Forces and Newton IIPhysics 131Hasbrook 210October 16th 2015Abstract: In this experiment we did two different experiments to tests Newton’s first and second laws. By mapping the forces that were applied to a ring, we could see that the ring was not accelerating when all the forces added up to zero. This shows Newton’s first law. Next we attached a hanging mass to a cart on a frictionless surface measured its acceleration. Questions1. The acceleration for both the golf ball and the hanging mass were constant. The golf ball was in free fall because no force besides gravity was acting upon it. The hanging mass was not in free fall because it was hanging from a sting. It had another force besides gravity acting upon it.2. The ring in the middle is not accelerating. When the forces all equal zero the ring will not accelerate. There is gravity acting upon the ring pushing it downwards. The ring isn’t in free fall because it has tension forces acing uponit from the springs. The horizontal and vertical components of the ring wouldnot change if you dropped the board. The ring is being held in place by the tension forces.3. We can find the magnitude and direction of vector A by using the formula Acos(theta)+B=C. We know that B has a magnitude of 4 and it’s a 30° angle . After solving the x coordinate we have to solve for the y coordinate. This is done by the equation Asin(theta)=4. Now that we have those numbers we candivide 4/4.535 and set it equal to tan(theta). This would give us a direction of41.4 degrees. To find the magnitude of the vector at 41.4 ° we would use the equation Asin(41.4)= 4 resulting in A equaling 6.05 which is the magnitude ofvector A.4. Free Body Diagrams Cart Hanging MassFnFtFtFgFg5. The expression of net force on the glider is to the left because the cart is moving in that direction. The net force of the hanging mass is down because it is moving downwards. The net force of the system as a whole is down because the weight is being pulled down the most. The tension of the string isequal to the normal force of the hanging mass and the tension force of the cart, resulting in a cancelation of the two. 6.Data sheet for Forceshanging mass m mass ratio acceleration a error Free Fall acc. g5mg 57.42 0.18 (+/-) 8.7E-4 10.3420mg 15.105 0.67 (+/-) .0065 10.1240mg 8.053 1.08 (+/-).04 8.7 (Average) 9.72 (Standard deviation)0.890168523 Force Board:Vector Magnitude AngleXComponent Y ComponentA 4.9 244 -2.15 -4.40B 7.84 0 7.84 0.00C 7.84 140 -6.01=(-.32) 5.04 =(.64)The percent error for the measured for the average value of g is .8163%. This is found by subtracting the theoretical force (9.8) from the measured force (9.72) this gives you the error. Next you have to divide the error by the theoretical force (9.8) and multiply it by 100. A possible error that might havealtered the results of the experiment is that the air pushing on the care to make it move on a frictionless surface was moving the cart. Another error that could have been a mechanical error such as the track not being flat or being held at an angle. Another error could have been with the way the stringwas swinging. If the mass was swinging it could have been pulling the cart in different directions on the track. 7. If the string were to be cut then there would be no tension force acing upon the care or the hanging mass. The acceleration and velocity of the mass fallingdownwards would increase and the acceleration and velocity of the cart would slow down because it has no force other than gravity pulling it down so it would not be accelerating on the x-axis anymore. Conclusion After conducting this experiment and analyzing the data, we can conclude that Newton’s first law is true because the ring was not accelerating because the forces on it added up to about zero. Objects only accelerate when the forces on them are not zero. We have also learned that he object is net force of the cart was about equal to the mass times the acceleration. This proves Newton’s second