Darren Luna
Grrade-12 STEM 1
A Newton’s Second Law
A Post- Lab Report
I. Introduction
Newton’s second law states that the acceleration of an object defends upon two
variables, the net force acting on the objects and the mass of the object. The acceleration of the
body is directly proportional to the net force acting on the body and inversely proportional to
the mass of the body.
The example to know and understand the newton’s second law is the dynamic cart
apparatus. A dynamic cart apparatus is used in studies of the many different principles of
physics including energy, forces, speed, velocity, acceleration, momentum, inertia and
Newton’s law of motion. The dynamic cart apparatus served as the practical platformed for
exploring the principles of newton’s second law. The relationship of this are through systematic
experimentation.
By doing this experiment, we can understand the importance of the dynamic cart
apparatus by knowing the newton’s second law. Doing several trials, there are the crucial part
the sting it must 80cm long while testing the string is cut off because the ring is so heavy and
the impact of acceleration of the cart. Patience and attention to details are needed.
II. Materials and Methods
A. List of the materials
-1 dynamics cart (w/o spring)
-1 electronic digital stopwatch
-1 stand bases
-2 rails (rods undersized on ends)
-1 leveling pad
-1 stopper wheel
-5 disk masses, 50 grams each
-5 ring masses, 3 grams each
-1 string, approx. 80 cm long
-1 meter tape
-1 piece modeling clay (friction clay)
-1 single pan balance
-1 piece plastic hose, 2 meters
B. Methods
B.1
1. Prepare the setup as shown in Figure 1, previous page. Set a certain distance on the rail
(approx. equal to the height of table) for the cart to travel. Bring the cart to the front end of
the rail. If the string touches the floor, shorten it enough to make it hang freely. Record this as
d in Table 1.
2. Measure the total mass: mass of cart plus mass of the five ring masses combined. Record
as 𝑚𝑡𝑜𝑡𝑎𝑙.
3. Put all 5 rings on top of the cart (inserting these thru a vertical rod) and pull the cart back
to the start position. Study Figure 1 again.
4. Friction between the setup's moving parts should be compensated. This can be achieved by
suspending some amount of clay on the free end of the string. To determine the right amount
of clay to use, push the cart slightly. If the cart moves with a steady speed until the opposite
end, the amount of clay is enough. If too much clay is suspended, you will see that the cart
moves with an increasing speed. If the cart does not move at all, or if the cart stops
momentarily, there is too little clay used.
5. Now, transfer one ring mass from the cart and suspend it on the free end of the string. Its
weight when suspended provides the net force. Pull the cart back to starting position.
6. Release the cart freely (do not push) and observe its motion. Measure the time the cart
takes to travel the distance you set along the rail. Do this for 3 trials.
7. Continue transferring one ring mass after another from the cart to the free end of the string.
Do step 6 every time a ring mass is transferred. You may not need to adjust the amount of
friction compensation clay every time you transfer one ring mass from the cart to the string
end. You might do so when 4 or 5 rings have been transferred.
8. Record all measurements in their respective places on Table 1.
9. Calculate the following quantities and enter values in Table 1. a. Acceleration of
the cart: 𝑎= 2𝑑𝑡2⁄
b. The ratio 𝐹𝑛𝑒𝑡𝑎⁄
10. Plot a graph of net force 𝐹𝑛𝑒𝑡 vs. acceleration 𝑎. Calculate slope of this graph.
Use a separate sheet of graphing paper.
B.2
1. Without a load on the cart, and no suspended ring mass on the string, compensate friction
by doing step No. 4 of Experiment A.
2. Now, suspend 3 ring masses on the free end of the string. Their combined weight when
suspended provides the net force (approximately 0.09 N).
3. Release the cart freely and measure the time it takes to travel the distance you set along the
rail (the same distance you set in Experiment A). Do this for 3 trials. Enter the results in
Table 2.
4. This time, put one (1) 50 g cylindrical mass on top of the cart, and remove all suspended
ring masses from the string. Adjust the friction compensation clay again, as you did in
Experiment A, step No.
5. Suspend the three ring masses back on the string and do step No. 3, Experiment B.
6. Load another 50 g cylindrical mass on the cart, and remove all suspended ring masses and
adjust the amount of friction compensation clay. Suspend the ring masses back and do step
No. 3 of Experiment B again.
7. Repeat step No. 6 of Experiment B for each of the remaining cylindrical masses.
8.Complete Table 2.
9. Plot the following graphs: acceleration a vs. mass m and acceleration a vs. inverse of mass
1/m. Use a separate sheet of graphing paper.
III. Result and Discussion
A. Experiment A-Force and Acceleration Data 1
𝐹𝑛𝑒𝑡 (𝑁)
𝑡 (𝑠)
𝑡𝑎𝑣𝑒 (𝑠)
𝑡2 (𝑠2)
𝑎 (𝑚/𝑠2)
1.50s
2.25s2
2.1 m/s2
0.01 kg
0.65s
0.42s2
3.8 m/s2
0.02 kg
0.56s
0.31s2
5.2 m/s2
0.01 kg
0.42s
0.18s2
8.9 m/s2
0.01 kg
0.26s
0.068s2
24 m/s2
0.0063 kg
𝐹𝑛𝑒𝑡/a
1.41s
0.03
1.43s
1.65s
0.65s
0.06
0.72s
0.59s
0.56s
0.09
0.59s
0.53s
0.41s
0.12
0.40s
0.44s
0.28s
0.15
0.28s
0.26s
Average= 0.01 kg
In the Experiment A-Force and Acceleration Data 1, the mass was measured and the
time recorded to converted to the average of time. The 0.03N and the average is 1.50s,0.06N
have a time 0.06s, in 0.09N has 0.56s of time, 0.12N have a 0.42s, and last is 0.15N has 0.26s
of time. The all averaged was multiplied by itself to get the answer of t2. To get the acceleration
the formula is a=2d/t2, after we graph the data for the acceleration to analyzed the data.
