Presentation is loading. Please wait.

Presentation is loading. Please wait.

Class Notes 4.20.11 WPE Problems Equation Sheet Retake TEST by Thursday 4.21.11 Hw LATE -50% Twin.

Similar presentations


Presentation on theme: "Class Notes 4.20.11 WPE Problems Equation Sheet Retake TEST by Thursday 4.21.11 Hw LATE -50% Twin."— Presentation transcript:

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

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.waiter who carried a tray full of meals above his head by one arm straight across the room at constant speed To Do Work, Forces Must Cause Displacements

4 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) To Do Work, Forces Must Cause Displacements

5 Work W=Fd CHANGES ENERGY FROM ONE FORM TO ANOTHER W = ΔPE and W = ΔKE ΔPE = PE f - PE i ΔKE = KE f - KE i 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 = PE f - PE i ΔKE = KE f - KE i Conditions for USE: If you know____, then you can find ____ and …

8 Energy KE, PE g or PE sp 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 PE g

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 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 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 PE f -PE i = KE f -KE i (Since initially at rest, KE i = 0 and cancels. Since the final height is 0, PE f = 0 and cancels.) ΔPE = ΔKE Loses Same amount of PE as the KE gains so… - PE i = KE f

16 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 J Example from

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

18 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 mv 2 P=W/t P=1000J / 20sec=50 Watts


Download ppt "Class Notes 4.20.11 WPE Problems Equation Sheet Retake TEST by Thursday 4.21.11 Hw LATE -50% Twin."

Similar presentations


Ads by Google