1. Chapter 5 Work and Energy

2. Definition of Work
There is a difference between the ordinary definition of work and the scientific definition of work
Ordinary Definition: To do something that takes physical or mental effort
Scientific definition: Work is equal to the magnitude of the applied force times the displacement of an object

3. Is it Scientific Work?
A student holds a chair at arm’s length for several minutes.
Is a force being applied? Yes
Does the object that the force is being applied to have a displacement? NO
Therefore, the student hasn’t done work in the scientific sense.

4. What is necessary for work to be done?
A force that causes displacement of an object does work on the object
Therefore, work is not done on an object unless the object is moved because of the action of a force
Work is being done!!!

5. What is necessary for work to be done?
Work is done only when components of a force are parallel to a displacement
When the force on an object and the object’s displacement are perpendicular, no work is done
No Work was Done!!!
Displacement
Force

6. What about forces at an angle?
Only the component of force that is parallel to the direction of the object’s displacement does work.
Example: A person pushes a box across a frictionless floor

7. FBD For the Box
What part of the applied force is parallel to the displacement? FN
Displacement of Box
Fapp,x
Fapp,y
Fapp Fg

8. So what work was done?
W= Fapp,xd
Fapp,x = Fcosθ
Therefore, W=Fdcosθ

9. General Equation For Work Done
Work= Component of Force that does the work x displacement x cos (angle between the force vector and the displacement)

10. Net Work done by a Constant Force
If there are many constant forces acting on the object, you can find the net work done by finding the net force acting on the object
Net work= Net force x displacement x cos of the angle between them

11. Angles between vectors
Vector Orientation
Angle between them
Cos(θ)
Θ= 90
Cos(90) =0
Θ= 0
Cos(0) = 1
Θ= 180
Cos(180) = -1

12. Units for Work The unit for Work is the Joule I J= 1 Nm
One Joule = One Newton x One meter

13. The sign of work is important
Work is a scalar quantity, but it can be positive or negative
Work is negative when the force is in the direction opposite the displacement
For example, the work done by the frictional force is always negative because the frictional force is opposite the displacement

14. Sample Problem p. 193 # 10
A flight attendant pulls her 70 N flight bag a distance of 253 m along a level airport floor at a constant velocity. The force she exerts is 40.0 N at an angle of 52.0° above the horizontal. Find the following:
The work she does on the flight bag
The work done by the force of friction on the flight bag
The coefficient of kinetic friction between the flight bag and the floor

15. FBD
Fapp FN
Fapp,y Ff
Fapp,x Fg

16. Work done by flight attendant
Only the component of Fapp that is parallel to the displacement does work.
Fapp,x is parallel to the displacement
Fapp,x = 40cos52 = N
Remember that in the W=Fdcosθ equation, θ represents the angle between the force vector and the displacement vector
W= (24.63 N)(253 m) cos(0)= 6230 J
Fapp,x d
Θ = 0°

17. Work done by friction
The bag is moving at constant velocity, so what is Ff?
Ff = Fapp,x= 40cos(52)= N
W= Fdcosθ= (24.62N)(253m)(cos180)
= J Ff d
Θ =180°

18. Find μk What is FN? Fn + Fapp,y = Fg
Fn = Fg - Fapp,y= 70N- 40sin(52)= N

19. Energy- Section 5.2 p. 172
Kinetic Energy- The energy of an object due to its motion
I Have Kinetic
Energy
I don’t Have Kinetic
Energy

20. Kinetic Energy Depends on Speed and Mass
Kinetic energy = ½ x mass x speed2
The unit for KE is Joules (J)

21. Sample Problem p. 173
A 7.00 kg bowling ball moves at 3.0 m/s. How much kinetic energy does the bowling ball have? How fast must a 2.45 g tennis ball move in order to have the same kinetic energy as the bowling ball?

