1. Class Notes 4.20.11 WPE Problems Equation Sheet
Bouncing ball lab is due
Thursday by 3pm.
Staple all 3 packets together in order
Class Notes
WPE Problems
Equation Sheet
List for sports bus for Twins
Retake TEST by Thursday
Hw LATE -50%
Twin Towers worksheet Late -50%

2. Work
Object

3. To Do Work, Forces Must Cause Displacements
Let's consider Scenario C above in more detail. Scenario C involves a situation similar to the waiter who carried a tray full of meals above his head by one arm straight across the room at constant speed. It was mentioned earlier that the waiter does not do work upon the tray as he carries it across the room. The force supplied by the waiter on the tray is an upward force and the displacement of the tray is a horizontal displacement. As such, the angle between the force and the displacement is 90 degrees.

4. To Do Work, Forces Must Cause Displacements
the waiter does not do work upon the tray as he carries it across the room.
The force supplied by the waiter on the tray is an upward force and the displacement of the tray is a horizontal displacement. (constant speed)

5. Involves a force and movement in the direction of the force.
Work W=Fd
CHANGES ENERGY FROM ONE FORM TO ANOTHER
W = ΔPE and W = ΔKE
ΔPE = PEf - PEi
ΔKE = KEf - KEi
Involves a force and movement in the direction of the force.
EXAMPLE:
A force of 20 newtons pushing an object 5 meters in the direction of the force does 100 joules of work.

6. Work: Force up by person to gain PE
Work: Force down by gravity to change PE into KE

7. Work = Change in Energy page 6 space 4 Fill in the rest on your own!!
All Labels are Joules
W = ΔPE & W = ΔKE
ΔPE = PEf - PEi
ΔKE = KEf - KEi
Conditions for USE:
If you know____, then you can find ____ and …

8. Energy KE, PEg or PEsp Energy is the capacity for doing work.
Can be due to an objects position or motion
Can exert a force that move an object a distance or deforms an object.
You must have energy to accomplish work - it is like the "currency" for performing work.
Example:
To do 100 joules of work, you must expend 100 joules of energy

9. Mechanical work and Energy work is done upon an object whenever a force acts upon it and changes the speed or changes the height

10. Mechanical work and Energy change in Speed means Change KE or change in Height means Change PEg

11. Power P = W / time The rate of doing work or the rate of using energy,
Or we can say it is how fast work is done
Example:
If you do 100 joules of work in one second (using 100 joules of energy), the power is 100 watts

12. Variables for power How fast work is done Force, distance, time

13. POWER

14. Gravitational Potential energy
Work and Energy Problems Work and energy are Scalar, so we do not use +/- on numbers for direction
Gravitational Potential energy
(height)
Work
Changes
KE to PE
and
PE to KE
Kinetic energy
(speed)
Power
How fast work is done

15. Loses Same amount of PE as the KE gains so…
A ball starts from rest on top of a tall pillar and falls to the ground below. Assume the effect of air resistance is negligible.   
 ΔPE = ΔKE
PEf -PEi = KEf -KEi
(Since initially at rest, KEi = 0 and cancels. Since the final height is 0, PEf = 0 and cancels.)
Loses Same amount of PE as the KE gains so…
- PEi = KEf

16. Example from
6. A 50-kg platform diver hits the water below with a kinetic energy of 5000 Joules. The height (relative to the water) from which the diver dove was approximately ____ meters. *the potential energy change *the work done.
5000 J
5000 J

17. W = 5000 J transferring ΔPE & ΔKE
Work and Energy Problems Work and energy are Scalar, so we do not use +/- on numbers for direction
PEi = mgh
PEi = 5000J
ΔPE = J
5000J=50(9.8)h
h=5000 /490
h= 10.2 m
KEf = 1/2 mv2
KEf = 5000 J
ΔKE = 5000J
W=ΔKE or
W=ΔPE
W = 5000 J transferring ΔPE & ΔKE

18. the potential energy change the work done the power delivered
Example from
7. Using 1000 J of work, a small object is lifted from the ground floor to the third floor of a tall building in 20 seconds. What power was required in this task?
the potential energy change
the work done
the power delivered

19. Work and Energy Problems Work and energy are Scalar, so we do not use +/- on numbers for direction
PE = mgh
ΔPE = +1000J
Moved UP
W=F*d
W = 1000 J transferred to PE
KE = 1/2 mv2
P=W/t
P=1000J / 20sec=50 Watts