work and energy dr. robert mackay clark college. introduction what is energy? what are some of the...
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Work and EnergyWork and Energy
Dr. Robert MacKay
Clark College
Introduction Introduction
What is Energy? What are some of the different forms of
energy? Energy = $$$
Overview Overview Work (W) Kinetic Energy (KE)
Potential Energy (PE) All Are measured in Units of Joules (J) 1.0 Joule = 1.0 N m
W KE
PE
Overview Overview
Work Kinetic Energy Potential Energy
W KE
PE Heat LossHeat Loss
Heat Loss
Crib SheetCrib Sheet
W FD// (JOULES)
KE 12mv2
GPE mgh
SPE 1
2kx 2 F kx
P W
t(J / sWatt )
Wnet KE
E KE PE
E f E0 WNC
Work and EnergyWork and Energy
Work = Force x distance W = F d Actually Work = Force x Distance parallel to force
d=4.0 m
F= 6.0 N
W= F d = 6.0 N (4.0m) = 24.0 J
Work and EnergyWork and Energy
Work = Force x Distance parallel to force
d= 8.0 m
F= 10.0 N
W = ?
Work and EnergyWork and Energy
Work = Force x Distance parallel to force
d= 8.0 m
F= 10.0 N
W = 80 J
Work and EnergyWork and Energy
Work = Force x Distance parallel to force
d= 8.0 m
F= - 6.0 N
W= F d = -6.0 N (8.0m) =-48 J
Work and EnergyWork and Energy
Work = Force x Distance parallel to force
d= 6.0 m
F= - 5.0 N
W= F d = ? J
Work and EnergyWork and Energy
Work = Force x Distance parallel to force
d= 6.0 m
F= - 5.0 N
W= F d = -30 J
Work and EnergyWork and Energy
Work = Force x Distance parallel to force
d= 6.0 m
F= ? N
W= 60 J
Work and EnergyWork and Energy
Work = Force x Distance parallel to force
d= 6.0 m
F= 10 N
W= 60 J
Work and EnergyWork and Energy
Work = Force x Distance parallel to force
d= ? m
F= - 50.0 N
W= 200 J
Work and EnergyWork and Energy
Work = Force x Distance parallel to force
d= -4.0 m
F= - 50.0 N
W= 200 J
Work and EnergyWork and Energy
Work = Force x Distance parallel to force
d= 8.0 m
F= + 6.0 N
W= 0(since F and d are perpendicular
PowerPower
Work = Power x time 1 Watt= 1 J/s 1 J = 1 Watt x 1 sec 1 kilowatt - hr = 1000 (J/s) 3600 s = 3,600,000 J Energy = $$$$$$ 1 kW-hr = $0.08 = 8 cents
Power Worktime
J / s
PowerPower 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.
PowerPower 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.
PowerPower Energy = Power x time E =P t [ kW-hr=(kW) hr] or [ J=(J/s) s= Watt * sec ]
PowerPower 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, & sor kW-hr, kW, hr
PowerPower 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, & sor kW-hr, kW, hr
Energy=0.1 kWatt (24 hrs)=2.4 kWatt-hr
PowerPower 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, & sor kW-hr, kW, hr
PowerPower 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, & sor kW-hr, kW, hr
PowerPower 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?
PowerPower 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
MachinesMachines
Levers D =8 md = 1 m
f=10 NF=?
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.
MachinesMachines
Levers D =8 md = 1 m
f=10 NF=?
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.
MachinesMachines
Pulleys
Dd
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.
MachinesMachines
Pulleys
Dd
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.
MachinesMachines
Pulleys
Dd
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.
MachinesMachines
Hydraulic machine
Dd
fF
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.
MachinesMachines
Hydraulic machine
Dd
fF
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.
EfficiencyEfficiency
Effeciency Energyout
Energyin
Ein
Eout
Eloss
EfficiencyEfficiency
?in
out
Energy
EnergyEfficiency
Ein = 200 J
Eout= 150 J
Eloss= ?
EfficiencyEfficiency
?in
out
Energy
EnergyEfficiency
Ein = 200 J
Eout= 150 J
Eloss= 50J
=0.75=75%
Two Machines e1 and e2 Two Machines e1 and e2 connected to each other in seriesconnected to each other in series
Two Machines e1 and e2Two Machines e1 and e2
Eout=eff (Ein)=0.5(100J)=50J
Two Machines e1 and e2Two Machines e1 and e2
Two Machines e1 and e2Two Machines e1 and e2
Total efficiency when 2 machines are connected one after the other is etot=e1 (e2)
Kinetic Energy, KEKinetic Energy, KE
KE =1/2 m v2
m=2.0 kg and v= 5 m/sKE= ?
