Carsten Institute Force and Acceleration Newton Second Law Lab Report pls finish this lab asap possible…………………………………………….

Carsten Institute Force and Acceleration Newton Second Law Lab Report pls finish this lab asap possible…………………………………………………………………………………………………………………………………. Force and Acceleration—Newton’s Second Law
(Summer 2020 Mechanics Remote Lab)
Backgroun
ds
→
Newton’s Second Law says that the net force Σ F acting on any object equals the product of
the object’s mass, m and acceleration →
a. Algebraically this is written
F
pulle
y
T2
ma
(1)
y
Tension force
acting on cart
Cart, mass
M
Low friction track
x
T2
clamp attached to
side of lab bench
Hanging mass, m
Figure 1. Experimental arrangement, showing a cart accelerated by tension in string.
For the geometry in Figure 1, we apply Newton’s second law to the hanging mass m and the
moving cart mass M. Summing the forces on the hanging mass (m) we get, in the y direction
ΣFy = T1 – mg = – ma1
(2)
where T1 and a1 are tension force and acceleration of the hanging mass. We assign positive in the “up”
direction.Summing the forces on the rolling mass (M)
ΣFx = T2 – f = Ma2
(3)
where f is friction force between the cart and the track and a2is the cart’s acceleration. Here we
assume a2 and T2 are positive to the left.
If we assume that the string has negligible mass and does not stretch and that the pulley is
ideal, we get T1 = T2=T and the Ia1I= Ia2I = IaI.
Subtracting equation 2 from 3 to eliminate T, the acceleration (to the left for mass M, down for mass m)
is:
1
The resulting
acceleration:
(mg – f) = a(m + M)
(4)
f
M)
(5)
=m
(m+
M)
− (m+
2
where f is the friction between the cart and the track, within the wheels of the cart, etc.
m
If we neglect friction, eq. 5 simply becomes: = (m+M)g
Substitute a in eq.3, and solve for T ,
Ma, we get
(6)
T=
= mM g
(m+
M)
(7)
Notice that here all masses move with acceleration, so the tension T is a variable and not equal to the
weight of the hanging mass.
Note: The
ratio
mM
(m+M) referred here as ‘mass factor’ determines the magnitude of net force and resulting
acceleration of the cart.
In practice based on Eq. 6, a series of cart acceleration data for various combination of hanging mass,
m, and cart’s mass M, can be used to deduce the value of acceleration due to gravity g.
(Watch video tutorial at
https://www.youtube.com/watch?v=vaz8J5jBSQQ for visual
demonstration and verification of Newton’s second law.)
Experiment
An in-class experiment can be performed using the equipment and setup shown in Figure 1. A smart
cart embedded with force and motion sensors is employed. A computerized system and software
program (e.g. DataStudio or Capstone) is used to acquired data of the force, position as a function of
time during cart’s along the track.
For motion with constant acceleration, the position versus time data is displayed in graph form, as
shown below, typically:
3
b
a
Figure 2. A sample of graph display of position versus time for motion with constant
acceleration. The segment a b describes variation of the cart’s position with time along the
horizontal track.
4
The program also provides the corresponding force, cart’s velocity and acceleration versus time graphs
for analysis. The cart’s accelerations for each run is found from the slope of the velocity vs. time graph
and directly from the acceleration graph for motion segment ab. See sketches of graph displays below.
Position
Time
Velocity
Time
Acceleration
time
(Following tables are samples of raw data taken from a certain in-class experiment based on the
setup shown in Fugure1. We will use the data for analyses and results for our lab report.)
To verify Newton’s second Law two experiments are conducted:
1. To test that the acceleration is proportional to force. The cart’s acceleration is measured for
several hanging mass, m, (which produces tension force on a string pulling the cart, as
recorded in run 1 to 7, In Table 1 below.
2. To test that for a constant net force, the cart’s acceleration is inversely proportional to the
moving mass, the hanging mass is kept at 0.20 kg, and the resulting acceleration is
measured as more mass is added to the cart, as recorded in run 7 to 9 In Table 1 below.
5
Data, Analysis, and Results
Table 1 Newton’s Second Law Data- Measured Cart Acceleration Versus Changing Force and
Changing Mass
Run
1
2
3
4
5
6
7
8
9
Hanging
Mass, m
(kg)
0.05
0.06
0.07
0.08
0.09
0.10
0.20
0.20
0.20
Cart Mass,
M
(kg)
0.84
0.84
0.84
0.84
0.84
0.84
1.34
1.84
2.34
Mass Factor
0.047
0.056
0.065
0.073
0.081
0.089
0.174
0.018
0.018
6
Measured
acceleration
(m/s2)
0.59
0.68
0.85
0.94
0.98
1.09
1.25
1.13
0.95
Analysis (assignment)
(Calculate Mass factor and include in Table1;
Plot graph of measured Acceleration versus Mass Factor. Based on Eq. 6 the slope should
equal to g ; Compare your result with the expected, find % difference from the expected
value of 9.8 m/ s2;
etc.,)
Discussion:
(Do your result support Newton’s second Law of Motion? Justify your answer;
Apply what you learn from this experiment setup to explain the extreme situations:
What would be the cart’s acceleration if the hanging mass is very small compared to the
cart’s mass?; What would be the cart’s acceleration if the hanging mass is very large
compared to the cart’s mass? Etc.)
Conclusion
(Complete the analysis/assignment and the rest of the report for submission.)
7
Newton’s Second Law
ΣF = Fnet = ma
Lab 4: Force Diagram
T2
f
T1
Mg
mg
T = Force of Tension
M = mass of cart
m = hanging mass
f = Force of Friction
ΣFx = T2 – f = Ma2
ΣFy = T1 – mg = – ma1
ΣFx = T2 – f = Ma2
ΣFy = T1 – mg = – ma1
Assuming the string to be of negligible mass and will not stretch then T1 =T2 = T
Assuming the pulley is ideal then the cart and the hanging mass accelerate together at a m/s2
T – f = Ma
T – mg = – ma
Using algebra and substituting for T gives us the relationship:
(mg – f) = a(m + M) solving for a gives: a = gm /(m+M) − f /(m+M)
Graphing a vs m/(m+M) gives a slope of g and a y-intercept which can be used to solve for the frictional force
Neglecting friction gives a = gm /(m+M)
Graphing a vs m/(m+M) gives a slope of g
Tips for Completing the Worksheet for Lab 4
– Fill in the mass factor on the given data table
– Use the entire data table to create your graph, a linear trend, and find an experimental g.
– Note whether the R2 parameter for the trendline is close to 1
– Answer all the questions (in blue) on the worksheet.
– Comment on in your discussion whether the data supports Newton’s Second Law
Does the data reflect F a a and m a 1/a
– Use an additional sheet to write the discussion and conclusion. These can be hand written but
the content should be what would typically go in a lab report. It should be at least 1 paragraph.
– The percent difference formula is a little bit different from percent error.
Percent difference = (difference between two values)/(mean of the two values) x 100%

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