# Newton Lab Report

I. Introduction: The purpose of this lab report is to differentiate between of Newton’s Third Law and Newton’s Second Law. Newton’s Third Law states that all forces come in pairs and that the two forces in a pair act on different objects and are equal in strength and opposite in direction. Newton’s Second Law states that the acceleration of an object is proportional to the net force and inversely proportional to the mass of the object being accelerated. Using calculation equations for acceleration, force, and percent error one will be able to distinguish and evaluate the relationship between the two laws.
II. Procedures of experiment: All groups had to complete two types of labs. The first lab our group had to complete was completed like so: using a car one had to tie a piece of string approximately 80 centimeters in length to a toy car on a ramp. This string was then feed one top of a super frictionless wheel. The end of the rope that had the end nearest to the ground was tied off with a 0. 2kilogram weight. The car was then pulled back by a participant until it reached 0. 8 meters and was let go. A second participant would then record the time it took for the car to reach the end of the ramp.
This was repeated a few times. One would add 0. 5 kilogram weight to the car and the time was then recorded in a similar fashion. These times would be recorded into a table and would be used as raw data. The second part of the lab was different in that: a group had to tie a small scale to opposite end of the rope where the weight was located. Immediately following that scale, one would then tie a toy car. After the toy car another scale would be attached; likewise another car would follow. One student would then pull the whole system back before the back end of the second car would touch the wall provided by the ramp.

The calculations for the second part of the lab would have been nearly impossible unless one understood that FT= -Fg=F1+F2. With this in mind, it was possible to ascertain that understand thatF2=-F1+Fg. Knowing this, one would be able to understand that the acceleration on the system is the same throughout. Also, it is extremely important that one must correctly change units into for the needed equations. Otherwise, almost all of the equations will be void. mass of the car | acceleration of car (m/s2)| F=(m1+m2)a (Newtons)| Fnet= (m2)(9. 80m/s2) (Newtons)| Percent error/difference| 0. 261kg (x1 car)| 1. 84999| 0. 57534689| 0. 49| 17. 177| 0. 763kg (x1 car + 500g)| 0. 720688| 1. 50404187| 1. 47| 2. 3157734| This is the first table representing the acceleration of the car, the force of the car as well as the percent error. The percent error for this particular section of the lab seems relatively low. One should see that the acceleration of the second car is significantly slower than the first car; this is because car one weighted les s than the second. It should also be seen that the force of the car with the extra mass is nearly three times that of the car without the extra mass. mass of the car | acceleration of car (m/s2)| F=(m1+m2)a (Newtons)| Fnet= (m2)(9. 0m/s2) (Newtons)| Percent error/difference| 0. 261kg (red car)| 1. 858237548| 0. 856647509| 1. 96| 56. 2934| 0. 261 kg (blue car)| 1. 858237548| 0. 856647509| 1. 96| 56. 2934| This is the second table representing the acceleration, force, and percent error of both of the cars tied together. The percent error for this particular section of the lab seems relatively high. One should see that the acceleration and forces are exactly indistinguishable. This is because they are tied together forcing the cars and scales to have the same acceleration and force etc; moreover, it is because of Newton’s Third Law.
Additionally here are some sample equations I utilized during the lab report and calculations. Manipulating some of the equations was tricky especially for percent error. Substituting the accepted value with an equation is very smart thinking. percent error=(m2(9. 80m/s2))-experimental value(m2(9. 80m/s2)) (100) 56. 2934944%=1. 96-(0. 856647509)(1. 96) (100) ForceMass=acceleration 0. 485N0. 261kg=1. 858237548 m/s2 Distance= 12(acceleration)(time in seconds)2 2(Distance)(t)2= acceleration 2(. 8)(0. 93)2= 1. 849924847m/s2 V. Conclusion: There are several errors that could have occurred during this lab.
One of them might have been that I could have corrupted my calculations. I am not to entirely sure with how I calculated my percent error and therefore I might have a lower percent error yet I would not even know it or vise versa. A way to correct this problem for the future is by asking for more assistance from others in the class. I am more than sure that others who grasp the concept easier than me would be more than willing to aid me. Another error that could have occurred was that of miss reading the scales for the force in the second lab.
Although important to read the scales as accurately and as quickly as possible, one could only do so much. There was only less than a split second to read the correct or desired reading from the scale. One way to correct this error in the future is to have a larger group work on the same lab. Therefore all the students in the group could work together to figure out a solution as a team rather than an individual effort. The group would learn better as a result because there would be hands to complete the work and more brains to understand that work that is being recorded.
All thought the margin of error was higher for this lab, it is understandable because it was nearly impossible to record the force of Newtons desired in the split seconds one had. With all the information above, it is clear that there is a tie between Newton’s Third Law and his Second Law. It is possible to understand that Newton’s Second Law deals with changes in state of motion while Newton’s Third Law deals with the relation between forces On my honor, I have neither given nor received unauthorized aid on this assignment.

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