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**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%

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Work Object

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**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.

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**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)

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**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.

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**Work: Force up by person to gain PE **

Work: Force down by gravity to change PE into KE

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**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 …

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**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

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

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

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**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

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

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POWER

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**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

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**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

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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

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**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

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**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

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**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

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