Physics Essay Example Describing the Experiment Based on Newton's Second Law of Motion

Published: 2019-11-25
Physics Essay Example Describing the Experiment Based on Newton's Second Law of Motion
Type of paper:  Essay
Categories:  Physics
Pages: 4
Wordcount: 922 words
8 min read
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Purpose of Experiment

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The purpose of this experiment is to use Newton's second law of motion to find the predicted acceleration of given mass of objects. From the predicted acceleration, the gravitational pull the object experiences can then be obtained. All the measurements are calculated based on the working of an Atwood machine.

Introduction

Newtons second law of motion predicts that the acceleration a of a given object is directly proportional to the net force the object experiences (Thornton et al., 57). Based on the same law, it can then be deduced that the acceleration is indirectly proportional to the mass (m) of the object in question. By using the Atwood machine, one can determine both theoretically and experimentally the values of the acceleration the object experiences as the basis for testing the validity of the second law of motion. This is normally done by using different acceleration data from the Atwood machine and comparing with the theoretical value based on the fact that Force is a product of mass and the acceleration of the object. It is also prudent that in carrying out the experiment, the masses that should be used should be almost equal to ensure that that the acceleration in question is slowed down. This helps one to obtain the value of g with high precision.

Procedure

The Atwood Machine was set up as shown in the figure below and everything checked to see whether the masses used were equal to 250grams on either side of the systems balance. Using the meter stick, the total distance travelled by the left hand side of the hanger was determined.

Different masses on the left side were placed and the observations noted on both the left and right hand side of the system and the time taken by M2 to drop a distance of 68.8 centimeters measured from the stopwatch recorded.

Figure 1 showing the set-up of an Atwood Machine

Observation

It is observed that when small masses are placed on the left side of the system, the system experiences a state of imbalance thus causing it to drop. Since the whole system is connected with a string, the right hand experiences a tension force that occurs on the string thus causing it to rise as a counter-action towards the drop on the left hand side.

It is also noted that the time taken to drop by the given object decreases with the increase in mass of M2 in comparison to the mass of M1which is held constant as 250grams.

Determination of the Different values of a

Using the displacement of M2 and the time taken for it to drop, one can obtain the different accelerations for experienced by the object. The formula used to obtain the acceleration is as shown below:

D = 12at2......Eqn.1

Where D is the displacement or distance that M2 drops in unites of meters (m)

a=acceleration of M2 in meters per second squared m/s2

t= time in seconds (s) taken by the object of mass M2 to drop to a distance given by the displacement of M2

For the case of M2 as 260 grams, the acceleration can be calculated as follows:Distance=0.688m

t=2.71 seconds

Making a the subject in Eqn.1 we obtain the following equation:

a=2Dt2.Eqn.2

a in this case will be obtained as: =(2x0.688m)/(2.71s)2

=0.1874 m/s2

For the other cases where different times are recorded for the different masses of M2, the accelerations can be obtained using the same approach. The table below shows the different accelerations obtained for the different masses:

Trial Mass of M2 (kg) Time to drop (s) Displacement (m) Acceleration (m/s2)

1. 0.2600 2.7100 0.6880 0.1874

2. 0.2700 1.9400 0.6880 0.3656

3. 0.2800 1.6100 0.6880 0.5308

4. 0.2900 1.4100 0.6880 0.6921

5. 0.3000 1.2800 0.6880 0.8398

Determination of the value of g for each trial

g is related to a by the formula below:

a=m2-m1m2+m1g.Eqn.3

Where a is the acceleration and g is the acceleration due to gravity in the units of m/s2

For the first case where the mass of M2 is 0.2600kg and that of M1 is 0.2500kg, the acceleration due to gravity is obtained as follows:

g=m2+m1m2-m1a

= ((0.2600+0.2500)/0.2600-0.2500))x0.1874m/s2

=9.5554m/s2

M2(kg) M1(kg) Acceleration (m/s2) Gravitational acceleration (m/s2)

0.2600 0.2500 0.1874 9.5554

0.2700 0.250 0.3656 9.5058

0.2800 0.2500 0.5308 9.3782

0.2900 0.2500 0.6921 9.3435

0.3000 0.2500 0.8389 9.2382

Based on the values of acceleration due to gravity obtained, the average value of g can be obtained by finding the mean of all g obtained. It is found as follows:

Average g= (9.5554+9.5058+9.3782+9.3435+9.2382)/5

=9.4043m/s2

Comparison with the Existing Value of g

The calculated value of g was found to be 9.4043m/s2 which is less than the theoretical value of g in literature noted as 9.81m/s2. It is less by 0.4057 which is an error of almost 4% of the theoretical value.

Conclusion

It is evident that by using the Atwood machine, one can obtained the value of acceleration of the different masses and use it to find the gravitational pull they experience due to gravity. This goes in hand to prove the fact that force is directly proportional to the acceleration due to gravity as predicted by Newtons second law of motion. The experiment was successful as it was evident that an increase in the mass of M2 led to an increase in the acceleration. The acceleration is therefore dependent on the mass of the object in question. It is difficult to obtain the same value of acceleration due to gravity because of the errors involved. The objectives of the experiment were achieved.

Work Cited

Thornton, Stephen T., and Andrew Rex. Modern physics for scientists and engineers. Cengage Learning, 2012.

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