Download 05.TE.Newton`s Second Law

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Experiment 5: Newton’s Second Law
Introduction
In this laboratory experiment you will consider Newton’s second law of motion, which
states that an object will accelerate if an unbalanced force acts on it. Upon successful completion of
this experiment you should be able to
l. determine the acceleration of a system with a constant mass and a varying force;
2. calculate the mass of the system from experimental data by knowing that the slope of the
graph is 1/mass; and
3. calculate the percentage error between the experimental result and the accepted value for a
mass.
Newton’s second law of motion states the relationship that can be used to predict the change
of motion (acceleration) of an object when an unbalanced force is applied. The acceleration is
directly proportional to the force and inversely proportional to the mass of the object, or a = F/m.
Recall that the slope of a straight line is a constant obtained from
slope ( m ) =
∆y
.
∆x
This equation can be solved for y, with m representing the slope of the straight line. When the origin
is zero, the solved equation is as shown in figure 5.1.
yy ==mx
mx
y
xx
figure 5.1
From this, you can see that y is directly proportional to x; that is, any increase in x (the manipulated
variable) will result in a proportional increase in y (the responding variable).
Figure 5.2 shows three different relationships between three different sets of x and y
variables. Each slope has a different proportionality constant (m1, m2, and m3) and each relationship
is represented by the equation y = mx. In each case the proportionality constant in the equation is
equal to the slope of the line.
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m
m11
m2
2
m3
3
y
x
figure 5.2
In this experiment, you will determine the acceleration of an object by varying the force and
keeping the mass constant. From Newton’s second law, F = ma, you know that
acceleration =
Force
mass
or a =
F
m
1
or a =   ( F ).
 m
This equation is in the same form as the equation for a straight line (when the origin is equal to zero),
or
y = mx
and
1
a =   ( F)
 m
where y = acceleration, x = force, and the slope (m) is equal to 1/m.
Again, recall that the slope of a straight line is determined from
slope ( m ) =
∆y
.
∆x
If you plot acceleration (m/s2) on the y-axis and force (N) on the x-axis, the units for the slope would
be
y = m / s2
x = newtons
m / s2
kg ⋅ m / s2
=
m
s2
×
s2 kg ⋅ m
=
=
1
.
kg
The slope therefore is the reciprocal of the mass, or 1/kg. As shown in figure 5.3, this is just what
you would expect for
1
a =   ( F ).
 m
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acceleration (m/s 2 )
y
mx
y = mx
x
a = (1/m)(F)
Force (newtons)
figure 5.3
Procedure
1. The method of measuring cart or glider velocity may vary with the laboratory setup. You might
use a system of a horizontally moving cart attached to weights on a hanger through a string over a
pulley. As the hanger falls, it exerts a horizontal force of F = mg on the cart. Varying the mass on
the hanger will thus vary the horizontal force on the cart. Other equipment with other means of
measuring velocity might be used in your laboratory, such as a cart with wheels and the use of
photogates and computer software. If so, the use of the apparatus will be explained by your
instructor.
air track
tape
cart and masses
ink jet
hanging
mass
Force = ? kg
9.8 m/s 2
figure 5.4
2. Tie a nylon string onto the cart. The string should hang about 30 cm down from the pulley when
the cart is in the starting position. A weight hanger is attached to the hanger string.
3. Arrange five 50.0 g masses on the cart so they are evenly distributed. You will make six
separate runs. On the first run the force on the cart will be the mass of the hanger only times g. For
each consecutive run a 50 g mass is removed from the cart and placed on the hanger. This
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keeps the mass of the system constant but varies the force on the cart. (Note: There are different
ways to define a system, and for this lab the system is defined as the cart, masses, and hanger).
According to the definitions used, the mass of the system is kept constant (while varying the force)
by moving masses from the cart and placing them on the hanger. You could add masses to the
hanger from outside the system defined, but that is a different experiment.
4. For each of the six runs, calculate the force on the cart from the mass of the hanging masses and
hanger in kg times 9.8 m/s2. The force will therefore be calculated in newtons. Record the
force for each run in the appropriate data table.
5. Measure the position of the dots on the tape for each of the six runs. The timer will be set so the
dots are 0.05 seconds apart. Carefully measure the distances between the dots and use the change
in distance and the time elapsed to calculate the velocity, change in velocity, and acceleration
for each run. Record these quantities in the appropriate data tables.
6. Find the average value of the acceleration (a) for each run. Note that the average value must be
converted to m/s2 so the units will be consistent with the units of a newton.
7. Record the average acceleration and the calculated force for each run in the summary data table
on page 62. Record the mass of the system (cart, hanger, and masses) in the summary data table.
8. Use the average acceleration and the force values in the summary data table for data points on a
graph, with the acceleration in m/s2 on the y-axis and the force in newtons on the x-axis. You
will have six data points, with each data point corresponding to a run. Calculate the slope of the
line (numbers and units) and write it somewhere on the graph.
Results
1. Using the reciprocal of your calculated slope as the experimental value of the mass of the system
and the measured mass of the system (cart, hanger, and masses) as found by a balance as the
accepted value, calculate the experimental error. Show your calculations here.
Mass
exp
% error =
=
1
slope
=
1
1

