9.
Record the time 3 times for
m
2
=7g, 12g, 17g, 22g, and 27g by using the short silver mass
(5g) and the tall silver mass.

Lab Data
m
2
t
1
(s)
t
2
(s)
t
3
(s)
t avg
(s)
A=2d/t
2
(m/s
2
)
F=m
2
*(g-A)
(N)
1
0.007
1.7810
1.7837
1.7958
1.7868
0.376
0.066
2
0.012
1.4694
1.3920
1.4036
1.4217
0.594
0.111
3
0.017
1.1522
1.1536
1.1509
1.1522
0.904
0.151
4
0.022
1.0211
1.0147
1.0129
1.0162
1.162
0.190
5
0.027
0.9256
0.9337
0.9316
0.9303
1.386
0.227
Computations and Graphs
t avg =
t
1
+
t
2
+
t
3
3
a =
2
d
t avg
2
F = m
2
*(9.8– a)
Percent Difference =
⃒
0.191
−
S
lop
e
0.191
× 100% =
⃒
0.191
−
0.1547
⃒
0.191
× 100% = 19%
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
0
0.05
0.1
0.15
0.2
0.25
F vs A
A=2d/t2 (m/s2)
F=m2*(g-A) (N)
F = mA+b
m (Slope): 0.1547 N/m/s
2

b (Y-intercept): 0.01210 N
Correlation: 0.9980
RMSE: 0.004661 N
Summary
In my case, I could have improved on the measurements for points #2, since these data points are
not on the best fit line.
The percent difference is equal to 19%. It’s low, because my partner and I were accurate in our
measurements. The more the slope is closer to 0.191, the smaller the percentage difference.
Questions
3.
The acceleration of an object as produced by a net force is directly proportional to the
magnitude of the net force, in the same direction as the net force, and inversely
proportional to the mass of the object.
The force acting on the body is equal to the product of the mass of the body on the
acceleration reported by this force.
This experiment proves the second law of Newton’s 2nd Law of dynamics. We knew the
mass of the body, found out the average time for which the body overcomes a certain
distance, learned the value of acceleration. And knowing of acceleration and body
weight, we easily found out the force.