1. NEWTON’S LAWS OF MOTION
MIT C.P. PHYSICS
NEWTON’S LAWS OF MOTION

2. Law of Universal Gravity
Sir Isaac Newton
Law of Universal Gravity

3. Lab: Hey Newton, Pass the Water!

4. Lab: Hey Newton, Pass the Water!
Data Chart
Volume
(mL)
Finish
Order

5. Newton’s Laws of Motion
First Law of Motion
An object at rest will stay at rest unless acted upon by an outside force.
An object in motion will stay in motion unless acted upon by an outside force.

6. Newton’s 1st Law of Motion
Inertia

7. Inertia increases with mass.

8. Lab: Inertia-It Makes Cents!

10. Lab: Inertia Ball

11. Lab: Inertia Ball

13. Lab: Inertia Ball
Data Chart
Pull
Slowly
Quickly

14. Lab: Inertia Ball
Analysis
Pull
Slowly
Quickly

15. Applications

16. Applications

17. Lab: Newton’s 2nd Law of Motion
Interactive

18. Lab: Newton’s 2nd Law of Motion
Data Chart
Trial #
Mass of Wagon
(g)
Hanging Mass
Acceleration
(m/s/s) 1
100 2 3 4 5
200 6
400 7
600 8
800 9
1000

19. Lab: Newton’s 2nd Law of Motion
Calculation Chart
Trial #
Mass
(kg) F
(N)
Acceleration
(m/s/s) 1 2 3 4 5 6 7 8 9

20. Lab: Newton’s 2nd Law of Motion
Formulas
Mass (kg) = mass of wagon/1000
Force (N) = hanging mass X

21. Newton’s Laws of Motion
2nd Law of Motion
The sum of all forces acting on an object, F(net), equals the object’s mass times its acceleration.
F(net) = ma

22. Newton’s 2nd Law of Motion
Small F(net), small m = acceleration
Small F(net), big m = no acceleration

23. Sample Problem 1
Two students are sitting across each other in lunch. Student A slides a 2.2kg plate of food towards student B. If the net force is 1.6N to the right, what will be the plate’s acceleration?
Step 1
F(net) = ma
Step 2
1.6N = (2.2 kg)(a)
Step 3
a = 0.73 m/s/s

24. Sample Problem 2
The net force acting on a model airplane is 7.0N. The plane accelerates at a rate of 2.2 m/s/s. What is the mass of the model airplane?
Step 1
F(net) = ma
Step 2
7.0N = (m)(2.2m/s/s)
Step 3
m = 3.2 kg

25. Lab: Football Physics

26. Lab: Football Physics Data Chart Player Weight (N) 36 m Speed (m/s)
36 m Time
(s)
Nate
833
16.0
4.51
“T.D.”
735
16.4
4.40
Bubba
911
15.0
4.82
Tony
825
15.3
4.71
Tiny
1010
14.7
4.90
Mike
931
15.7
4.60

27. Lab: Football Physics Calculation Chart Player Mass (kg) Acceleration
(m/s/s)
F(net)
(N)
Nate
“T.D.”
Bubba
Tony
Tiny
Mike

28. Lab: Football Physics
Analysis
Linemen 1. 2.
Backs
Ends

29. Lab: The Elevator Lab

32. If all the forces acting on an object are balanced, the object moves at a constant velocity.

33. Lab: The Elevator Lab
Data Chart
Procedure #
Force
(N) 1 2 3 4 5

34.  = Balanced  = Unbalanced
Lab: The Elevator Lab
Calculation Table
F(g)
(N)
Mass
(kg) a
(m/s/s)
F(app)   1 2 3 4 5
 = Balanced  = Unbalanced

35. Sample Problem 1
A student has a mass of kg. He wants to lose weight so he stands on a bathroom scale while in an elevator. He presses the up button causing him to ascend at an acceleration of 4.00 m/s/s. What does the bathroom scale read on the way up?
Step 1
Determine the F(g).
F(g) = mg
F(g) = (200. kg)(9.81m/s/s)
F(g) = 1960 N

36. Sample Problem 1 Step 2 Determine the F(app). F(app) = ma
F(app) = (200. kg)(4.00 m/s/s)
F(app) = 800.N
Step 3
Determine the F(net).
F(net) = F(g) + F(app)
F(net) = 1960N N
F(net) = 2760N

37. Sample Problem 2
A student has a mass of kg. He wants to lose weight so he stands on a bathroom scale while in an elevator. He presses the down button causing him to descend at an acceleration of 4.00 m/s/s. What does the bathroom scale read on the way down?
Step 1
Determine the F(g).
F(g) = mg
F(g) = (200. kg)(9.81m/s/s)
F(g) = 1960 N

38. Sample Problem 2 Step 2 Determine the F(app). F(app) = ma
F(app) = (200. kg)(-4.00 m/s/s)
F(app) = -800.N
Step 3
Determine the F(net).
F(net) = F(g) + F(app)
F(net) = 1960N + (-800.N)
F(net) = 1160N

39. Free Fall & Terminal Velocity

40. Terminal Velocity

41. Sample Problem 1
At a point during a jump, a 89.1 kg skydiving dog experience a F(air) of 400.N Calculate the net force on the skydiving dog.
Step 1
F(net) = F(g) + F(air)
Step 2
F(net) = (89.1kg)(9.81m/s/s) + (-400N)
Step 3
F(net) = 474N

42. Sample Problem 2
At a point during a jump, a 89.1 kg skydiving dog experience a F(air) of 400.N Calculate the skydiving dog’s rate of acceleration.
Step 1
F(net) = ma
Step 2
474 = (89.1kg)a
Step 3
a = 5.32 m/s/s

43. Feather & Elephant Challenge

44. Feather & Elephant Challenge

45. Newton’s Laws of Motion
3rd Law of Motion
For every action, there is always an opposite but equal reaction.

48. Applications

49. Action-Reaction Pairs

50. Action-Reaction Pairs

51. Which Rope Is Being Pulled With More Force?
Same!

52. Identify 6 Pairs of Action-Reaction

55. Test Review Game Added Points Score(Single) Score(Double) 1 3000 15002
3500
1750 3
4000
2000 4
4500
2250 5
5000
2500 6
5500
2750 7
6000 8
6500
3250 9
7000 10
7500
3750