1. Applications of Newton’s Laws
Chapter 6
Applications of Newton’s Laws

2. Midterm # 1
Average = Standard Deviation = 3.10 Well done!

3. Who Wins?
Kevin Parks (5’6”, 90kg) takes a handoff from Michael Rocco, but the hole closes as he hits the line. He is pinned between to O-linemen behind him, and 3 D-linemen ahead... and everyone is pushing, lifting, grabbing, and kicking. His acceleration is 1 m/s2 upfield.
What is the net force on KP?
a) I need to know a lot more about angles, masses of the line men, etc.
b) I could answer this if only I had the initial velocity
c) I would also need to know what planet this happened on.
d) 90 N, upfield.
e) This isn’t the Tech game, is it?

4. Who Wins?
Kevin Parks (5’6”, 90kg) takes a handoff from Michael Rocco, but the hole closes as he hits the line. He is pinned between to O-linemen behind him, and 3 D-linemen ahead... and everyone is pushing, lifting, grabbing, and kicking. His acceleration is 1 m/s2 upfield.
What is the net force on KP?
a) I need to know a lot more about angles, masses of the line men, etc.
b) I could answer this if only I had the initial velocity
c) I would also need to know what planet this happened on.
d) 90 N, upfield.
e) This isn’t the Tech game, is it?
If you know the acceleration of a body, you know the direction of the net force. If you also know the mass, then you know the magnitude.

5. Will It Budge?
a) moves to the left, because the force of static friction is larger than the applied force
b) moves to the right, because the applied force is larger than the static friction force
c) the box does not move, because the static friction force is larger than the applied force
d) the box does not move, because the static friction force is exactly equal the applied force
e) The answer depends on the value for μk.
A box of weight 100 N is at rest on a floor where μs = A rope is attached to the box and pulled horizontally with tension T = 30 N. Which way does the box move? T m
Static friction (s = 0.4 )

6. Will It Budge?
a) moves to the left, because the force of static friction is larger than the applied force
b) moves to the right, because the applied force is larger than the static friction force
c) the box does not move, because the static friction force is larger than the applied force
d) the box does not move, because the static friction force is exactly equal the applied force
e) The answer depends on the value for μk.
A box of weight 100 N is at rest on a floor where μs = A rope is attached to the box and pulled horizontally with tension T = 30 N. Which way does the box move?
The static friction force has a maximum of sN = 40 N. The tension in the rope is only 30 N. So the pulling force is not big enough to overcome friction. T m
Static friction (s = 0.4 )
Follow-up: What happens if the tension is 35 N? What about 45 N?

7. Springs
Hooke’s law for springs states that the force increases with the amount the spring is stretched or compressed:
The constant k is called the spring constant.

8. Springs
Note: we are discussing the force of the spring on the mass. The force of the spring on the wall are equal, and opposite.

9. Springs and Tension a) (L1 + L2) / 2 b) L1 or L2, whichever is smaller
A mass M hangs on spring 1, stretching it length L1
Mass M hangs on spring 2, stretching it length L2
Now spring1 and spring2 are connected end-to-end, and M1 is hung below. How far does the combined spring stretch?
a) (L1 + L2) / 2
b) L1 or L2, whichever is smaller
c) L1 or L2, whichever is bigger
d) depends on which order the springs are attached
e) L1 + L2

10. Springs and Tension a) (L1 + L2) / 2 b) L1 or L2, whichever is smaller
A mass M hangs on spring 1, stretching it length L1
Mass M hangs on spring 2, stretching it length L2
Now spring1 and spring2 are connected end-to-end, and M1 is hung below. How far does the combined spring stretch?
a) (L1 + L2) / 2
b) L1 or L2, whichever is smaller
c) L1 or L2, whichever is bigger
d) depends on which order the springs are attached
e) L1 + L2 W
Fs=T

11. Springs and Tension a) (L1 + L2) / 2 b) L1 or L2, whichever is smaller
A mass M hangs on spring 1, stretching it length L1
Mass M hangs on spring 2, stretching it length L2
Now spring1 and spring2 are connected end-to-end, and M1 is hung below. How far does the combined spring stretch?
a) (L1 + L2) / 2
b) L1 or L2, whichever is smaller
c) L1 or L2, whichever is bigger
d) depends on which order the springs are attached
e) L1 + L2 W
Fs=T
Spring 1 supports the weight.
Spring 2 supports the weight.
Both feel the same force, and
stretch the same distance as before.