B. Experiment B-Mass and Acceleration Data 2
𝑙𝑜𝑎𝑑
(kg)
𝑚𝑡𝑜𝑡𝑎𝑙
(kg)
0
435kg
0.05
485kg
0.1O0
535kg
0.15
585kg
0.20
625kg
0.25
685kg
𝑡 (𝑠)
0.63s
0.65s
0.72s
0.75s
0.66s
0.75s
0.75s
0.75s
0.78s
0.76s
0.78s
0.81s
0.82s
0.85s
0.84s
0.87s
0.85s
0.87s
𝑡𝑎𝑣𝑒 (𝑠)
𝑡𝑎𝑣𝑒2
(𝑠2)
𝑎 (𝑚/𝑠2)
𝑚𝑋𝑎 (𝑁)
1𝑚⁄
(1𝑘𝑔⁄)
0.41s
0.42s2
1.57 m/s2
683N
435kg
0.50s
0.52s2
1.40 mls2
679 N
485kg
0.56s
0.59s2
1.30 m/s2
696 N
535kg
0.59s
0.62s2
1.31 m/s2
766 N
585kg
0.69s
0.69s2
1.20 m/s2
762 N
625kg
0.72s
0.72s2
1.18 m/s2
808 N
685kg
In the Experiment B-Mass and Acceleration Data 2 the relationship of m and a is
inverse proportional when comparing the all values in the table, where higher mass
corresponded to lower acceleration and vice versa, the product it remains consistent for each
set of trials in the table. From the table there is direct relationship between the mass and the
acceleration. The comparison of mXa and the Fnet this alignment between the calculated force
and the net force of newton’s second law.
C. Graph
C.1 Experiment A-Force and Acceleration Data 1
Fnet vs. a
0,18
0,16
0,14
Fnet(N)
0,12
0,1
0,08
0,06
0,04
0,02
0
0
5
10
15
20
25
30
a (m/s2)
In the graph Experiment A-Force and Acceleration showed the direct proportional
between the two variables. The relationship exists between Fnet and a, the graph showed us a
positive slope, which indicates the force and acceleration are proportional.
C.2 Experiment B-Mass and Acceleration Data 2
a vs. m
1,8
1,6
1,4
𝑎 (𝑚/𝑠2)
1,2
1
0,8
0,6
0,4
0,2
0
435
485
535
585
625
685
m(kg)
In the a vs. m graph showed the inverse proportional, the relationship between m and a they are
inverse proportional. If a mass increased, acceleration decreased, and the vice versa assuming a constant
force.
C.3 Experiment B-Mass and Acceleration Data 2
a vs. 1/m (1kg)
1,8
1,6
𝑎 (𝑚/𝑠2)
1,4
1,2
1
0,8
0,6
0,4
0,2
0
435
485
535
1/m(1kg)
585
625
685
In the graph showed the same result of graph of a vs. m because when you multiplied the
total mass to 1/m (1kg) the answer will be the same that why the graph is inverse proportional.
IV. Conclusion
Newton's second law provides the explanation for the behavior of objects upon which the
forces do not balance. The law states that unbalanced forces cause objects to accelerate with
an acceleration that is directly proportional to the net force and inversely proportional to the
mass. Through this lab, we have found out that acceleration and mass are directly proportional,
which is what Newton had stated, and this is seen through both the theoretical and experimental
acceleration vs mass graphs. We also found that acceleration and Fnet are proportional, and this
is seen through both the theoretical and experimental graphs as well. Both these finding make
sense through what we know about forces. F=ma describes both these relationships and we
know that as something obtains more mass, more force is required to move it due to Fa,
therefore acceleration would be low. If something is lighter. Fs is smaller, so if the same amount
of force that was applied to the heavier object was applied to the lighter object, the force would
make the lighter object acceleration significantly. In the other graph there are inverse
proportional like the a vs. m graph showed the inverse proportional, the relationship between
m and a they are inverse proportional. If a mass increased, acceleration decreased, and the vice
versa assuming a constant force, and the graph showed the same result of graph of a vs. m
because when you multiplied the total mass to 1/m (1kg) the answer will be the same that why
the graph is inverse proportional.
Exercises/Problems:
1. A net force of 10 N is continuously applied on a 2 kg object. Find its acceleration.
Answer: 5m/s2
2. A 1.200 kg jeepney rests on a level highway. The driver starts the engine and the jeepney
moves a distance of 5 meters in 2 seconds. How much net force (assuming to be constant)
was exerted on the jeepney?
Answer: 1.500N
3. Both a 10-wheeler truck and an automobile accelerate at 3 𝑚𝑠2⁄.Which of the two requires
more net force and why?
Answer. The 10 wheeler truck required more net force that the automobile because according
to newton’s second law force is directly proportional to mass and the acceleration, the one
with greater mass required more force to accelerate at the same rate.
4. A cyclist is pedaling his bicycle along a level road and their combined mass is 70 kg. For
whatever reason, the cyclist stopped pedaling and thereafter, the bicycle traveled 25 meters
more for 10 seconds before it finally stopped by “itself.” Calculate the friction force between
the bicycle's wheels and the road.
Answer: 686N
5. A rocket fired into space has a continuous net force applied with its mass continuously
decreasing. Describe the motion of the rockAnswer: The rocket mass are decreases to
conserve the momentum. That means is the rocket will experience the acceleration in the
direction of applied force. In otherwise the rocket will accelerate and move faster into space
due to the continuous force and the decreasing mass.