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The following data contains the information about m1 and m2, the measured acceleration, the calculated acceleration and the net force

The graph of net force and measured acceleration is plotted as below

And the slope of this line should be equal to the total mass of 2 objects, which is unchanged during this experiment.

The error of the measured acceleration and the calculated acceleration for each run

Run #1:

Run #2:

Run #3:

Run #4:

Preliminary assignment:

1) From the two equations of motion for free body diagram of block 1 and block 2 in Atwood’s machine, we can deduce that:

Since the tension is the same, by adding those 2 equations we shall end up with

Hence we have the equation (5.2)

From this equation (5.2) we can substitute back into either equation of free body diagram to solve for T, let’s use the equation for block 1,

2)

We can use the equation for calculation. Note that and

Hence

and

1.1900000000000004 0.8500000000000002 0.53 0.2 678.3 484.5 302.10000000000002 114 (
)
(
)
3102600.85484.5
+=
(
)
(
)
3002709.8
0.52
300270
–
=
+
(
)
(
)
3002700.53302.1
+=
(
)
(
)
2902809.8
0.17
290280
–
=
+
(
)
(
)
2902800.20114.0
+=
1.191.20
100%.83%
1.20
–
´=
0.850.86
100%1.16%
0.86
–
´=
0.530.52
100%1.92%
0.52
–
´=
0.200.17
100%17.6%
0.17
–
´=
11
maTmg
=-
22
mamgT
=-
(
)
(
)
1212
mamaTmgmgT
Þ+=-+-
(
)
(
)
(
)
1212
mmaTmgmgT
Þ+=-+-
(
)
(
)
1221
mmammg
Þ+=-
(
)
21
12
mmg
a
mm
–
Þ=
+
(
)
(
)
1
21
1
12
Tmag
mmg
Tmg
mm
Þ=+
–
æö
Þ=+
ç÷
+
èø
2112
1
1212
mmmm
Tmg
mmmm
æö
-+
Þ=+
ç÷
++
èø
2
1
12
2
m
Tmg
mm
æö
Þ=
ç÷
+
èø
12
12
2
mm
Tg
mm
Þ=
+
(
)
(
)
3202509.8
1.20
320250
–
=
+
(
)
21
12
mmg
a
mm
–
=
+
1
110.00.1
mg
=±
2
175.00.1
mg
=±
(
)
(
)
21
175.0110.00.10.165.00.265.00.31%
mmgg
-=-±+=±=±
(
)
(
)
12
175.0110.00.10.1285.00.2285.00.07%
mmgg
+=+±+=±=±
(
)
(
)
(
)
(
)
21
2
12
659.8
.31%.07%2.24.38%/
285
mmg
ams
mm
–
Þ==±+=±
+
(
)
(
)
3202501.19678.3
+=
(
)
(
)
3102609.8
0.86
310260
–
=
+