12. Recall -- Instantaneous acceleration
Velocity vector is always in the direction of motion; acceleration vector can points in the direction velocity is changing:

13. Circular Motion and Centripetal Force
An object moving in a circle must have a force acting on it; otherwise it would move in a straight line!
The net force must have a component pointing to the center of the circle
centripetal force
This force may be provided by the tension in a string, the normal force, or friction, or....

14. Circular Motion and Centripetal Acceleration
An object moving in a circle must have a force acting on it; otherwise it would move in a straight line.
If the speed is constant, the direction of the acceleration (which is due to the net force) is towards the center of the circle.
The magnitude of this centripetal component of the force is given by:
For circular motion problems, it is often convenient to choose coordinate axes with one pointing along the direction of this centripetal force a

15. Circular Motion
An object may be changing its speed as it moves in a circle; in that case, there is a tangential acceleration as well:

16. Missing Link a b c d e
A Ping-Pong ball is shot into a circular tube that is lying flat (horizontal) on a tabletop. When the Ping-Pong ball leaves the track, which path will it follow?

17. Missing Link a b c d e
A Ping-Pong ball is shot into a circular tube that is lying flat (horizontal) on a tabletop. When the Ping-Pong ball leaves the track, which path will it follow?
Once the ball leaves the tube, there is no longer a force to keep it going in a circle. Therefore, it simply continues in a straight line, as Newton’s First Law requires!
Follow-up: What physical force provides the centripetal force?

18. Tetherball
In the game of tetherball, the struck ball whirls around a pole. In what direction does the net force on the ball point?
a) toward the top of the pole
b) toward the ground
c) along the horizontal component of the tension force
d) along the vertical component of the tension force
e) tangential to the circle
Answer: c W T

19. Tetherball
In the game of tetherball, the struck ball whirls around a pole. In what direction does the net force on the ball point?
a) toward the top of the pole
b) toward the ground
c) along the horizontal component of the tension force
d) along the vertical component of the tension force
e) tangential to the circle W T
The vertical component of the tension balances the weight. The horizontal component of tension provides the necessary centripetal force that accelerates the ball to towards the pole and keeps it moving in a circle. W T

21. Examples of centripetal force
no friction is needed to hold the track!
when

22. A hockey puck of mass m is attached to a string that passes through a hole in the center of a table, as shown in the figure. The hockey puck moves in a circle of radius r. Tied to the other end of the string, and hanging vertically beneath the table, is a mass M. Assuming the tabletop is perfectly smooth, what speed must the hockey puck have if the mass M is to remain at rest? 22

23. necessary centripetal force:
A hockey puck of mass m is attached to a string that passes through a hole in the center of a table, as shown in the figure. The hockey puck moves in a circle of radius r. Tied to the other end of the string, and hanging vertically beneath the table, is a mass M. Assuming the tabletop is perfectly smooth, what speed must the hockey puck have if the mass M is to remain at rest?
necessary centripetal force:
Only force on puck is tension in the string!
To support mass M, the necessary tension is:

24. Circular motion and apparent weight
Car at the bottom of a dip
Note: at any specific point, any curve can be approximated by a portion of a circle
This normal force is the apparent, or perceived, weight

