Transcript Work Energy - Dr. Robert MacKay

Work and Energy
Dr. Robert MacKay
Clark College
Introduction
What is Energy?
 What are some of the different forms of
energy?
 Energy = $$$

Overview
Work (W)
Kinetic Energy (KE)
Potential Energy (PE)
 All Are measured in Units of Joules (J)
 1.0 Joule = 1.0 N m

W
KE
PE
Overview

Work Kinetic Energy
Potential Energy
Heat Loss
W
Heat Loss
KE
PE
Heat Loss
W  FD//
Crib Sheet
(JOULES)
1 2
KE  m v
2
GPE  m gh
1 2
SPE  kx
F  kx
2
W
P
(J / s  Watt)
t
Wne t  KE
E  KE  PE
E f  E 0  W NC
Work and Energy
Work = Force x distance
W=Fd
 Actually
 Work = Force x Distance parallel to force

d=4.0 m
F= 6.0 N
W= F d
= 6.0 N (4.0m)
= 24.0 J
Work and Energy

Work = Force x Distance parallel to force
d= 8.0 m
F= 10.0 N
W=?
Work and Energy

Work = Force x Distance parallel to force
d= 8.0 m
F= 10.0 N
W = 80 J
Work and Energy

Work = Force x Distance parallel to force
d= 8.0 m
F= - 6.0 N
W= F d
= -6.0 N (8.0m)
=-48 J
Work and Energy

Work = Force x Distance parallel to force
d= 6.0 m
F= - 5.0 N
W= F d
=?J
Work and Energy

Work = Force x Distance parallel to force
d= 6.0 m
F= - 5.0 N
W= F d
= -30 J
Work and Energy

Work = Force x Distance parallel to force
d= 6.0 m
F= ? N
W= 60 J
Work and Energy

Work = Force x Distance parallel to force
d= 6.0 m
F= 10 N
W= 60 J
Work and Energy

Work = Force x Distance parallel to force
d= ? m
F= - 50.0 N
W= 200 J
Work and Energy

Work = Force x Distance parallel to force
d= -4.0 m
F= - 50.0 N
W= 200 J
Work and Energy

Work = Force x Distance parallel to force
d= 8.0 m
F= + 6.0 N
W= 0
(since F and d are perpendicular
Power
Work
 J / s
Power 
time
Work = Power x time
 1 Watt= 1 J/s
 1 J = 1 Watt x 1 sec
 1 kilowatt - hr = 1000 (J/s) 3600 s = 3,600,000 J
 Energy = $$$$$$
 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

d=1m
D =8 m
Levers
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=1m
D =8 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.
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
Work in = Work out
f
D
=
F
d
D
f
d
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
Work in = Work out
f
D
=
F
d
D/d = 4 so F/f = 4
If F=200 N
f=?
D
f
d
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
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
D
f
d
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
F
f
d
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
F
f
D
d
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?
Efficiency
Energy out
Effeciency 
Energy in
Eout
Ein
Eloss
Efficiency
Energy out
Efficiency 
?
Energy in
Eout= 150 J
Ein = 200 J
Eloss= ?
Efficiency
Energy out
Efficiency 
 ? =0.75=75%
Energy in
Eout= 150 J
Ein = 200 J
Eloss= 50J
Two Machines e1 and e2
connected to each other in series
Two Machines e1 and e2
Eout=eff (Ein)=0.5(100J)=50J
Two Machines e1 and e2
Two Machines e1 and e2
Total efficiency when 2 machines are connected one after
the other is etot=e1 (e2)
Kinetic Energy, KE
 KE
=1/2 m v2
m=2.0 kg and v= 5 m/s
KE= ?
Kinetic Energy
 KE
=1/2 m v2
m=4.0 kg and v= 5 m/s
KE= ?
m=2.0 kg and v= 5 m/s
KE= 25 J
Kinetic Energy
 KE
=1/2 m v2
m=4.0 kg and v= 5 m/s
KE= 50J
m=2.0 kg and v= 5 m/s
KE= 25 J
Kinetic Energy
 KE
=1/2 m v2
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 v2
m=2.0 kg and v= 5 m/s
KE= 25 J
m=2.0 kg and v= 10 m/s
KE= 100J
1
2
KE  mv
2
if v  2v
1
1
2
2
KE  m2v   4  mv 
2
2

Double speed and KE increases by 4
Kinetic Energy
 KE
 if
=1/2 m v2
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 v2

F=ma
Work Energy Theorm
K
=1/2 m v2

F=ma
Fd=m a d

Work Energy Theorm
 KE



=1/2 m v2
F=ma
F d =m a d
F d = m (v/t) [(v/2)t]
Work Energy Theorm
K
E=1/2 m v2

F=ma
Fd=mad
F d = m (v/t) [(v/2)t]
W = 1/2 m v2



Work Energy Theorm
 KE





=1/2 m v2
F=ma
Fd=mad
F d = m (v/t) [(v/2)t]
W = 1/2 m v2
W = ∆ KE
Work Energy
W = ∆KE

How much work is required to stop a 2000
kg car traveling at 20 m/s (45 mph)?
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 v2
=-1/2(2000 kg)(20 m/s)2
= - 1000kg (400 m 2 /s 2)
= - 400,000 Joules
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
W = ∆KE
v = 20 m/s
d=? m
Same Friction Force
v = 10 m/s
d= 15 m
Work Energy
W = ∆KE
v = 20 m/s
d=60m
(4 times 15m)
Same Friction Force
v = 10 m/s
d= 15 m
Potential Energy, PE
•
•
•
•
•
Gravitational Potential Energy
Springs
Chemical
Pressure
Mass (Nuclear)
• Measured in Joules
Potential Energy, PE
The energy required to put
something in its place (state)
•
•
•
•
•
Gravitational Potential Energy
Springs
Chemical
Pressure
Mass (Nuclear)
Potential Energy

Gravitational Potential Energy = weight x height
PE=(mg) h
4.0 m
m = 2.0 kg
Potential Energy
PE=(mg) h
PE=80 J
m = 2.0 kg
4.0 m
K=?
Potential Energy to Kinetic Energy
PE=(mg) h
m = 2.0 kg
K E= 0 J
PE=40 J
2.0 m
1.0 m
KE=?
Conservation of Energy
Energy can neither be created nor destroyed only
transformed from one form to another
Total Mechanical Energy, E = PE +K
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 Energy
PE=100 J
K=0J
m = 1.02 kg
(mg = 10.0 N)
PE = 75 J
K = 25 J
10.0 m PE = 50 J
K = 50 J
PE= 25 J
PE = 0 J
Constant E
{E = K + PE}
Ef = Eo
K= ?
K=?
No friction
No Air resistance
Conservation of Energy
PE=100 J
m = 2.0 kg
K=0 J
Constant E
{E = K + PE}
U}
Ef=Eo
5.0 m
No friction
PE = 0 J
K=?
Conservation of Energy
PE =100 J
m = 2.0 kg
K=0J
Constant E
{E = K + PE}
U}
Ef=Eo
5.0 m
No friction
v=?
K = 100 J