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Work and Energy Dr. Robert MacKay Clark College

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Introduction What is Energy? What are some of the different forms of energy? Energy = $$$

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Overview Work (W) Kinetic Energy (KE) Potential Energy (PE) All Are measured in Units of Joules (J) 1.0 Joule = 1.0 N m W KE PE

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Overview Work Kinetic Energy Potential Energy W KE PE Heat Loss

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Crib Sheet

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Work and Energy Work = Force x distance W = F d Actually Work = Force x Distance parallel to force d=4.0 m F= 6.0 N W= F d = 6.0 N (4.0m) = 24.0 J

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Work and Energy Work = Force x Distance parallel to force d= 8.0 m F= 10.0 N W = ?

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Work and Energy Work = Force x Distance parallel to force d= 8.0 m F= 10.0 N W = 80 J

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Work and Energy Work = Force x Distance parallel to force d= 8.0 m F= - 6.0 N W= F d = -6.0 N (8.0m) =-48 J

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Work and Energy Work = Force x Distance parallel to force d= 6.0 m F= - 5.0 N W= F d = ? J

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Work and Energy Work = Force x Distance parallel to force d= 6.0 m F= - 5.0 N W= F d = -30 J

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Work and Energy Work = Force x Distance parallel to force d= 6.0 m F= ? N W= 60 J

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Work and Energy Work = Force x Distance parallel to force d= 6.0 m F= 10 N W= 60 J

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Work and Energy Work = Force x Distance parallel to force d= ? m F= - 50.0 N W= 200 J

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Work and Energy Work = Force x Distance parallel to force d= -4.0 m F= - 50.0 N W= 200 J

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Work and Energy Work = Force x Distance parallel to force d= 8.0 m F= + 6.0 N W= 0 (since F and d are perpendicular

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Power Work = Power x time 1 Watt= 1 J/s 1 J = 1 Watt x 1 sec 1 kilowatt - hr = 1000 (J/s) 3600 s = 3,600,000 J Energy = $$$$$$ 1 kW-hr = $0.08 = 8 cents

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Power Work = Power x time W=P t [ J=(J/s) s= Watt * sec ] work = ? when 2000 watts of power are delivered for 4.0 sec.

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Power Work = Power x time W=P t [ J=(J/s) s= Watt * sec ] work = 8000J when 2000 watts of power are delivered for 4.0 sec.

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Power Energy = Power x time E =P t [ kW-hr=(kW) hr] or [ J=(J/s) s= Watt * sec ]

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Power Energy = Power x time How much energy is consumed by a 100 Watt lightbulb when left on for 24 hours? What units should we use? J,W, & s or kW-hr, kW, hr

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Power Energy = Power x time How much energy is consumed by a 100 Watt lightbulb when left on for 24 hours? What units should we use? J,W, & s or kW-hr, kW, hr Energy=0.1 kWatt (24 hrs)=2.4 kWatt-hr

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Power Energy = Power x time What is the power output of a duck who does 3000 J of work in 0.5 sec? What units should we use? J,W, & s or kW-hr, kW, hr

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Power Energy = Power x time What is the power output of a duck who does 3000 J of work in 0.5 sec? power=energy/time =3000 J/0.5 sec =6000 Watts What units should we use? J,W, & s or kW-hr, kW, hr

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Power Energy = Power x time E =P t [ kW-hr=(kW) hr] Energy = ? when 2000 watts (2 kW) of power are delivered for 6.0 hr. Cost at 8 cent per kW-hr?

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Power Energy = Power x time E =P t [ kW-hr=(kW) hr] Energy = 2kW(6 hr)=12 kW-hr when 2000 watts (2 kW) of power are delivered for 6.0 hr. Cost at 8 cent per kW-hr? 12 kW-hr*$0.08/kW-hr=$0.96

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Machines Levers D =8 m d = 1 m f=10 N F=? Work in = Work out f D = F d The important thing about a machine is although you can increase force with a machine or increase distance (or speed) with a machine you can not get more work (or power) out than you put into it.

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Machines Levers D =8 m d = 1 m f=10 N F=? Work in = Work out 10N 8m = F 1m F = 80 N The important thing about a machine is although you can increase force with a machine or increase distance (or speed) with a machine you can not get more work (or power) out than you put into it.

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Machines Pulleys D d f F Work in = Work out f D = F d The important thing about a machine is although you can increase force with a machine or increase distance (or speed) with a machine you can not get more work (or power) out than you put into it.

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Machines Pulleys D d f F Work in = Work out f D = F d D/d = 4 so F/f = 4 If F=200 N f=? The important thing about a machine is although you can increase force with a machine or increase distance (or speed) with a machine you can not get more work (or power) out than you put into it.

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Machines Pulleys D d f F Work in = Work out f D = F d D/d = 4 so F/f = 4 If F=200 N f = 200 N/ 4 = 50 N The important thing about a machine is although you can increase force with a machine or increase distance (or speed) with a machine you can not get more work (or power) out than you put into it.

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Machines Hydraulic machine D d f F Work in = Work out f D = F d if D=20 cm, d =1 cm, and F= 800 N, what is the minimum force f? The important thing about a machine is although you can increase force with a machine or increase distance (or speed) with a machine you can not get more work (or power) out than you put into it.