 2.07 

kg 
(0.500 − 0.484 ) kg
0.500 kg
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= 0.484 kg
× 100 = 3 %
2. How do you know if the acceleration was proportional to the force? (Hint: see figure 5.3)
The graph of a vs. F as linear and the y-intercept was approximately zero.
3. How would you know if the acceleration was inversely proportional to the mass?
You could vary the mass and keep the force constant. A plot of a vs 1/m should yield a
straight line with no intercept if acceleration is inversely proportional to mass.
4. Would the presence of an additional mass in the cart during all the runs increase, decrease, or not
affect the slope? How could you use this experimental set-up to find the value of the additional
mass that was in the cart?
The slope would decrease if there were an additional mass in the cart. If the mass is
unknown, compare the value of the mass determined from the slope to the mass of the known
quantities. The difference will be the unknown mass.
5. Suppose the string breaks as the cart accelerates one-third of the way across the track. What is the
acceleration of the cart for the remaining length of track? Explain.
F = ma and if there is no net force acting on the cart, then a = 0. The velocity would remain
constant for the rest of the data.
6. Was the purpose of this lab accomplished? Why or why not? (Your answer to this question should
be reasonable and make sense, showing thoughtful analysis and careful, thorough thinking.)
(The students should have verified F = ma. They should mention that their data agrees with
Newton's 2nd law.)
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Invitation to Inquiry
1. Design a demonstration of Newton’s second law of motion using a student on ice skates (or
roller blades) and a spring scale. The student on skates or roller blades should hold the hook on one
end of the scale while you pull the hook on the other end, pulling with a sufficient force to exactly
maintain a constant force. Calculate the acceleration by using the equation F = ma, then measure the
acceleration a second way by marking distances and measuring the change of velocity.
2. Repeat the demonstration with a more massive student on ice skates or roller blades.
Compare the difference, if any, needed to maintain a constant force on a more massive moving
object.
3. Make a velocity vs. time graph for the application of a constant force on the first student,
and another graph for the more massive student. What characteristic of the graph changes?
4. Using any everyday materials you wish, design a demonstration to prove that a moving
object with zero net force moves at a constant velocity.
5. For the demonstration in Invitation 1 (constant force) and the demonstration in Invitation 4
(zero force) sketch three graphs to predict (a) force vs. time; (b) acceleration vs. time; and, (c) force
vs. acceleration. Do several demonstrations to show your six predictions are correct.
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Data Table 5.1
Run 1
d
(cm)
∆d
(cm)
∆t
(s)
v=∆d/∆t
(cm/s)
∆v
(cm/s)
a = ∆v/∆t
(cm/s 2 )
1
2
3
4
5
Force = ____________(kg)(9.8 m/s2 ) = ____________________N
Average acceleration = ___________(total) / __________(number) = ________cm/s2
2
Average acceleration = _______________ m/s
Data Table 5.2
Run 2
d
(cm)
∆d
(cm)
∆t
(s)
v=∆d/∆t
(cm/s)
∆v
(cm/s)
a = ∆v/∆t
(cm/s 2 )
1
2
3
4
5
Force = ____________(kg)(9.8 m/s2 ) = ____________________N
Average acceleration = ___________(total) / __________(number) = ________cm/s2
2
Average acceleration = _______________ m/s
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Data Table 5.3
Run 3
d
(cm)
∆d
(cm)
∆t
(s)
v=∆d/∆t
(cm/s)
∆v
(cm/s)
a = ∆v/∆t
(cm/s 2 )
1
2
3
4
5
Force = ____________(kg)(9.8 m/s2 ) = ____________________N
Average acceleration = ___________(total) / __________(number) = ________cm/s2
2
Average acceleration =
Data Table 5.4
m/s
Run 4
d
(cm)
∆d
(cm)
∆t
(s)
v=∆d/∆t
(cm/s)
∆v
(cm/s)
a = ∆v/∆t
(cm/s 2 )
1
2
3
4
5
Force = ____________(kg)(9.8 m/s2 ) = ____________________N
Average acceleration = ___________(total) / __________(number) = ________cm/s2
A
l
ti
/
60
2
Data Table 5.5
Run 5
d
(cm)
∆d
(cm)
∆t
(s)
v=∆d/∆t
(cm/s)
∆v
(cm/s)
a = ∆v/∆t
(cm/s 2 )
1
2
3
4
5
Force = ____________(kg)(9.8 m/s2 ) = ____________________N
Average acceleration = ___________(total) / __________(number) = ________cm/s2
Aver ge
2
eler tion =
Data Table 5.6
m/s
Run 6
d
(cm)
∆d
(cm)
∆t
(s)
v=∆d/∆t
(cm/s)
∆v
(cm/s)
a = ∆v/∆t
(cm/s 2 )
1
2
3
4
5
Force = ____________(kg)(9.8 m/s2 ) = ____________________N
Average acceleration = ___________(total) / __________(number) = ________cm/s2
l
i
61/
2
Data Table 5.7
Run
Summary
Average Acceleration
(m/s2)
1
2
3
4
5
6
62
Force
(N)
Mass of System
(kg)
Acceleration (m/s 2)
0
1
2
3
4
5
6
0
0.5
•
1.0
•
1.5
•
Force (N)
2.0
•
Acceleration vs. Force
2.5
•
3.0
•
Slope
=
∆a
∆ F = 2.07
Integrated Science
kg
1