22. Solve the Problem KE= ½ mv^2= ½ (7kg)(3m/s)^2= 31.5 J
How fast must 2.45 g ball move to have the same KE?
Convert g to kg  2.45 g = kg
Solve for v

23. Work-Kinetic Energy Theorem
The net work done by a net force acting on an object is equal to the change in kinetic energy of the object
You must include all the forces acting on the object for this to work!

24. Sample problem p. 176 #2
A 2000 kg car accelerates from rest under the actions of two forces. One is a forward force of 1140 N provided by the traction between the wheels and the road. The other is a 950 N resistive force due to various frictional forces. Use the work-KE theorem to determine how far the car must travel for its speed to reach 2.0 m/s.

25. What information do we have?
M= 2000 kg
Vi= 0m/s
Vf= 2 m/s
Ffriction= 950 N
Fforward= 1140 N

26. What is the Work-Ke Theorem?
Remember that:
So Expand the equation to this:

27. Solve for Fnet
Fnet= Fforward- Ffriction= 190 N forward
Rearrange the equation for d and plug in values
Why is θ= 0 in the denominator? Because the net force is in the same direction as the displacement.

28. Potential Energy  Section 5.2
Potential Energy is stored energy
There are two types of PE
Gravitational PE
Elastic PE

29. Gravitational Potential Energy
Gravitational Potential Energy (PEg) is the energy associated with an object’s position relative to the Earth or some other gravitational source

30. Elastic Potential Energy
Elastic Potential Energy (PEelastic) is the potential energy in a stretched or compressed elastic object.
k= spring (force) constant
X= displacement of spring

31. Displacement of Spring

32. Sample Problem p. 180 #2
The staples inside a stapler are kept in place by a spring with a relaxed length of m. If the spring constant is 51.0 N/m, how much elastic potential energy is stored in the spring when its length is m?

33. What do we know?
K = 51.0 N/m
We need to get x, in order to use the equation for elastic PE
X is the distance the spring is stretched or compressed
Relaxed length is m, stretched length is How much was it stretched?
m= m= x

34. Solve the problem

35. Conservation of Energy – 5.3
The total amount of energy in the universe is a constant
So we say that energy is conserved
From IPC: The Law of Conservation of Energy: Energy can neither be created nor destroyed

36. Mechanical Energy
There are many types of energy (KE, PE, Thermal, etc)
We are concerned with Mechanical Energy
Mechanical Energy is the sum of kinetic energy and all forms of potential energy
ME= KE + PE

37. Conservation of ME
In the absence of friction, mechanical energy is conserved
When friction is present, ME can be converted to other forms of energy (i.e. thermal energy) so it is not conserved.

38. Expanded Form of Conservation of ME
Without elastic PE
With elastic PE

39. Practice Problem p. 185 #2
A 755 N diver drops from a board 10.0 m above the water’s surface. Find the diver’s speed 5.00 m above the water’s surface. Find the diver’s speed just before striking the water.

40. What do we know? W= 755 N Initial height = 10 m Vi= 0 m/s
There is no elastic PE involved.

41. Solve part a.
What is the diver’s speed 5.0 m above the water’s surface?
M= Weight/g=76.96 kg
Vi= 0m/s
Initial height = 10 m
Final Height = 5 m

42. Rearrange equation and solve for vf
Vi = 0 m/s

43. Second Part What is the diver’s speed just before striking the water?
M= Weight/g=76.96 kg
Vi= 0m/s
Initial height = 10 m
Final Height = 0 m

44. Finish the Problem
Vi = 0 m/s
hf = 0

45. Power- Section 5.4 Power:The rate at which work is done
The unit for power is Watts
1 Watt = 1 J
1 s

46. Alternate Form for Power

47. Sample Problem (Not in book)
At what rate is a 60 kg boy using energy when he runs up a flight of stairs 10 m high in 8.0 s?
Time = 8 s
What is work done?
W=Fdcos(θ)

48. Solve the Problem
What force does the boy apply to get himself up the stairs?
F= Weight= mg= N
d= 10m
W= Fdcos(θ)=588.6N(10m)(cos(0))
W=5886 J
P=W/t = 5886 J/ 8s= Watts