1. QQ: What is the weight of a 65 kg. person on the Earth?
What is the weight of a 65 kg. person on the moon where the force of gravity is
1.6 N kg-1?
3. A boxer punches a sheet of tissue paper
with mass of kg in midair.
In only 0.05 seconds, it
gains a speed of 30 m/s.
Find the force the boxer
places on the paper.

2. QQ: N What is the weight of a 65 kg. person on the Earth? m = 65 kg
g = 9.8 m/s2
w = fg = m*g
fg = 65 kg*9.81 m/s2
fg = 637 kg * m/s2 N

3. What is the weight of a 65 kg
What is the weight of a 65 kg. person on the moon where the force of gravity is
1.6 N kg-1?
M = 65 kg g = 1.6 N kg-1
W = fg = m*g
W = fg = 65 kg * 1.6 N/kg
fg = 104 kg N/kg
= N
= 1.6 N/kg

4. 3. A boxer punches a sheet of tissue paper
with mass of .001 kg in midair.
In only 0.05 seconds, it
gains a speed of 30 m/s.
Find the force the boxer
places on the paper.
m = .001 kg t = 0.05 s s = 30 m/s
f = ma
f = .001kg * 600 m/s2
f = .6 kg*m/s2
a = s/t
a = 30 m/s
.05 s
= 600 m/s2
= 30 m
0 .05 s2

5. Newton’ s Three Laws of Motion

6. Let’s all jump!

7. Tension is a force that is distributed along a linear object like a rope, string, wire, cable, chain, etc.
It is transmitted when the ‘linear object’ is stretched by other forces which act at opposite ends and pull in opposite directions.
Any tension force generated is evenly distributed along the length of the linear object.

8. In this example, even as it approaches its breaking point, it continues to hold.
The forces are balanced and
in equilibrium. Logically,
this means there must be some force holding it together.
The tension at the bottom equals the for weight of the kitten. The tension of the rope below the fray must be equal to the tension below it only in the opposite direction.

9. Forces of tension are acting at the same time as other forces.

10. These forces are of equal magnitude so the forces are balanced.
By pulling on the rope from opposite ends and in opposite directions, each team applies force to the rope. The greater the opposing pulling forces, the greater the magnitude of tension force along the rope.
These forces are of equal magnitude
so the forces are balanced.

11. Tension is measured in Newtons.
The smaller the opposing pulling forces, the lesser the ‘magnitude’ of the tension force per unit area of the rope.
Tension is measured in Newtons.

12. Lab Activity: Tension in a String

13. .

14. .

15. online problems

16. Two 100-N weights are attached to a spring scale as shown
Two 100-N weights are attached to a spring scale as shown. Does the scale read zero, 100 N, or 200 N -- or some other reading?

17. The scale reads the tension in the string
The scale reads the tension in the string. The tension in the string is 100 N. This is the force the string must exert up on either of the 100-N weights at either end of the string.

18. Nothing is moving, nothing is accelerating, so the net force on the spring is zero. Likewise, the net force on either of the 100-N weights is also zero. But that is another question. The spring scale does not measure the net force. The spring scale simply measures the tension, the magnitude of the force exerted by the string.

19. Concept Question 1
Why are we able to walk?

20. Concept Question Answer
We walk forward because when one foot pushes backward against the ground, the ground pushes forward on that foot.
Force exerted
on the person’s foot by the ground. Fpg
Force exerted on the ground
by person’s foot. Fgp
Fgp=-Fpg

21. Concept Question 2
What makes a car go forward?

22. Concept Question answer
By Newton’s third law, the ground pushes on the tires in the opposite direction, accelerating the car forward.

23. Concept Question
Which is stronger, the Earth’s pull on an orbiting space shuttle or the space shuttle’s pull on the earth?

24. Concept Question Answer
According to Newton’s Third Law, the two forces are equal and opposite. Because of the huge difference in masses, however the space shuttle accelerates much more towards the Earth than the Earth accelerates toward the space shuttle.
a = F/m

25. Problem 1
What force is needed to accelerate the 60 kg cart at 2m/s^2?

26. How to solve Problem 1 Force = mass times acceleration F = m * a
What force is needed to accelerate the 60kg cart at 2 m/s^2?
Force = mass times acceleration
F = m * a
F = 60kg * 2m/s^2
F = 120kgm/s^2
Kgm/s^2 = Newton
Newton = N
F =120 N

27. Problem 2
A force of 200 N accelerates a bike and rider at 2 m/s^2. What is the mass of the bike and rider?

28. How to solve Problem 2 F = ma therefore:m =F/a m = 200N/2m/s^2
A force of 200 N accelerate a bike and rider at 2m/s^2. What is the mass of the bike and rider?
F = ma therefore:m =F/a
m = 200N/2m/s^2
N= kgm/s^2 so when divide your answer will be kg left.
m = 100kg

29. Why can you exert greater force on the pedals of a bicycle if you pull up on the handlebars?
The handlebars then pull down on you, somewhat as if someone were pushing down on your shoulders. This lets you exert a greater downward force on the pedals.