1. Newton’s Laws

2.  An object is moved with F1, with a certain m1 and a1. The same force is applied to m2 and a2. A2 is twice as large as A1. If you add m2 and m1 together and apply the same force, what is the acceleration?

3.  F1=m1a1 ◦ Plug in simple numbers for m and a ( I chose 1) ◦ F1= (1)(1)=1 F1= m2 a2 You know the force is the same, 1. And you also know that a2 is twice as large, so it would equal 2 in the insistence 1=m2 * (2) M2 = ½ Then you have to add m1 and m2 to find the acceleration for when both are added and the force stays the same so: F=(m1 +m2) (a3) 1 = (1 + 1/2) (a3) 1/(1/2) = a3=.666 or 2/3

4.  An African elephant can reach heights of 13 feet and possess a mass of as much as 6000 kg. Determine the weight of an African elephant in Newtons and in pounds. (Given: 1.00 N =.225 pounds)

5.  6000kg (9.8) = 58800 N  58800 N (.225 lbs/ 1 N) = 13230 lbs

6.  According to the National Center for Health Statistics, the average mass of an adult American male is 86 kg. Determine the mass and the weight of an 86-kg man on the moon where the gravitational field is one-sixth that of the Earth.

7.  Mass = 86 kg ◦ Stays the same, is constant  Weight = 140 N ◦ 86 * (9.8/6) = 140 N

8.  Ethan is dragging a bag of grass from the garage to the street on the evening before garbage pick- up day. Use the free body diagram to determine the net force, mass, and acceleration acting upon the bag. The values of the individual forces are:  F grav = F norm = 60.5 N F app = 40.2 N F frict = 5.7 N.

9.  F net : 16 N, right ◦ Fy cancel out ◦ Fx = +182 + -166 = 16N m: 21.1 kg Take normal force, 207 N / 9.8 = 21.1 a: 0.76 m/s/s, right Take the force in x direction, and use F=ma 16N = 21.1Kg (a) a =.76

10.  It's Friday night and Skyler has been assigned the noble task of baby-sitting Casey, his 2-year old brother. He puts a crash helmet on Casey, places him in the red wagon and takes him on a stroll through the neighborhood. As Skyler starts across the street, he exerts a 52 N forward force on the wagon. There is a 24 N resistance force and the wagon and Casey have a combined weight of 304 N. Construct a free body diagram depicting the types of forces acting upon the wagon. Then determine the net force, mass and acceleration of the wagon.

11.  F net : 28 N, forward ◦ Fx = 52 + - 24 = 28 m: 25.7 kg 304 – 52 = 252 N by wagon 252/ 9.8 = 25.7 F=ma 28N= 25.7kg(a) a=1.09 m/s2