Download sample lab report

Title: If we break the 2nd law will the Physics police get us?
Purpose:
This lab had two purposes. The first was to determine the relationship of the total mass of a system and it’s
acceleration when the unbalanced force is held constant. The second was to determine the relationship of the
net force and the acceleration of the system when the total mass is kept constant.
Apparatus:
Track, cart, string, assorted masses, mass hanger, pulley, balance, computer and photogate.
computer
track
cart
string
pulley
photogate
adjusting foot
lab table
hanger and mass
Procedure:
Preliminary:
1. Set up the apparatus as shown in the diagram above, except do not hook the string to the cart yet.
2. Using the adjusting foot, set the angle of the track such that the cart alone, when given a small tap toward
the pulley, keeps moving along the track at a constant speed. (This balances out the friction force.) Then
hook the string to the cart.
3. Activate the “smart pulley” program in Logger Pro. Adjust the photogate so the spokes in the pulley cut
through the photogate beam.
Part 1: Varying total mass with a constant net force.
1. Determine the mass of the cart and record it in the data table.
2. Pick an amount of mass for the hanger and mass and don’t change it. We used 55 grams.
3. Hold the cart at the end of the track farthest from the pulley. Start the program and release the cart.
4. Stop the timer after the hanger gets to the floor.
5. Use the computer to plot velocity versus time. Take the slope of the velocity time graph. Record that
acceleration in your data table.
6. Repeat run and record acceleration. If it is close move on, if not, repeat again.
7. Repeat steps 3-6 for five more trials but add 200 to 400 grams to the cart before each trial.
Part 2: Varying net force with a constant total mass.
1. Place 5 grams on the hanger. (Hanger + mass = 10 grams)
2. Place all of the other masses in the set (200 grams) on the cart.
3. Repeat the timing and the acceleration data as before.
4. Repeat five more trials but move 10 grams from the top of the cart to the hanger before the next trial, and 50
grams for each of the four after that. For the last trial all of the masses should be on the hanger.
Data:
Defining variables:
mass cart = mass of the cart and everything on top of it
mass hanger = mass of the hanger and everything stacked on it
total mass = mass of cart, hanger and any masses on them (also known as mass of system)
net force = the weight of the hanger and mass on it
mass of the cart alone: 500.8 grams
mass of the hanger alone: 10.0 grams
Part 1:
mass cart (g)
mass hanger (g)
500.8
995.9
1,195.9
1,509.0
1,692.0
1,992.0
55.0
55.0
55.0
55.0
55.0
55.0
Part 2:
mass cart (g)
700.8
690.8
640.8
590.8
540.8
500.8
Evaluation of data
Part 1
total mass (kg)
0.5558
1.0509
1.2509
1.5470
1.7470
2.0470
Part 2:
total mass (kg)
0.7108
0.7108
0.7108
0.7108
0.7108
0.7108
mass hanger (g)
10.0
20.0
70.0
120.0
170.0
210.0
net force (N)
.539
.539
.539
.539
.539
.539
net force (N)
.098
.196
.686
1.176
1.666
2.058
acceleration (
m
s2
)
acceleration (
.939
.506
.424
.347
.310
.266
acceleration (
m
s2
)
m
s2
)
.941
.506
.428
.345
.310
.268
m
s2
)
acceleration (
0.121
0.252
0.949
1.600
2.229
2.829
0.121
0.248
0.951
1.600
2.301
2.832
acceleration (
m
s2
inverse mass (1/kg)
)
.940
.506
.426
.346
.310
.267
acceleration (
1.799
.95157
.79942
.64641
.57241
.48852
m
s2
)
0.121
0.250
0.950
1.600
2.300
2.830
Average acceleration:
accelerati on 1  accelerati on 2
2
.248 sm2  .252 sm2
2
 .250 sm2
Inverse mass = 1/mass
1
1/.5558kg = 1.799 kg
total mass 
 1kg 
mass cart  mass hanger
 500.8g  55.0g  555.8g
g
  .5558kg
1000 kg
 1000g 
 1kg 
N
N
net force  mass of hanger in kg  9.80 kg
 55.0g
 9.80 kg  .539N
 1000g 


Above is a plot of acceleration versus mass when the net force was kept constant.
