1 Class Notes 4.20.11 WPE Problems Equation Sheet Bouncing ball lab is dueThursday by 3pm.Staple all 3 packets together in orderClass NotesWPE ProblemsEquation SheetList for sports bus for TwinsRetake TEST by ThursdayHw LATE -50%Twin Towers worksheet Late -50%
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=FdCHANGES ENERGY FROM ONE FORM TO ANOTHERW = ΔPE and W = ΔKEΔPE = PEf - PEiΔKE = KEf - KEiInvolves 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 JoulesW = ΔPE & W = ΔKEΔPE = PEf - PEiΔKE = KEf - KEiConditions 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 motionCan 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 doneExample: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
14 Gravitational Potential energy Work and Energy Problems Work and energy are Scalar, so we do not use +/- on numbers for directionGravitational Potential energy(height)WorkChangesKE to PEandPE to KEKinetic energy(speed)PowerHow 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 = ΔKEPEf -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 from6. 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 J5000 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 directionPEi = mghPEi = 5000JΔPE = J5000J=50(9.8)hh=5000 /490h= 10.2 mKEf = 1/2 mv2KEf = 5000 JΔKE = 5000JW=ΔKEorW=ΔPEW = 5000 J transferring ΔPE & ΔKE
18 the potential energy change the work done the power delivered Example from7. 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 changethe work donethe power delivered
19 Work and Energy Problems Work and energy are Scalar, so we do not use +/- on numbers for direction PE = mghΔPE = +1000JMoved UPW=F*dW = 1000 J transferred to PEKE = 1/2 mv2P=W/tP=1000J / 20sec=50 Watts
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