every other mark and calculated its distance in relative to our origin and recorded our measurements on our Excel program. We the found the distance (Δx)between a previous point to the next point on excel using a similar equation as:
Δx= Pn - P(n-1) where n cannot = 0
Figure 1. Side view of the apperatus.
As well as mid- interval time and speed with:t(mid)= t(n) + (1/120)
V(mid)= Δx/(1/60)
respectively. During the lab, we had made some
numerical errors in our Excel spreadsheet. But thanks
to Prof. Wolf we had it revised and corrected, as
seen in our revised fig 4. From there, we graphed our
scatter plot of our values of Mid Interval Speed (m/s)
for each Mid Interval Time (s), also seen in fig 4.
You'll also notice that the is a linear line along the
approximate slope of the scatter plot which we put
there with Excel. We then found our slope,
acceleration of Gravity (g), to be 937.76 cm/(s^2).
As a Class:
We (as a class) record each group's calculation of
the acceleration of Gravity (g), then found the
average out of those values and got:
g(avg)= 948.41 cm/(s^2)
numerical errors in our Excel spreadsheet. But thanks
to Prof. Wolf we had it revised and corrected, as
seen in our revised fig 4. From there, we graphed our
scatter plot of our values of Mid Interval Speed (m/s)
for each Mid Interval Time (s), also seen in fig 4.
You'll also notice that the is a linear line along the
approximate slope of the scatter plot which we put
there with Excel. We then found our slope,
acceleration of Gravity (g), to be 937.76 cm/(s^2).
As a Class:
We (as a class) record each group's calculation of
the acceleration of Gravity (g), then found the
average out of those values and got:
g(avg)= 948.41 cm/(s^2)
Figure 2. A view of the weight and electric
generator with the strip along the column.
With our average acting as somewhat of a baselineof our value of g, we needed to find the range from
our values. So with Excel we measured how fareach value deviated from the average value of
g and then squared the difference following:
x = the value of a given g
= the average value of gs= Standard Deviation
When the formula above, we find that our Standard
Deviation is ±8.710 cm/(s^2), seen in fig 5. Thus,
we have found our value of gravity to be
g= 948.41 ± 8.710 cm/(s^2).
Figure 3. Measuring the distinct marks
from our 0 to record each point.
Figure 4. Our revise Excel Spreadsheet with our graph of Mid Interval Speed (m/s) throughout Mid Interval Time (s) with a linear- fit line labeled as our calculated slope (g).
Analysis:
Now if you notice in fig 4. that the scatter plot has been linear fitted with a constant acceleration our line is
approximated with an correlation of (R^2)= .998 is:
y = 937.76x + 30.094
Now, if this linear graph is as accurate as can be then we should see that the average of all the values of velocity should equal the middle value on our line. So let's check:
y (avg)= (54 + 66+ 90...+ 228) = 139.5cm/s
12
12
(Mid point of y):
y ((.125+ .108)/2) = y (.1165) = 937.76(.1165) + 30.094 = 139.34cm/s
And if you find their correlation between y(avg) and y(mid) you find that y(mid) is
.998< y(mid)< .999 in correlation to y(avg).
Figure 5. Shows the calculated values of the acceleration of Gravity (g) along with a class average, deviation from average (dev from avg) and the squared value of those deviations (dev squared), and our calculated standard deviation (std dev).
Conclusion:
In Conclusion, our results of the acceleration of gravity (g) from our slope g in our velocity vs. time graph is lower than the accepted value of g which is 981 cm/(s^2) by getting our average over the accepted then multiply and subtract by 100% :
g (us)= 937.76 cm/(s^2) g(acp)= 981 cm/(s^2)
g(us) = (937.76) = .9559(100%) - 100% = -4.407%
g(acp) (981)
So our results compare to the accepted value of g was off by -4.407% most likely due to systematic error with the apparatus rubbing when the weight dropped causing friction. Also, since we did not drop the weight ourselves to witness we don't really know if there was an issue when the weight was dropped and out of our class, our results were the lowest and one of the farthest deviation from the average calculated value of g. Overall, our best estimate would be g = 945 cm/(s^2) ± 12%.
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