1. Lesson 1: Newton’s First Law
Unit 4: Newton’s Laws
Lesson 1: Newton’s First Law

2. Newton’s First Law (The Law of Inertia)
An object remains at rest or in uniform motion (constant speed in a straight line) unless acted upon by a net unbalanced force.
Example 1: A car approaches a curve covered in black ice, which reduces friction to zero. What will happen next?
FREE BODY DIAGRAM:
10 min
normal force
The car will move in a straight line since there are no unbalanced forces on it.
force of gravity

3. Example 2 (Try it!): A box is pulled along a rough floor by a horizontal force of 15 N at a constant speed. Draw a free body diagram and find the friction force.
FREE BODY DIAGRAM:
Normal force
P = 15 N f
(Force of friction)
10 min
Force of gravity
f = 15 N
(must be balanced)

4. Inertia
Inertia is the tendency of objects to resist changes in their states of motion (i.e. their tendency to follow Newton’s 1st Law).
The greater the mass of an object, the more inertia it has.

5. Brain Break!

6. Think-Pair-Share!
Does 2 kg of apples have twice the inertia or half the inertia of 1 kg of apples?
If the pen on your desk is at rest, can you say that no forces are acting on it?
If the forces acting on a pen are balanced, is it correct to say that the pen is at rest?
10 min

7. Think-Pair-Share!
If you place a ball in the centre of a wagon and then quickly push the wagon forward, in what direction does the ball appear to go?
You are travelling in a school bus. The driver has to apply the brakes suddenly to prevent an accident. Describe how your body would move in response to this sudden braking.
10 min

8. Think-Pair-Share!
While travelling in Africa, you are chased by a very large elephant. Would it make more sense to run in a straight line to get away, or in a zigzag motion?
A hockey puck moving across an ice rink eventually comes to a stop. Does this prove that Newton’s 1st Law does not apply to all situations?
10 min

9. Lesson 2: Newton’s Second Law
Unit 4: Newton’s Laws
Lesson 2: Newton’s Second Law

10. Newton’s Second Law
If a net force “F” acts on an object of mass “m”, the object accelerates in the direction of the net force such that F=ma
Example 1: A sled of mass 50 kg is pushed across a frictionless ice surface with a force of 220 N. If it starts from rest, find the acceleration and the velocity after 10 seconds. N
Unbalanced!
Fnet = 220 N
220 N
10 min mg

11. Example 2: A 561 kg rocket has an upward thrust force of 16500 N
Example 2: A 561 kg rocket has an upward thrust force of N. It starts from rest on the ground. Find its height after 10 seconds. T
5 min mg

12. Example 3 (Try it!): A 50 kg sled is pulled with a force of 250 N and accelerates at 1.5 m/s2. Find the force of friction. N f
250 N mg
10 min

13. Brain Break!

14. Example 4 (Try it!): A 530 kg snowmobile is pushed forwards by a force of 2500 N. There is a friction force of 750 N and air resistance of 300 N. Find the acceleration. N f
P = 2500 N A
10 min mg

15. Practice Time!
Do #1-3 on the Newton’s Second Law Problem Set.
20 min

16. Lesson 3: Practice with Newton’s Second Law
Unit 4: Newton’s Laws
Lesson 3: Practice with Newton’s Second Law

17. there is a 50 N friction force  =0.3
Practice #1: Find the acceleration of a 12 kg sled pulled with a force of 120 N if
there is no friction
there is a 50 N friction force
 =0.3 N f
120 N mg
20 min

18. Brain Break!

19. Find: a) the thrust required.
Practice #2: A 7000 kg jet accelerates forwards at 2.0 m/s2 against a 25000N drag force.
Find: a) the thrust required.
b) the acceleration if the thrust force is N.
L = lift
T = thrust
D = drag mg
15 min

20. Practice #3: A helicopter moving upwards at 10 m/s slows down and stops in 2.5 seconds. Find the tension in the rope supporting the 20 kg package during this time.
20kg
T = tension
20kg mg
10 min

21. Homework
Do #6 and 7 on the Newton’s Second Law Problem Set.
10 min

22. Lesson 4: More Practice with Newton’s Second Law
Unit 4: Newton’s Laws
Lesson 4: More Practice with Newton’s Second Law

23. b) the reading if the acceleration is 2.0 m/s2 upwards.
Practice #4: Being a curious individual, Jacob Z. decides to conduct an experiment by riding an elevator while standing on a scale. Jacob has a mass of 75 kg and his scale is calibrated in Newtons.
Find: a) the scale reading if the elevator is moving upwards at a constant speed of 1.0 m/s.
b) the reading if the acceleration is 2.0 m/s2 upwards.
c) the reading if the elevator is moving downwards and speeding up at m/s2.
d) the acceleration if the reading is 650 N.
e) the acceleration if the reading is 800 N.
N (scale reading)
30 min
mg = (75)(9.8)
= 735 N

24. Either going up and slowing down, or going down and speeding up
5 min

25. Brain Break!

26. Practice #5: A sled of mass 15 kg accelerates from rest to 15 m/s over 30 m. The pulling force is 75 N; find the coefficient of friction. N
P = 75 N f mg
N = mg
10 min

27. Homework
Do #4, 5, and 8 on the Newton’s Second Law Problem Set. (You have now been assigned all of the questions.)
5 min

28. Lesson 5: Newton’s Third Law
Unit 4: Newton’s Laws
Lesson 5: Newton’s Third Law

29. Newton’s 3rd Law
“Every action force is accompanied by an equal and opposite reaction force.”
In other words, if A pushes or pulls B, then B pushes or pulls A in the opposite direction with the same amount of force.
5 min

30. The ball pushes back on the bat with 250 N [East].
Try This: A baseball player hits a ball with a bat. The force on the ball during contact is 250 N [West]. What is the reaction force?
The ball pushes back on the bat with 250 N [East].
5 min

31. Try This: Your head slowly sinks into a soft pillow
Try This: Your head slowly sinks into a soft pillow. Show all the action-reaction pairs in a diagram. Fp Fg Rp Rg
10 min

32. Brain Break!

33. Action/Reaction Problems
Example: Two boxes (10 kg and 15 kg) are tied together by a string. The 15 kg box is pulled forwards by a force of 50 N. If there is no friction, find:
the acceleration of the boxes.
the force exerted by a the 10 kg box on the 15 kg box.
15 kg
10 kg
50 N b)
10 kg S
F = ma
S = (10)(2)
= 20 N
15 min OR a)
25 kg
15 kg
F = ma
50-S = (15)(2)
S = 20 N
50 N
F = ma
50 = 25a
a = 2 m/s2 S
50 N

34. the force exerted by a the 10 kg box on the 15 kg box.
Your turn!
Repeat the previous problem, but with a coefficient of friction of 0.1 this time…
15 kg
10 kg
50 N
Find:
the acceleration.
the force exerted by a the 10 kg box on the 15 kg box. a) N
f = (mg)
= (0.1)(25)(9.8)
= 24.5 N
F = ma
= 25a
a = 1.02 m/s2
25 kg f
50 N mg
20 min b) N
F = ma
S = (10)(1.02)
S = 20 N
f = (mg)
= (0.1)(10)(9.8)
= 9.8 N
10 kg f S mg

35. Unit 4: Newton’s Laws
Lesson 6: Review