final report
Cebu Institute of Technology-University
Physics Department
Name: Abeto, Mark V.Exp#/Group#/Time: 5 / 6 / W -7:3010:30TRIAL12345TRIAL12345
Date Submitted: March 13, 2013Instructor: Mrs. Luzpura Abellana
I. Title: Newton's Second Law of Motion: Simple Atwood's Machine
II. Objectives:
a. To study and verify Newton's Second Law of Motion
b. To measure the acceleration due to gravity
III. Apparatus and Materials:
Motion Sensor, PASCAR, Linear Track Apparatus, Pulley, Weight hanger, Slotted Masses: 2pcs. 5 grams, 2 pcs. 10 grams, 1pc. 100 grams; string
IV. Data and Results: Part1. Constant Mass and Increasing Net Force m1 (g)m2 (g)M(g)a(m/s2 )450150.32599.8609.80.612250.53859.8600.000350.75959.8750.749450.96109.9610.816551.180010.180.388 Part2. Constant Net Force and Increasing Mass m1 (g)m2 (g)M(g)a(m/s2 )gexpt’l%error257750.31959.9059.81.07106250.29559.7520.48988750.27709.6951.07009250.26309.7310.78409750.25159.8100.1020
V. Computation and Analysis:
1. By the use of appropriate formulas, and using the data obtained acceleration due to gravity and percentage of error.
2. How does the change in the hanging mass(m) affect the value of acceleration?
The mass affects acceleration. For objects with different masses (not objects with changing mass because F=ma does not apply to these situations, such as a rocket) an equivalent force will accelerate objects with greater mass at a lower rate. The lower the mass the greater the acceleration for a given force.
3. In the second part of the experiment, are the values of acceleration affected by the change in the value of the total moving mass(M)? explain.
If the force applied is constant, then yes, because "F = m * a"
Force equals mass times acceleration.
If mass changes, and force stays constant then acceleration must be affected.
4. In each case of the experiment, how do mass, net force and acceleration relate to each other?
Net force, mass, and acceleration are related to each other, because the product of mass and acceleration is directly proportional to the net force.
VI. Conclusion: Questions and Problems:
1. Is the weight of the body the same as its mass?
Mass is a measurement of how much matter is in an object; weight is a measurement of how hard gravity is pulling on that object.
2. A cord passes over a weightless and frictionless pulley. Masses 200 and 300 grams are attached to the end of the cord. Find the distance the masses the masses will move during the fifth second after they had started.
a=[(M2–M1)/(M1+M2)]*g a=g/5 =(a/2)(52–42)
=(g/10)(25-16)
=8.82 m
3. Two bodies having masses m1= 30 grams and m2= grams are attached to the ends of a string of negligible mass and suspended from light frictionless pulley. Find the acceleration of the bodies and the tension in the string.
T-m2g=-m2a
T-m1g=m2a
m2g-m1g=-m1a-m2a a=(m2–m1)g/(m2+m1) a=(0.04-0.03)(9.8)/(0.04+0.03) a=1.4m/s2 Generalization:
The use of Atwood's machine in illustrating the principles of the application of Newton's second law to a simple physical system, whose connected parts have both rotational and translational motion, is shown. A diagram which shows the forces acting on all the component masses is drawn, and this picture facilitates the setting-up of the appropriate F = ma and τ = Iα equations. The moment of inertia of the pulley, the friction in its bearings, and the mass of the string are included in the accelerating system in a natural way. The observed acceleration of the hanging masses can be compared with that computed by this method.
gexpt’l%error
Physics Department
Name: Abeto, Mark V.Exp#/Group#/Time: 5 / 6 / W -7:3010:30TRIAL12345TRIAL12345
Date Submitted: March 13, 2013Instructor: Mrs. Luzpura Abellana
I. Title: Newton's Second Law of Motion: Simple Atwood's Machine
II. Objectives:
a. To study and verify Newton's Second Law of Motion
b. To measure the acceleration due to gravity
III. Apparatus and Materials:
Motion Sensor, PASCAR, Linear Track Apparatus, Pulley, Weight hanger, Slotted Masses: 2pcs. 5 grams, 2 pcs. 10 grams, 1pc. 100 grams; string
IV. Data and Results: Part1. Constant Mass and Increasing Net Force m1 (g)m2 (g)M(g)a(m/s2 )450150.32599.8609.80.612250.53859.8600.000350.75959.8750.749450.96109.9610.816551.180010.180.388 Part2. Constant Net Force and Increasing Mass m1 (g)m2 (g)M(g)a(m/s2 )gexpt’l%error257750.31959.9059.81.07106250.29559.7520.48988750.27709.6951.07009250.26309.7310.78409750.25159.8100.1020
V. Computation and Analysis:
1. By the use of appropriate formulas, and using the data obtained acceleration due to gravity and percentage of error.
2. How does the change in the hanging mass(m) affect the value of acceleration?
The mass affects acceleration. For objects with different masses (not objects with changing mass because F=ma does not apply to these situations, such as a rocket) an equivalent force will accelerate objects with greater mass at a lower rate. The lower the mass the greater the acceleration for a given force.
3. In the second part of the experiment, are the values of acceleration affected by the change in the value of the total moving mass(M)? explain.
If the force applied is constant, then yes, because "F = m * a"
Force equals mass times acceleration.
If mass changes, and force stays constant then acceleration must be affected.
4. In each case of the experiment, how do mass, net force and acceleration relate to each other?
Net force, mass, and acceleration are related to each other, because the product of mass and acceleration is directly proportional to the net force.
VI. Conclusion: Questions and Problems:
1. Is the weight of the body the same as its mass?
Mass is a measurement of how much matter is in an object; weight is a measurement of how hard gravity is pulling on that object.
2. A cord passes over a weightless and frictionless pulley. Masses 200 and 300 grams are attached to the end of the cord. Find the distance the masses the masses will move during the fifth second after they had started.
a=[(M2–M1)/(M1+M2)]*g a=g/5 =(a/2)(52–42)
=(g/10)(25-16)
=8.82 m
3. Two bodies having masses m1= 30 grams and m2= grams are attached to the ends of a string of negligible mass and suspended from light frictionless pulley. Find the acceleration of the bodies and the tension in the string.
T-m2g=-m2a
T-m1g=m2a
m2g-m1g=-m1a-m2a a=(m2–m1)g/(m2+m1) a=(0.04-0.03)(9.8)/(0.04+0.03) a=1.4m/s2 Generalization:
The use of Atwood's machine in illustrating the principles of the application of Newton's second law to a simple physical system, whose connected parts have both rotational and translational motion, is shown. A diagram which shows the forces acting on all the component masses is drawn, and this picture facilitates the setting-up of the appropriate F = ma and τ = Iα equations. The moment of inertia of the pulley, the friction in its bearings, and the mass of the string are included in the accelerating system in a natural way. The observed acceleration of the hanging masses can be compared with that computed by this method.
gexpt’l%error