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LAB REPORT FOR M5 EXPERIMENT: NEWTON’S 2
nd
LAW
The premise for this lab was to investigate Newton’s second law of motion that states
a
F
m
=
. To do this we
used an apparatus that looks like this:
Using these measurements and performing the
experiment well, we can calculate acceleration
of the system as well as the acceleration due to
gravity.
VI2:
In V2 we obtained a series of values of
t
∆
and
s
∆
for fixed values of M
2
and M
T
. This was
done by varying
. My data for V2 is tabulated
below:
(cm)
1
t
∆
(s)
2
t
∆
(s)
3
t
∆
(s)
4
t
∆
(s)
t
∆
(s)
t
∆
σ
(s)
(
29
2
t
∆
(s
2
)
(
29
2
d
d
s

+
=
∆
(cm)
5
0.1222
0.1226
0.1221
0.1238
0.12267
5
0.00078
0
0.01505
0.1926
10
0.2375
0.2384
0.2468
0.2399
0.24065
0
0.00421
8
0.05791
0.7180
20
0.4611
0.4546
0.4441
0.4440
0.45095
0
0.00839
8
0.20336
2.5403
30
0.6929
0.6877
0.6946
0.6969
0.69302
5
0.00391
0
0.48028
5.1472
Mass on hanger, M = 10.0 g
M
1
= M(glider) + M(5 cm flag) + M(10 cm flag) + M(20 cm flag) + M(30 cm flag) = 547.8 g
M
2
= M + M(weight hanger) = 15.5 g
Start position of glider rear end on the air track: L = 120.2 cm
Final position of glider rear end on the air track: L’ = 90.2 cm
d = L – L’ = 30.0 cm
From the data in the table I was able to create a graph of
s
∆
vs.
(
29
2
t
∆
. The data points create a straight line with
a slope that is equal to acceleration/2 from which I was able to calculate acceleration. The graph below was
created using EXCEL and the equation of the line of best fit was found using the LINEST function. EXCEL
returned the following results:
(
29
2
t
∆
s
∆
0.01505
0.19259
3
0.05791
0.71796
7
1
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View Full Document0.20336
2.54033
3
0.48028
5.14718
6
Slope
11.0054
4
0
Intercept
Uncertainty in
Slope
0.37743
7
N/A
Uncertainty in
Intercept
Ave. time squared vs. delta s
y = 11.005x
0
1
2
3
4
5
6
0 0.1 0.2 0.3 0.4 0.5 0.6
Ave. time squared (sq. seconds)
Delta s (cm
From equation 1b in the manual we can show the equation of the line in terms of
(
29
2
t
∆
s
∆
and
a
. The equation
of this line is:
(
29
2
2
t
a
s
∆
=
∆
where y =
s
∆
, x =
(
29
2
t
∆
and slope =
a
/2. Using the slope calculated by EXCEL I was
able to calculate acceleration as well as the uncertainty in acceleration.
Slope of line on graph = 11.005
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