25. Going in Circles I
You’re on a Ferris wheel moving in a vertical circle. When the Ferris wheel is at rest, the normal force N exerted by your seat is equal to your weight mg. How does N change at the top of the Ferris wheel when you are in motion?
a) N remains equal to mg
b) N is smaller than mg
c) N is larger than mg
d) none of the above

26. Going in Circles I
You’re on a Ferris wheel moving in a vertical circle. When the Ferris wheel is at rest, the normal force N exerted by your seat is equal to your weight mg. How does N change at the top of the Ferris wheel when you are in motion?
a) N remains equal to mg
b) N is smaller than mg
c) N is larger than mg
d) none of the above
You are in circular motion, so there has to be a centripetal force pointing inward. At the top, the only two forces are mg (down) and N (up), so N must be smaller than mg.
Follow-up: Where is N larger than mg? 26

27. Vertical circular motion
Centripetal acceleration must be
vertical (up) C A B
horizontal
vertical (down)
Condition for falling: N=0
at C:
(note: before falling, apparent weight is in the opposite direction to true weight!)
So, as long as:
at the top, then N>0 and pointing down. 27

28. Work and Kinetic Energy
Chapter 7
Work and Kinetic Energy

29. Work Done by a Constant Force
The definition of work, when the force is parallel to the displacement:
SI unit: newton-meter (N·m) = joule, J 29

30. Working Hard... or Hardly Working
Atlas holds up the world.
Sisyphus pushes his rock up a hill.
Answer: b
(b)
Who does more work?
(a)

31. Working Hard... or Hardly Working
Atlas holds up the world.
Sisyphus pushes his rock up a hill.
Answer: b
(b)
With no displacement, Atlas does no work
(a)

32. Forces not along displacement
If the force is at an angle to the displacement: 32

33. Friction and Work I
A box is being pulled across a rough floor at a constant speed. What can you say about the work done by friction?
a) friction does no work at all
b) friction does negative work
c) friction does positive work
Answer: b

34. Friction and Work I
A box is being pulled across a rough floor at a constant speed. What can you say about the work done by friction?
a) friction does no work at all
b) friction does negative work
c) friction does positive work f N mg
Displacement
Pull
Friction acts in the opposite direction to the displacement, so the work is negative. Or using the definition of work (W = F (Δr)cos ), because  = 180º, then W < 0.

35. Friction and Work II Can friction ever do positive work? a) yes b) no
Answer: a

36. Friction and Work II
Can friction ever do positive work?
a) yes
b) no
Consider the case of a box on the back of a pickup truck. If the box moves along with the truck, then it is actually the force of friction that is making the box move. ... but .... the friction isn’t really doing the work, it is only transmitting the forces to the box, while work is done by the truck engine.
My view: the language is confusing so I’m not interested in arguing the point.

37. Convenient notation: the dot product
The work can also be written as the dot product of the force and the displacement:
vector “dot” operation: project one vector onto the other

38. Force and displacement
The work done may be positive, zero, or negative, depending on the angle between the force and the displacement: 38

39. Sum of work by forces = work by sum of forces
If there is more than one force acting on an object, the work done by each force is the same as the work done by the net force:

40. Units of Work
Lifting 0.5 L H2O up 20 cm = 1 J

41. Play Ball!
In a baseball game, the catcher stops a 90-mph pitch. What can you say about the work done by the catcher on the ball?
a) catcher has done positive work
b) catcher has done negative work
c) catcher has done zero work
Answer: b

42. Play Ball!
In a baseball game, the catcher stops a 90-mph pitch. What can you say about the work done by the catcher on the ball?
a) catcher has done positive work
b) catcher has done negative work
c) catcher has done zero work
The force exerted by the catcher is opposite in direction to the displacement of the ball, so the work is negative. Or using the definition of work (W = F (Δr)cos  ), because  = 180º, then W < Note that the work done on the ball is negative, and its speed decreases.
Follow-up: What about the work done by the ball on the catcher?

