Justin Geeslin
Eng Physics I - M 1:00
Experiment Six – Newton’s Second Law
Introduction
This experiment demonstrates the effects of Newton’s Second Law of motion. The following procedure correlates the effects of a varying mass on the force of a system. The force is then displayed upon the system. Force in a system is equal to the mass times the objects acceleration. In this case, the acceleration is, once again, caused by gravity. The object accelerating is one of varying mass on a frictionless plane.
Procedure
The setup includes a level track for a toy car to travel across. Attached to the car is a string that runs parallel to the track. At the end of track, it loops over a pulley and supports a number of different weights. These weights will be accelerated by gravity causing the car to accelerate down the track. The car to will carry a varying number of weights. This procedure will be repeated six times. Each time mass will be shifted from the car to the hanging end of the string. This will cause the acceleration to increase with each trial. It is important to note that the mass of the car itself apart from any added weight would affect the car’s acceleration thus, the car’s weight must be measured and taken into account. The photogates along the track will detect the car’s acceleration. The photogates should be .5m apart and, the car should be released right before the first photogate. It is important to note that simply adding mass to one end of the string while keeping the mass of the car constant would not cause the acceleration to change because the weight(s) would be accelerated by gravity at the same rate regardless of mass. The total mass of each trial will be constant because mass never leaves the system. It is only shifted from the car to the end of the string. The trial data should be as follows:
Trial | Added Mass | Total Mass | V1 | V2 | A=(v22-v21)/2x | F=mg |
1 | 10g | 366g | 0 | .282 | .079 | .098 |
2 | 20g | 366g | 0 | .425 | .184 | .196 |
3 | 40g | 366g | 0 | .605 | .366 | .392 |
4 | 60g | 366g | 0 | .726 | .527 | .588 |
5 | 80g | 366g | 0 | .816 | .655 | .784 |
6 | 100g | 366g | 0 | .980 | .960 | .980 |
Weight of the Car : 0.266kg
V1 is the initial velocity v2 is the velocity as recorded by the photogates. Acceleration is calculated by finding the average velocity (v22-v21) then dividing by twice the distance between the photogates (2(.5)). In the next column, the net force for the system is calculated. The acceleration due to gravity (9.8 m/s2) is multiplied by the added mass (.01, .02, .04, etc.)
Questions
Yes, the net force on an object and the object’s acceleration are directly proportional. This is proven by the equation F = ma.
The units of a force/acceleration graph when dealing with the Newton would be kg/m/s.
An equation that relates force, mass, and acceleration is simply Force = mass x acceleration.