The fact that the graph above has the shape of a hyperbola gives an indication that the relationship is some type
of inverse. So a test graph was plotted where the inverse of the mass was calculated and then acceleration
versus inverse mass was plotted.
This produced a linear graph. The slope of the line is 0.514
m
s2
1
kg
which is 0.514
m
s2
 kg
which is 0.514
1
kgm
s2
.
This makes the mathematical equation for the relationship of acceleration and mass:
a  (0.514
kgm
s2
1
)
m
or a 
0.514 kgs2m
m
. In the whiteboarding session I learned a
same as a newton of force, so the equation can be written as a  (0.514N)
1
m
or a 
intercept has been dropped because it should be zero as explained in the conclusion.
kgm
s2
is the
0.514N
. The
m
Then I plotted acceleration versus net force.
The graph is a straight line. The slope of the line is 1.38
m
s2
N
.
which is 1.38 Nm
s2
)Fnet .
This makes the mathematical equation for the relationship of acceleration and force: a  (1.384 Nm
s2
The unit
m
s2
N
is the same as
m
s2
kgm
s2
due to the definition of the newton which simplified becomes
1
. So I can write
kg
1
)Fnet . Once again, the intercept has been dropped because it should be zero as
my equation as a  (1.38 kg
explained in the conclusion.
Conclusion:
In part 1 of the lab I conclude that when the net force on a system is constant, the acceleration of the
system is inversely proportional to the mass of the system. My justification for this is that when I graphed my
data it formed a hyperbola, indicating some type of inverse relationship. To get a straight line graph when
dealing with these two variables, I needed to plot acceleration versus the inverse of the mass. My general form
k
for my equation is: a =
, or k = ma. The slope of the graph is the value for k. In the post lab discussion we
m
learned that the units of the slope, kg2m are the same thing as a newton. In my case that would make my slope
s
0.514 N, which is within 5% of my calculated net force value of 0.539 N. Because of this, I believe the general
equation is a =
Fnet
, or in words, the acceleration is equal to the net force divided by the total mass. Since my
m
slope is the net force, if I would have hung more mass on the hanger at the beginning of the lab the relationship
of acceleration and mass would have stayed the same but the slope would have gotten larger. I would expect
the intercept of my inverse mass graph to be zero, because if the mass were close to infinite the acceleration
would be essentially zero. (The inverse of infinite mass would be zero.)
In part 2 of the lab I conclude that the acceleration on a system is directly proportional to the net force. My
justification for this is that when I graphed these variables I obtained a straight line. The general form of the
equation would be a = kF. If I take the units of the slope
m
Ns2
and substitute in
kgm
s2
for N, I get
1
kg
for the unit.
This would indicate the slope of the graph is the inverse of a mass. I happened to keep the total mass constant
1
in this part so I looked at that. My constant mass was 0.7108 kg, and the inverse of that is 1.407 kg
. This value
is within 1.6% of my slope of 1.384
or a =
m
Ns2
. Because of this, I believe the general equation is the same as in part 1,
Fnet
. I would expect the intercept of the graph to be zero, as if there is zero net force, the system should
m
experience no acceleration.
One new term that came up in the lab was net force, the vector sum of all of the forces acting on an object
or system. The other was defining the unit “newton” which is the force necessary to accelerate one kilogram of
mass at a rate of one meter per second squared.
As stated earlier, the slopes of my graphs were not exactly the theoretical values. My net force on the cart
was less than the weight force of the hanger. Possible reasons were not having the track at quite the correct
angle to balance the friction force. It also takes some force to cause the pulley to accelerate as it spins. It is
also possible that air resistance was of some importance in the lab.