43. Tension and Work
A ball tied to a string is being whirled around in a circle with constant speed. What can you say about the work done by tension?
a) tension does no work at all
b) tension does negative work
c) tension does positive work

44. Tension and Work
A ball tied to a string is being whirled around in a circle with constant speed. What can you say about the work done by tension?
a) tension does no work at all
b) tension does negative work
c) tension does positive work v T
No work is done because the force acts in a perpendicular direction to the displacement. Or using the definition of work (W = F (Δr)cos  ),  = 90º, then W = 0.
Follow-up: Is there a force in the direction of the velocity?

45. Work by gravity θ W = Fd = Fgx h W = mgh
A ball of mass m drops a distance h. What is the total work done on the ball by gravity?
W = Fd = Fgx h
W = mgh
Path doesn’t matter when asking “how much work did gravity do?” Only the change in height!
A ball of mass m rolls down a ramp of height h at an
angle of 45o. What is the total work done on the ball by gravity? h a θ
Fgx = Fg sinθ
h = L sinθ N Fg
W = Fd = Fgx L = (Fg sinθ) (h / sinθ)
W = Fg h = mgh

46. Motion and energy
When positive work is done on an object, its speed increases; when negative work is done, its speed decreases.

47. Kinetic Energy
As a useful word for the quantity of work we have done on an object, thereby giving it motion, we define the kinetic energy:

48. Work-Energy Theorem
Work-Energy Theorem: The total work done on an object is equal to its change in kinetic energy.
(True for rigid bodies that remain intact)

49. Lifting a Book FHAND r v = const a = 0 mg
You lift a book with your hand in such a way that it moves up at constant speed. While it is moving, what is the total work done on the book?
a) mg × r
b) FHAND × r
c) (FHAND + mg) × r
d) zero
e) none of the above mg r
FHAND
v = const
a = 0

50. Lifting a Book FHAND r v = const a = 0 mg
You lift a book with your hand in such a way that it moves up at constant speed. While it is moving, what is the total work done on the book?
a) mg × r
b) FHAND × r
c) (FHAND + mg) × r
d) zero
e) none of the above
The total work is zero because the net force acting on the book is zero. The work done by the hand is positive, and the work done by gravity is negative. The sum of the two is zero. Note that the kinetic energy of the book does not change either! mg r
FHAND
v = const
a = 0
Follow-up: What would happen if FHAND were greater than mg?

51. Kinetic Energy I
By what factor does the kinetic energy of a car change when its speed is tripled?
a) no change at all
b) factor of 3
c) factor of 6
d) factor of 9
e) factor of 12
Answer: d

52. Kinetic Energy I
By what factor does the kinetic energy of a car change when its speed is tripled?
a) no change at all
b) factor of 3
c) factor of 6
d) factor of 9
e) factor of 12
Because the kinetic energy is mv2, if the speed increases by a factor of 3, then the KE will increase by a factor of 9.

53. Slowing Down
If a car traveling 60 km/hr can brake to a stop within 20 m, what is its stopping distance if it is traveling 120 km/hr? Assume that the braking force is the same in both cases.
a) 20 m
b) 30 m
c) 40 m
d) 60 m
e) 80 m
Answer: e 53

54. Slowing Down
If a car traveling 60 km/hr can brake to a stop within 20 m, what is its stopping distance if it is traveling 120 km/hr? Assume that the braking force is the same in both cases.
a) 20 m
b) 30 m
c) 40 m
d) 60 m
e) 80 m
F d = Wnet = KE = 0 – mv2,
and thus, |F| d = mv2.
Therefore, if the speed doubles,
the stopping distance gets four times larger. 54

55. Work Done by a Variable Force
We can interpret the work done graphically:

56. Work Done by a Variable Force
If the force takes on several successive constant values:

57. Work Done by a Variable Force
We can then approximate a continuously varying force by a succession of constant values.

58. Work Done by a Variable Force
The force needed to stretch a spring an amount x is F = kx.
Therefore, the work done in stretching the spring is