Work and Energy Dr. Robert MacKay Clark College. Introduction What is Energy? What are some of the different forms of energy? Energy = $$$

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Presentation transcript:

Work and Energy Dr. Robert MacKay Clark College

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

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

Overview Work Kinetic Energy Potential Energy W KE PE Heat Loss

Crib Sheet

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

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

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

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

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

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

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

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

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

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

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

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

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.

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.

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

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

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

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

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

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?

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

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.

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.

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.

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.

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.

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.

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.

Efficiency E in E out E loss

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

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

Two Machines e1 and e2 connected to each other in series

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

Two Machines e1 and e2

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

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

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= ?

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

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= ?

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

Double speed and KE increases by 4

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

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

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

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]

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

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

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

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 = kg (400 m 2 /s 2 ) = - 400,000 Joules

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

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

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

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

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)

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

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

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

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

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

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

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