Kinetic EnergyKinetic Energy
KE =1/2 m v2m=2.0 kg and v= 5 m/sKE= 25 J
m=4.0 kg and v= 5 m/sKE= ?
Kinetic EnergyKinetic Energy
KE =1/2 m v2m=2.0 kg and v= 5 m/sKE= 25 J
m=4.0 kg and v= 5 m/sKE= 50J
Kinetic EnergyKinetic Energy
KE =1/2 m v2m=2.0 kg and v= 5 m/sKE= 25 J
m=2.0 kg and v= 10 m/sKE= ?
Kinetic EnergyKinetic Energy
KE =1/2 m v2m=2.0 kg and v= 5 m/sKE= 25 J
m=2.0 kg and v= 10 m/sKE= 100J
22
2
mv2
142vm
2
1KE
2v vif
mv2
1KE
Double speed and KE increases by 4
Kinetic EnergyKinetic Energy
KE =1/2 m v2
if m doubles KE doubles if v doubles KE quadruples if v triples KE increases 9x if v quadruples KE increases ____ x
Work Energy TheormWork Energy Theorm
KE =1/2 m v2
F = m a
Work Energy TheormWork Energy Theorm
K =1/2 m v2
F = m a F d = m a d
Work Energy TheormWork Energy Theorm
KE =1/2 m v2
F = m a F d =m a d F d = m (v/t) [(v/2)t]
Work Energy TheormWork Energy Theorm
K E=1/2 m v2
F = m a F d = m a d F d = m (v/t) [(v/2)t] W = 1/2 m v2
Work Energy TheormWork Energy Theorm
KE =1/2 m v2
F = m a F d = m a d F d = m (v/t) [(v/2)t] W = 1/2 m v2
W = ∆ KE
Work EnergyWork Energy W = ∆KE
How much work is required to stop a 2000 kg car traveling at 20 m/s (45 mph)?
Work EnergyWork 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 v2
=-1/2(2000 kg)(20 m/s)2
= - 1000kg (400 m 2 /s 2) = - 400,000 Joules
Work EnergyWork 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 EnergyWork Energy W = ∆KE v = 20 m/s
d=? m
v = 10 m/s
d= 15 m
Same Friction Force
Work EnergyWork Energy W = ∆KE v = 20 m/s
d=60m(4 times 15m)
v = 10 m/s
d= 15 m
Same Friction Force
Potential Energy, PEPotential Energy, PE
• Gravitational Potential Energy Gravitational Potential Energy • SpringsSprings• ChemicalChemical• PressurePressure• Mass (Nuclear)Mass (Nuclear)
• Measured in JoulesMeasured in Joules
Potential Energy, PEPotential Energy, PE
• Gravitational Potential Energy Gravitational Potential Energy • SpringsSprings• ChemicalChemical• PressurePressure• Mass (Nuclear)Mass (Nuclear)
The energy required to put something in its place (state)
Potential EnergyPotential Energy
Gravitational Potential Energy = weight x height
PE=(mg) h
4.0 m
m = 2.0 kg
Potential EnergyPotential EnergyPE=(mg) h
4.0 m
m = 2.0 kg
K=?
PE=80 J
Potential Energy to Kinetic EnergyPotential Energy to Kinetic EnergyPE=(mg) h
2.0 m
m = 2.0 kg
KE=?
PE=40 J
1.0 m
K E= 0 J
Conservation of EnergyConservation 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=Eo
Conservation of EnergyConservation 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= ?
K = 50 J
K = 25 J Constant E{E = K + PE}
Ef = Eo
No frictionNo Air resistance
Conservation of EnergyConservation of Energy
5.0 m
m = 2.0 kgK=0 J
PE=100 J
PE = 0 J
K = ?
Constant E{E = K + U}Constant E{E = K + PE}Ef=Eo
No friction
Conservation of EnergyConservation of Energy
5.0 m
m = 2.0 kgK = 0 J
PE =100 J
v = ?
K = 100 J
Constant E{E = K + U}Constant E{E = K + PE}Ef=Eo
No friction