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Machines Hydraulic machine D d f F Work in = Work out f D = F d f 20 cm = 800 N (1 cm) f = 40 N if D=20 cm, d =1 cm, and F= 800 N, what is the minimum force f? The important thing about a machine is although you can increase force with a machine or increase distance (or speed) with a machine you can not get more work (or power) out than you put into it.

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Efficiency E in E out E loss

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Efficiency E in = 200 J E out = 150 J E loss = ?

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Efficiency E in = 200 J E out = 150 J E loss = 50J =0.75=75%

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Two Machines e1 and e2 connected to each other in series

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Two Machines e1 and e2 E out =eff (E in )=0.5(100J)=50J

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Two Machines e1 and e2

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Total efficiency when 2 machines are connected one after the other is e tot =e 1 (e 2 )

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Kinetic Energy, KE KE =1/2 m v 2 m=2.0 kg and v= 5 m/s KE= ?

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Kinetic Energy KE =1/2 m v 2 m=2.0 kg and v= 5 m/s KE= 25 J m=4.0 kg and v= 5 m/s KE= ?

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Kinetic Energy KE =1/2 m v 2 m=2.0 kg and v= 5 m/s KE= 25 J m=4.0 kg and v= 5 m/s KE= 50J

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Kinetic Energy KE =1/2 m v 2 m=2.0 kg and v= 5 m/s KE= 25 J m=2.0 kg and v= 10 m/s KE= ?

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Kinetic Energy KE =1/2 m v 2 m=2.0 kg and v= 5 m/s KE= 25 J m=2.0 kg and v= 10 m/s KE= 100J

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Double speed and KE increases by 4

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Kinetic Energy KE =1/2 m v 2 if m doubles KE doubles if v doubles KE quadruples if v triples KE increases 9x if v quadruples KE increases ____ x

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Work Energy Theorm KE =1/2 m v 2 F = m a

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Work Energy Theorm K =1/2 m v 2 F = m a F d = m a d

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Work Energy Theorm KE =1/2 m v 2 F = m a F d =m a d F d = m (v/t) [(v/2)t]

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Work Energy Theorm K E=1/2 m v 2 F = m a F d = m a d F d = m (v/t) [(v/2)t] W = 1/2 m v 2

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Work Energy Theorm KE =1/2 m v 2 F = m a F d = m a d F d = m (v/t) [(v/2)t] W = 1/2 m v 2 W = KE

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Work Energy Work Energy W = KE How much work is required to stop a 2000 kg car traveling at 20 m/s (45 mph)?

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Work Energy Work Energy W = KE How much work is required to stop a 2000 kg car traveling at 20 m/s (45 mph)? W= KE =-1/2 m v 2 =-1/2(2000 kg)(20 m/s) 2 = - 1000kg (400 m 2 /s 2 ) = - 400,000 Joules

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Work Energy Work Energy W = KE How much work is required to stop a 2000 kg car traveling at 20 m/s? If the friction force equals its weight, how far will it skid? W= K = - 400,000 Joules F=weight=mg=-20,000 N W=F d d=W/F=-400,000 J/-20,000N = 20.0 m

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Work Energy Work Energy W = KE v = 20 m/s d=? m v = 10 m/s d= 15 m Same Friction Force

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Work Energy Work Energy W = KE v = 20 m/s d=60m (4 times 15m) v = 10 m/s d= 15 m Same Friction Force

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Potential Energy, PE Gravitational Potential EnergyGravitational Potential Energy SpringsSprings ChemicalChemical PressurePressure Mass (Nuclear)Mass (Nuclear) Measured in JoulesMeasured in Joules

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Potential Energy, PE Gravitational Potential EnergyGravitational Potential Energy SpringsSprings ChemicalChemical PressurePressure Mass (Nuclear)Mass (Nuclear) The energy required to put something in its place (state)

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Potential Energy Gravitational Potential Energy = weight x height PE=(mg) h 4.0 m m = 2.0 kg

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Potential Energy PE=(mg) h 4.0 m m = 2.0 kg K=? PE=80 J

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Potential Energy to Kinetic Energy PE=(mg) h 2.0 m m = 2.0 kg KE=? PE=40 J 1.0 m K E= 0 J

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Conservation of Energy Total Mechanical Energy, E = PE +K Energy can neither be created nor destroyed only transformed from one form to another In the absence of friction or other non-conservative forces the total mechanical energy of a system does not change E f =E o

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Conservation of Energy 10.0 m m = 1.02 kg (mg = 10.0 N) K = 0 JPE=100 J PE = 75 J PE = 50 J PE = 0 J PE= 25 J K = ? K = 50 J K = 25 J Constant E {E = K + PE} E f = E o No friction No Air resistance

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Conservation of Energy 5.0 m m = 2.0 kg K=0 J PE=100 J PE = 0 J K = ? Constant E {E = K + U} Constant E {E = K + PE} E f =E o No friction

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Conservation of Energy 5.0 m m = 2.0 kg K = 0 J PE =100 J v = ? K = 100 J Constant E {E = K + U} Constant E {E = K + PE} E f =E o No friction

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